Just How Small Is the Proton?
Galaxy-Scale Bubbles Extending From the Milky Way
CERN Physicists Trap Antimatter Atoms
Sunday, November 21, 2010
Tuesday, November 16, 2010
20.4-20.7: Radioactive Rate of Decay-Nuclear Fission and Nuclear Fusion
Radioactive Rate of Decay:
half-life: the time it takes from one-half of the nuclei in a sample to decay. Independent of the amount of sample, denoted t½.
First Order Half Life:
t½ = 0.693 / k
Example:
k = 1.45 x 10-12 s-1 for Ta-183, find t½ in years.
t½ = 0.693 / 1.45 x 10-12 s-1
t½ = 4.78 x 1011 s
Now convert to years:
(4.78 x 1011 s) x (1h / 3600s) x (1 day / 24h) x (1 year / 365 days) = 1.52 x 104 years
Once we know the decay constant (k) for a radioactive isotope, we can calculate the fraction of the radioactive nuclei left after a given time by the following equation:
ln (Nt / No) = –kt
No = the number of nuclei in the original sample
Nt = the number of nuclei left after time t
Never really need to find Nt or No, it is the fraction that is important.
Example:
When t½ = 10.76 years, what is the fraction remaining after 25 years?
Remember k = 0.693 / t½:
k = 0.693 / (10.76 years)
ln (Nt / No) = –(0.693t / t½)
ln (Nt / No) = –(0.693x 25years / 10.76 years)
ln (Nt / No) = –1.61
(Nt / No) = e–1.61
(Nt / No) = 0.200
This means 20% is left.
Example:
How long until there is only 5% left?
(Nt / No) = 5% = 0.05
ln(0.05) = –(0.693t / 10.76 years)
t = 46.6 years
Radioactive Dating:
Carbon–14, t½ = 5730 years
The upper atmosphere is constantly bombarded by cosmic ray radiations from space, which creates high energy neutrons when it reacts with Nitrogen–14 (the most abundant nitrogen nuclide).
147N + 10n → 146C + 11H
Then goes into carbon dioxide:
146CO2 → plants → animals
There is a set amount of Carbon–14 per kg of living matter. When something dies, it no longer takes in Carbon–14. The Carbon–14 then decays:
146C → 147N + 0-1e
In living matter, we measure 15.3 disintegrations per kg of C per minute. Anything lower → must be dead. In this way, this ratio of carbon isotopes becomes a clock measuring the time since death of an organism.
Nt / No = (less than 15.3 dis/kg • min) / (15.3 dis/kg • min)
Example:
You want to carbon date a caveman's foot, where you observe (4.50 dis/kg • min). When did the caveman lose his foot?
ln(Nt / No) = –(0.693t / t½)
ln(4.50 / 15.3) = –(0.693t / 5730 years)
t = 10118 years
Applications of Radioactive Isotopes:
Chemical Analysis:
Radioactive Tracer: a very small amount of radioactive isotope added to a chemical, biological, or physical system to study the system.
Evidence of dynamic equilibrium can be seen using radioactive tracers. In a beaker, imagine a solution of PbI2 in contact with the solid. This beaker contains only iodine atoms with nonradioactive isotopes. Now add a radioactive iodide ion to the solution. The solution is still saturated, and the amount of solid remains constant. But, after time, the solid lead iodide, which was initially nonradioactive, is radioactive. Nonradioactive iodide ions in the solid have swapped with radioactive ions in solution.
Tracers are also used to determine chemical mechanisms, like photosynthesis.
Isotope Dilution: a technique to determine the quantity of a substance in a mixture or of the total volume of solution by adding a known amount of an isotope to it. After time, removal of a sample of the mixture and measurement of the fraction by which the isotope has diluted provides a way to determine the quantity of substance or volume of solution.
Neutron Activation Analysis: an analysis of elements in a sample based on the conversion of stable isotopes to radioactive isotopes by bombarding a sample with neutrons. Example: measuring trace amounts of arsenic in human hair. The arsenic is converted into a metastable nucleus via bombardment with a neutron. As the arsenic returns to its ground state, it emits gamma rays. These measurements can be interpreted to find who the hair belongs to, or how much arsenic is in it.
7533As + 10n → 76m33As
76m33As → 7533As + 00γ
Mass-Energy Equivalence:
When nuclei decay, they form products of lower energy. The change of energy is related to changes in mass, according to the mass-energy equivalence derived by Albert Einstein.
E = mc2
Where c = the speed of light = 3.00 x 108 m/s
E = energy
m = mass
For any mass, there is an associated energy. For any energy, there is an associated mass.
When the energy changes by an amount ΔE, the mass changes by an amount Δm.
ΔE = Δmc2
When energy is given off, mass decreases.
Example:
Chemical reaction:
When carbon burns in oxygen, it releases heat energy:
C (gr) + O2 (g) → CO2 (g); ΔH = –393.5 kJ
ΔE = –3.935 x 105 J (converted from kJ to J)
Recall that J = (kg•m2 / s2)
Δm = ΔE / c2
Δm = –(3.935 x 105 kg•m2 / s2) / (3.00 x 108 m/s)2
Δm = –4.37 x 10-12 kg
Mass changes in nuclear reactions are approximately a milion times larger per mole of reactant thatn those in chemical reaction.
Example:
Nuclear Reaction:
Consider the alpha decay of Uranium–238 to Thorium–234
23892U → 23490Th + 42He
Use the nuclear mass in amu to find the change in mass for this reaction. You would think the mass of the reactants is the same as the mass of the products, but not so. There is a mass loss.
23892U = 238.0003 amu
23490Th = 233.9942 amu
42He = 4.00150 amu
Δm = (233.9942 amu + 4.00150 amu) – (238.00003 amu)
Δm = – 0.0046 amu
(Assume 1 mole of isotope.)
Δm = – 0.0046 amu = – 0.0046g = – 4.6 x 10-6 kg
ΔE = Δmc2 = (– 4.6 x 10-6 kg)(3.00 x 108 m/s)2
ΔE = – 4.14 x 1011 J = – 4.14 x 108 kJ
Nuclear Fission: spontaneous break up of isotopes, naturally occurring, does not require energy input. Powers nuclear reactors and bombs.
Nuclear Fusion: Fusing of two nuclei into a larger nucleus. Takes GIGANTIC amounts of energy.
21H 31Th→ 42He + 10n
half-life: the time it takes from one-half of the nuclei in a sample to decay. Independent of the amount of sample, denoted t½.
First Order Half Life:
t½ = 0.693 / k
Example:
k = 1.45 x 10-12 s-1 for Ta-183, find t½ in years.
t½ = 0.693 / 1.45 x 10-12 s-1
t½ = 4.78 x 1011 s
Now convert to years:
(4.78 x 1011 s) x (1h / 3600s) x (1 day / 24h) x (1 year / 365 days) = 1.52 x 104 years
Once we know the decay constant (k) for a radioactive isotope, we can calculate the fraction of the radioactive nuclei left after a given time by the following equation:
ln (Nt / No) = –kt
No = the number of nuclei in the original sample
Nt = the number of nuclei left after time t
Never really need to find Nt or No, it is the fraction that is important.
Example:
When t½ = 10.76 years, what is the fraction remaining after 25 years?
Remember k = 0.693 / t½:
k = 0.693 / (10.76 years)
ln (Nt / No) = –(0.693t / t½)
ln (Nt / No) = –(0.693x 25years / 10.76 years)
ln (Nt / No) = –1.61
(Nt / No) = e–1.61
(Nt / No) = 0.200
This means 20% is left.
Example:
How long until there is only 5% left?
(Nt / No) = 5% = 0.05
ln(0.05) = –(0.693t / 10.76 years)
t = 46.6 years
Radioactive Dating:
Carbon–14, t½ = 5730 years
The upper atmosphere is constantly bombarded by cosmic ray radiations from space, which creates high energy neutrons when it reacts with Nitrogen–14 (the most abundant nitrogen nuclide).
147N + 10n → 146C + 11H
Then goes into carbon dioxide:
146CO2 → plants → animals
There is a set amount of Carbon–14 per kg of living matter. When something dies, it no longer takes in Carbon–14. The Carbon–14 then decays:
146C → 147N + 0-1e
In living matter, we measure 15.3 disintegrations per kg of C per minute. Anything lower → must be dead. In this way, this ratio of carbon isotopes becomes a clock measuring the time since death of an organism.
Nt / No = (less than 15.3 dis/kg • min) / (15.3 dis/kg • min)
Example:
You want to carbon date a caveman's foot, where you observe (4.50 dis/kg • min). When did the caveman lose his foot?
ln(Nt / No) = –(0.693t / t½)
ln(4.50 / 15.3) = –(0.693t / 5730 years)
t = 10118 years
Applications of Radioactive Isotopes:
Chemical Analysis:
Radioactive Tracer: a very small amount of radioactive isotope added to a chemical, biological, or physical system to study the system.
Evidence of dynamic equilibrium can be seen using radioactive tracers. In a beaker, imagine a solution of PbI2 in contact with the solid. This beaker contains only iodine atoms with nonradioactive isotopes. Now add a radioactive iodide ion to the solution. The solution is still saturated, and the amount of solid remains constant. But, after time, the solid lead iodide, which was initially nonradioactive, is radioactive. Nonradioactive iodide ions in the solid have swapped with radioactive ions in solution.
Tracers are also used to determine chemical mechanisms, like photosynthesis.
Isotope Dilution: a technique to determine the quantity of a substance in a mixture or of the total volume of solution by adding a known amount of an isotope to it. After time, removal of a sample of the mixture and measurement of the fraction by which the isotope has diluted provides a way to determine the quantity of substance or volume of solution.
Neutron Activation Analysis: an analysis of elements in a sample based on the conversion of stable isotopes to radioactive isotopes by bombarding a sample with neutrons. Example: measuring trace amounts of arsenic in human hair. The arsenic is converted into a metastable nucleus via bombardment with a neutron. As the arsenic returns to its ground state, it emits gamma rays. These measurements can be interpreted to find who the hair belongs to, or how much arsenic is in it.
7533As + 10n → 76m33As
76m33As → 7533As + 00γ
Mass-Energy Equivalence:
When nuclei decay, they form products of lower energy. The change of energy is related to changes in mass, according to the mass-energy equivalence derived by Albert Einstein.
E = mc2
Where c = the speed of light = 3.00 x 108 m/s
E = energy
m = mass
For any mass, there is an associated energy. For any energy, there is an associated mass.
When the energy changes by an amount ΔE, the mass changes by an amount Δm.
ΔE = Δmc2
When energy is given off, mass decreases.
Example:
Chemical reaction:
When carbon burns in oxygen, it releases heat energy:
C (gr) + O2 (g) → CO2 (g); ΔH = –393.5 kJ
ΔE = –3.935 x 105 J (converted from kJ to J)
Recall that J = (kg•m2 / s2)
Δm = ΔE / c2
Δm = –(3.935 x 105 kg•m2 / s2) / (3.00 x 108 m/s)2
Δm = –4.37 x 10-12 kg
Mass changes in nuclear reactions are approximately a milion times larger per mole of reactant thatn those in chemical reaction.
Example:
Nuclear Reaction:
Consider the alpha decay of Uranium–238 to Thorium–234
23892U → 23490Th + 42He
Use the nuclear mass in amu to find the change in mass for this reaction. You would think the mass of the reactants is the same as the mass of the products, but not so. There is a mass loss.
23892U = 238.0003 amu
23490Th = 233.9942 amu
42He = 4.00150 amu
Δm = (233.9942 amu + 4.00150 amu) – (238.00003 amu)
Δm = – 0.0046 amu
(Assume 1 mole of isotope.)
Δm = – 0.0046 amu = – 0.0046g = – 4.6 x 10-6 kg
ΔE = Δmc2 = (– 4.6 x 10-6 kg)(3.00 x 108 m/s)2
ΔE = – 4.14 x 1011 J = – 4.14 x 108 kJ
Nuclear Fission: spontaneous break up of isotopes, naturally occurring, does not require energy input. Powers nuclear reactors and bombs.
Nuclear Fusion: Fusing of two nuclei into a larger nucleus. Takes GIGANTIC amounts of energy.
21H 31Th→ 42He + 10n
20.2-20.3: Nuclear Bombardment Reactions-Radiations and Matter
Transuranium Elements:
DOES NOT INCLUDE URANIUM!!!
Includes elements with an atomic number greater than that of uranium (Z = 92), the naturally occurring element of greatest Z.
Radiation Counters:
Because nuclear radiations can ionize molecules and break chemical bonds, they adversely affect biological organisms. So it's good to know when they're there.
Geiger Counter: a kind of ionization counter used to count particles emitted by radioactive nuclei, consists of a metal tube filled with gas, such as argon. Can detect both alpha and beta particles, and under special circumstances, neutrons.
10n + 105B → 73Li + 42He
Scintillation Counter: a device that detects nuclear radiation from flashes of light generated in a material by the radiation. A phosphor is a substance that emits flashes of light when struck by radiation. Can detect both beta and gamma particles using a photomultiplier tube.
One e– can produce 106 e–.
This vintage reel on Scintillation Counters is way better than a diagram:
Radiation counters measure the number of nuclear disintegrations in a radioactive sample.
Activity: (activity of a radioactive source) is the number of nuclear disintegrations per unit time occurring in a radioactive material.
Curie: Denoted Ci, is a unit of activity equal to 3.700 x 1010 disintegrations per second.
Nuclear disintegrations / second = nuc / s = Ci
Biological Effects of Radiation Dosage:
Radiation dosage effects can be quite damaging. DNA is especially affected, which interferes with cell division.
rad → (radiation absorbed dose) the dosage of radiation that deposits 1 x 10-2 J of energy per kg of tissue.
Depends not only on the amount of energy deposited in the tissue, but also on the particle. Neutrons are more destructive than gamma rays of the same radiation dosage measured in rads.
rem → a unit of radiation dosage used to relate various kinds of radiation in terms of biological destruction. It equals the rad times a factor for the type of radiation, called the relative biological effectiveness, or RBE (basically, how efficient it is at killing you).
rems = rads x RBE → total damage
RBE = 1 for γ (gamma) and β (beta).
RBE = 5 for no
RBE = 10 for α (alpha)
Here's that movie mentioned in class:
It really is an awesome movie, go rent it!
Rate of Radioactive Decay:
The rate equation for radioactive decay has the same form as the rate law for a first order chemical reaction.
R = kNt
R = Rate = Activity
Nt = the number of radioactive nuclei at time t
k = radioactive decay constant, the rate constant for radioactive decay
Example:
A 2.50 mg sample of Tc-99 has an activity of 2.70 x 10-5 Ci and is decaying by beta emission. What is the decay constant?
R = kNt
First convert the activity to nuc/s:
R = 2.70 x 10-5 Ci x (3.700 x 1010 nuc/s / 1s) = 9.99 x 105 nuc/s
Now convert the given sample to Nt:
2.50 mg Tc-99 = 2.50 x 10-3g Tc-99
(2.50 x 10-3g Tc-99) x (1 mol Tc-99 / 99g Tc-99) x (6.023 x 1023 nuclei / 1 mol Tc-99) = 1.52 x 1019 nuclei = Nt
k = Rate / Nt
k = (9.99 x 105 nuc/s) / ( 1.52 x 1019 nuclei)
k = 6.57 x 10-14 s-1
Units for k will always be inverted seconds because it is first order.
DOES NOT INCLUDE URANIUM!!!
Includes elements with an atomic number greater than that of uranium (Z = 92), the naturally occurring element of greatest Z.
Radiation Counters:
Because nuclear radiations can ionize molecules and break chemical bonds, they adversely affect biological organisms. So it's good to know when they're there.
Geiger Counter: a kind of ionization counter used to count particles emitted by radioactive nuclei, consists of a metal tube filled with gas, such as argon. Can detect both alpha and beta particles, and under special circumstances, neutrons.
10n + 105B → 73Li + 42He
Scintillation Counter: a device that detects nuclear radiation from flashes of light generated in a material by the radiation. A phosphor is a substance that emits flashes of light when struck by radiation. Can detect both beta and gamma particles using a photomultiplier tube.
One e– can produce 106 e–.
This vintage reel on Scintillation Counters is way better than a diagram:
Radiation counters measure the number of nuclear disintegrations in a radioactive sample.
Activity: (activity of a radioactive source) is the number of nuclear disintegrations per unit time occurring in a radioactive material.
Curie: Denoted Ci, is a unit of activity equal to 3.700 x 1010 disintegrations per second.
Nuclear disintegrations / second = nuc / s = Ci
Biological Effects of Radiation Dosage:
Radiation dosage effects can be quite damaging. DNA is especially affected, which interferes with cell division.
rad → (radiation absorbed dose) the dosage of radiation that deposits 1 x 10-2 J of energy per kg of tissue.
Depends not only on the amount of energy deposited in the tissue, but also on the particle. Neutrons are more destructive than gamma rays of the same radiation dosage measured in rads.
rem → a unit of radiation dosage used to relate various kinds of radiation in terms of biological destruction. It equals the rad times a factor for the type of radiation, called the relative biological effectiveness, or RBE (basically, how efficient it is at killing you).
rems = rads x RBE → total damage
RBE = 1 for γ (gamma) and β (beta).
RBE = 5 for no
RBE = 10 for α (alpha)
Here's that movie mentioned in class:
It really is an awesome movie, go rent it!
Rate of Radioactive Decay:
The rate equation for radioactive decay has the same form as the rate law for a first order chemical reaction.
R = kNt
R = Rate = Activity
Nt = the number of radioactive nuclei at time t
k = radioactive decay constant, the rate constant for radioactive decay
Example:
A 2.50 mg sample of Tc-99 has an activity of 2.70 x 10-5 Ci and is decaying by beta emission. What is the decay constant?
R = kNt
First convert the activity to nuc/s:
R = 2.70 x 10-5 Ci x (3.700 x 1010 nuc/s / 1s) = 9.99 x 105 nuc/s
Now convert the given sample to Nt:
2.50 mg Tc-99 = 2.50 x 10-3g Tc-99
(2.50 x 10-3g Tc-99) x (1 mol Tc-99 / 99g Tc-99) x (6.023 x 1023 nuclei / 1 mol Tc-99) = 1.52 x 1019 nuclei = Nt
k = Rate / Nt
k = (9.99 x 105 nuc/s) / ( 1.52 x 1019 nuclei)
k = 6.57 x 10-14 s-1
Units for k will always be inverted seconds because it is first order.
20.1: Radioactivity-Nuclear Bombardment Reactions
Nuclear Stability:
Why don't all the protons in the nucleus repel each other and cause the nucleus to blow apart?
Nuclear Force: a strong force of attraction between nucleons that acts only at very short distances, overwhelms the electrostatic force (the positive charges repelling each other.)
Shell Model of the Nucleus: a nuclear model in which protons and neutrons exist in levels, or shells, analogous to the shell structure that exists for electrons in an atom.
First Test For Stability:
Magic Numbers: the number of nuclear particles in a completed shell of protons or neutrons. Think of the way full electron shells make noble gases stable, this is similar. If a nucleus has two magic numbers, it is very stable.
Magic Numbers for p+ → 2, 8, 20, 28, 50, 82, (114)
Magic Numbers for no → 2, 8, 20, 28, 50, 82, 126
Example:
42He
p+ → 2
no → 2
Both are magic numbers, very stable.
Example:
Which is more stable?
10251Sn or 10151Sn?
10251Sn → p+ = 51, no = 51
10151Sn → p+ = 51, no = 50
10151Sn has a magic number and 10251Sn doesn't.
Second Test for Stability:
Even/Odd Test:
Nuclei with an even number of protons and an even number of neutrons are very likely to be stable.
Nuclei with an odd number of protons and an odd number of neutrons are not at all likely to be stable.
Nuclei with variably even and odd numbers of protons or neutrons may or may not be stable.
Example:
Which is more likely to be stable, 6231Ga or 6432Ge?
6432Ge because it has both an even number of protons and an even number of neutrons.
Third Test for Stability:
Band of Stability: the region in which stable nuclides lie in a plot of number of protons against number of neutrons.
When it turns out to be unstable, what does the nucleus do about it?
The Six Types of Radioactive Decay:
1. Alpha Emission: emission of a 42He nucleus, or alpha particle from an unstable nucleus. Happens for very large nuclei.
Example:
22688Ra → 22286Rn + 42He
2. Beta Emission: emission of a high-speed electron (β–) from an unstable nucleus.
Example:
146C → 147N + 0-1β
Equivalent to the conversion of a neutron to a proton.
10n → 11p + 0-1e
3. Positron Emission: emission of a positron (β+, or 01e) from an unstable nucleus.
Example:
9543Tc → 9542Mo + 01e
Equivalent to the conversion of a proton to a neutron.
11p → 10n + 01e
Professor said a positron is an electron traveling backward in time, but I have no idea what that means.
4. Electron Capture: the decay of an unstable nucleus by capturing, or picking up, an electron from an inner orbital of an atom.
Example:
4019K + 0-1e → 4018Ar
In effect, a proton is changed to a neutron, as in positron emission:
11p → 01e- + 10n
5. Gamma Emission: emission from an excited nucleus of a gamma photon (denoted γ). Often, gamma emission occurs very quickly after radioactive decay. In some cases, however, an excited state has a significant lifetime before it emits a gamma photon.
Metastable Nucleus: a nucleus in an excited state with a lifetime of at least one nanosecond (10-9s). In time, the metastable nucleus decays by gamma emission.
Example:
99m43Tc → 9943Tc + 00γ
99m43Tc → the m denotes metastable, excited state for nucleus
9943Tc → ground state
6. Spontaneous Fission: the spontaneous decay of an unstable nucleus in which a heavy nucleus of mass number greater than 89 splits into lighter nuclei and energy is released.
Example:
23692U → 9639Y + 13633I + 410n
When A is greater than 89 → spontaneous fission
Recall the Band of Stability:
*To the left of the band, nuclides have a neutron to proton ratio larger than that needed for stability, so they tend to decay by beta emission.
*To the right of the band, nuclides have a smaller neutron to proton ratio that that needed for stability, so they tend to decay by either positron emission or electron capture.
*As the curve follows Z as it becomes larger than 83, decay is often by alpha emission.
Radioactive Decay Series: a sequence in which one radioactive nucleus decays to a second, which then decays to a third, and so forth. Eventually, a stable nucleus, which is an isotope of lead, is reached.
Each step will involve an alpha or a beta decay.
Example:
What is the nucleus formed from Uranium–238 after six alpha (42He) and two beta (0-1β) emissions?
23892U → □□□ + 642He + 20-1e
□□□ = 21482Pb
Nuclear Bombardment Reactions:
Alchemists → wanted to turn lead to gold and live forever
Transmutation: the change of one element to another by bombarding the nucleus of the element with nuclear particles or nuclei.
Notation:
Target Nucleus + Subatomic Particle → Product Nucleus + Subatomic Particle
Example:
147N +42He → 178O + 11H
147N(42He, 11H)178O
Example:
94Be +42He → 126C + 10n
94Be(42He, 10n)126C
Elements with large atomic numbers deflect alpha particles (large positive nucleus deflect positive alpha particles.) To shoot large particles into heavy nuclei, charged particles must be accelerated.
Particle Accelerator: a device used to accelerate electrons, protons, and alpha particles and other ions to very high speeds. The kinetic energies of these particles is units of electron volts.
Electron Volt: Denoted eV, the quantity of energy that would have to be imparted to an electron to accelerate it by one volt potential difference.
1 eV = 1.602 x 10-19 J
Cyclotron: a type of particle accelerator consisting of two hollow, semicircular metal electrodes called dees (because the shape resembles the letter D), in which charged particles are accelerated by stages to higher and higher kinetic energies.
Ions introduced at the center of the cyclotron are accelerated in the space between the two dees. A magnetic field keeps the ions moving in a spiral path.
The dees are connected to a high-frequency electric current that changes their polarity so that each time the ion moves in the space between the dees, it is accelerated. Then it leaves the cyclotron at high speed and hits its target.
This guy really loves cyclotrons:
Why don't all the protons in the nucleus repel each other and cause the nucleus to blow apart?
Nuclear Force: a strong force of attraction between nucleons that acts only at very short distances, overwhelms the electrostatic force (the positive charges repelling each other.)
Shell Model of the Nucleus: a nuclear model in which protons and neutrons exist in levels, or shells, analogous to the shell structure that exists for electrons in an atom.
First Test For Stability:
Magic Numbers: the number of nuclear particles in a completed shell of protons or neutrons. Think of the way full electron shells make noble gases stable, this is similar. If a nucleus has two magic numbers, it is very stable.
Magic Numbers for p+ → 2, 8, 20, 28, 50, 82, (114)
Magic Numbers for no → 2, 8, 20, 28, 50, 82, 126
Example:
42He
p+ → 2
no → 2
Both are magic numbers, very stable.
Example:
Which is more stable?
10251Sn or 10151Sn?
10251Sn → p+ = 51, no = 51
10151Sn → p+ = 51, no = 50
10151Sn has a magic number and 10251Sn doesn't.
Second Test for Stability:
Even/Odd Test:
Nuclei with an even number of protons and an even number of neutrons are very likely to be stable.
Nuclei with an odd number of protons and an odd number of neutrons are not at all likely to be stable.
Nuclei with variably even and odd numbers of protons or neutrons may or may not be stable.
Example:
Which is more likely to be stable, 6231Ga or 6432Ge?
6432Ge because it has both an even number of protons and an even number of neutrons.
Third Test for Stability:
Band of Stability: the region in which stable nuclides lie in a plot of number of protons against number of neutrons.
When it turns out to be unstable, what does the nucleus do about it?
The Six Types of Radioactive Decay:
1. Alpha Emission: emission of a 42He nucleus, or alpha particle from an unstable nucleus. Happens for very large nuclei.
Example:
22688Ra → 22286Rn + 42He
2. Beta Emission: emission of a high-speed electron (β–) from an unstable nucleus.
Example:
146C → 147N + 0-1β
Equivalent to the conversion of a neutron to a proton.
10n → 11p + 0-1e
3. Positron Emission: emission of a positron (β+, or 01e) from an unstable nucleus.
Example:
9543Tc → 9542Mo + 01e
Equivalent to the conversion of a proton to a neutron.
11p → 10n + 01e
Professor said a positron is an electron traveling backward in time, but I have no idea what that means.
4. Electron Capture: the decay of an unstable nucleus by capturing, or picking up, an electron from an inner orbital of an atom.
Example:
4019K + 0-1e → 4018Ar
In effect, a proton is changed to a neutron, as in positron emission:
11p → 01e- + 10n
5. Gamma Emission: emission from an excited nucleus of a gamma photon (denoted γ). Often, gamma emission occurs very quickly after radioactive decay. In some cases, however, an excited state has a significant lifetime before it emits a gamma photon.
Metastable Nucleus: a nucleus in an excited state with a lifetime of at least one nanosecond (10-9s). In time, the metastable nucleus decays by gamma emission.
Example:
99m43Tc → 9943Tc + 00γ
99m43Tc → the m denotes metastable, excited state for nucleus
9943Tc → ground state
6. Spontaneous Fission: the spontaneous decay of an unstable nucleus in which a heavy nucleus of mass number greater than 89 splits into lighter nuclei and energy is released.
Example:
23692U → 9639Y + 13633I + 410n
When A is greater than 89 → spontaneous fission
Recall the Band of Stability:
*To the left of the band, nuclides have a neutron to proton ratio larger than that needed for stability, so they tend to decay by beta emission.
*To the right of the band, nuclides have a smaller neutron to proton ratio that that needed for stability, so they tend to decay by either positron emission or electron capture.
*As the curve follows Z as it becomes larger than 83, decay is often by alpha emission.
Radioactive Decay Series: a sequence in which one radioactive nucleus decays to a second, which then decays to a third, and so forth. Eventually, a stable nucleus, which is an isotope of lead, is reached.
Each step will involve an alpha or a beta decay.
Example:
What is the nucleus formed from Uranium–238 after six alpha (42He) and two beta (0-1β) emissions?
23892U → □□□ + 642He + 20-1e
□□□ = 21482Pb
Nuclear Bombardment Reactions:
Alchemists → wanted to turn lead to gold and live forever
Transmutation: the change of one element to another by bombarding the nucleus of the element with nuclear particles or nuclei.
Notation:
Target Nucleus + Subatomic Particle → Product Nucleus + Subatomic Particle
Example:
147N +42He → 178O + 11H
147N(42He, 11H)178O
Example:
94Be +42He → 126C + 10n
94Be(42He, 10n)126C
Elements with large atomic numbers deflect alpha particles (large positive nucleus deflect positive alpha particles.) To shoot large particles into heavy nuclei, charged particles must be accelerated.
Particle Accelerator: a device used to accelerate electrons, protons, and alpha particles and other ions to very high speeds. The kinetic energies of these particles is units of electron volts.
Electron Volt: Denoted eV, the quantity of energy that would have to be imparted to an electron to accelerate it by one volt potential difference.
1 eV = 1.602 x 10-19 J
Cyclotron: a type of particle accelerator consisting of two hollow, semicircular metal electrodes called dees (because the shape resembles the letter D), in which charged particles are accelerated by stages to higher and higher kinetic energies.
Ions introduced at the center of the cyclotron are accelerated in the space between the two dees. A magnetic field keeps the ions moving in a spiral path.
The dees are connected to a high-frequency electric current that changes their polarity so that each time the ion moves in the space between the dees, it is accelerated. Then it leaves the cyclotron at high speed and hits its target.
This guy really loves cyclotrons:
20.1-Radioactivity
Nuclear Chemistry:
Radioactive Decay and Nuclear Bombardment Reactions:
Antoine Henri Becquerel discovered in 1896 that photographic plates developed bright spots when exposed to uranium minerals and concluded that they must give off radiation.
Terms:
α = alpha particle = positively charged, consist of helium-4 nuclei (nuclei with two protons and two neutrons)
β = beta particle = negatively charged, consist of high speed electrons
γ = gamma particle = no charge = a form of electromagnetic radiation similar to x-rays except higher in energy.
Nuclide Symbol: AZX
Example: 23892U → Uranium–238
X = element
Z = atomic number → number of protons → p+
A = mass number → number of protons and neutrons → no
number of neutrons = N = A – Z
A gives the mass of the isotope
Z gives the charge
isotopes → same Z, same element. Ex: 126C, 136C, 146C
nuclides → technically any combination 126C, 147N, 188O
*But now these terms are used interchangeably
Notation:
protons → 11H, 11p, p+
neutrons → 10n, no, n
electrons → 0– 1e, 0– 1β (high energy beta particle), β–
positron → 01e, 0+ 1β (like a positive electron), β+
gamma photon → 00γ, γ
alpha particle → 42He, α
deuterium → 21H, D, d (H plus one neutron)
tritium → 31H, t (H plus two neutrons)
Radioactive Decay: has one reactant, occurs naturally, no energy input needed. Usually, high energy particles come out.
Example:
Nuclear Equation:
23892U → 23490Th + 42He
These must be mass balanced and charge balanced (superscripts and subscripts on the left side of the equation equal the sum of the superscripts and subscripts on the right side of the equation.)
Nuclear Bombardment Reactions: has two reactants. Usually synthesized and energy input is required.
Example:
□□□ + 21H → 2411Na + 42He
□□□ = 2612Mg
Radioactive Decay and Nuclear Bombardment Reactions:
Antoine Henri Becquerel discovered in 1896 that photographic plates developed bright spots when exposed to uranium minerals and concluded that they must give off radiation.
Terms:
α = alpha particle = positively charged, consist of helium-4 nuclei (nuclei with two protons and two neutrons)
β = beta particle = negatively charged, consist of high speed electrons
γ = gamma particle = no charge = a form of electromagnetic radiation similar to x-rays except higher in energy.
Nuclide Symbol: AZX
Example: 23892U → Uranium–238
X = element
Z = atomic number → number of protons → p+
A = mass number → number of protons and neutrons → no
number of neutrons = N = A – Z
A gives the mass of the isotope
Z gives the charge
isotopes → same Z, same element. Ex: 126C, 136C, 146C
nuclides → technically any combination 126C, 147N, 188O
*But now these terms are used interchangeably
Notation:
protons → 11H, 11p, p+
neutrons → 10n, no, n
electrons → 0– 1e, 0– 1β (high energy beta particle), β–
positron → 01e, 0+ 1β (like a positive electron), β+
gamma photon → 00γ, γ
alpha particle → 42He, α
deuterium → 21H, D, d (H plus one neutron)
tritium → 31H, t (H plus two neutrons)
Radioactive Decay: has one reactant, occurs naturally, no energy input needed. Usually, high energy particles come out.
Example:
Nuclear Equation:
23892U → 23490Th + 42He
These must be mass balanced and charge balanced (superscripts and subscripts on the left side of the equation equal the sum of the superscripts and subscripts on the right side of the equation.)
Nuclear Bombardment Reactions: has two reactants. Usually synthesized and energy input is required.
Example:
□□□ + 21H → 2411Na + 42He
□□□ = 2612Mg
19.9-19.11: Electrolysis of Molten Salts-Stoichiometry of Elecrolysis
Electrolysis: the process of producing a chemical change in an electrolytic cell, which is an electrochemical cell in which an electric current drives an otherwise non-spontaneous reaction.
Downs Cell: a commercial electrochemical cell used to obtain sodium metal by the electrolysis of molten sodium chloride.
The half-reactions are:
Na+ (l) + e– → Na (l)
Cl– (l) → (½)Cl2 (g) + e–
Where the reduction of Na+ to Na occurs at the cathode and the oxidation of Cl– to Cl2 occurs at the anode. No water is used.
CaCl2 is added to reduce the melting point of NaCl from 801oC to 580oC. And the electrodes do not need to be separated because the non-spontaneous reaction does not want to proceed anyway. This is the only way to commercially produce Na(s).
The cell reaction is:
Na+ (l) + Cl– (l) → Na (l) + (½)Cl2 (g)
Overvoltage:
“The presence of water in a solution of sodium chloride must be examined in respect to its reduction and oxidation in both electrodes. Usually, water is electrolysed as mentioned in electrolysis of water yielding gaseous oxygen in the anode and gaseous hydrogen in the cathode. On the other hand, sodium chloride in water dissociates in Na+ and Cl– ions, cation, which is the positive ion, will be attracted to the cathode (+), thus reducing the sodium ion. The anion will then be attracted to the anode (–) oxidizing chloride ion.
The following half reactions describes the process mentioned:
1. Cathode: Na+(aq) + e– → Na(s) (E° = –2.71 V)
2. Anode: 2 Cl–(aq) → Cl2(g) + 2 e– (E° = +1.36 V)
3. Cathode: 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq) (E° = –0.83 V)
4. Anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e– → (E° = +1.23 V)
Reaction 1 is discarded as it has the most negative value on standard reduction potential thus making it less thermodynamically favorable in the process.
When comparing the reduction potentials in reactions 2 and 4, the reduction of chloride ion is favored. Thus, if the Cl– ion is favored for reduction, then the water reaction is favored for oxidation producing gaseous oxygen, however experiments show gaseous chlorine is produced and not oxygen.
Although the initial analysis is correct, there is another effect that can happen, known as the overvoltage effect. Additional voltage is sometimes required, beyond the voltage predicted by the E°cell. This may be due to kinetic rather than thermodynamic considerations. In fact, it has been proven that the activation energy for the chloride ion is very low, hence favorable in kinetic terms. In other words, although the voltage applied is thermodynamically sufficient to drive electrolysis, the rate is so slow that to make the process proceed in a reasonable time frame, the voltage of the external source has to be increased (hence, overvoltage.)
Finally, reaction 3 is favorable because it describes the proliferation of OH– ions thus letting a probable reduction of H+ ions less favorable an option.
The overall reaction for the process according to the analysis would be the following:
Anode (oxidation): 2 Cl–(aq) → Cl2(g) + 2 e–
Cathode (reduction): 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
Overall reaction: 2 H2O + 2 Cl–(aq) → H2(g) + Cl2(g) + 2 OH–(aq)
As the overall reaction indicates, the concentration of chloride ions is reduced in comparison to OH– ions (whose concentration increases). The reaction also shows the production of gaseous hydrogen, chlorine and aqueous sodium hydroxide.”
Source
Electroplating of Metals:
Copper (II) ion leaves the anode and plates out on the cathode. Unreactive substances (such as gold) collect as anode mud under the anode (very valuable).
Stoichiometry of Electrolysis:
Ampere: Denoted A, is the SI base unit for current. The coulomb ( C ) is the SI unit of electric charge and is equivalent to an ampere-second.
Coulombs = Amperes x Seconds
(Electric Charge = Electric Current x Time Lapse)
and
Amperes = Coulombs / Seconds
One faraday (96,485 Coulombs) is equivalent to the charge on one mole of electrons.
Example:
System: Cu2+ (aq) + 2e– → Cu (s)
A constant current passes through this system for 6.50 hours and 6.75g Cu (s) was deposits. What was the current?
Amperes = Coulombs / Seconds
To get the electric current, convert Cu from grams to moles, use stoichiometry to determine moles of e– to moles of Cu, then use moles of e– to convert to Coulombs:
6.75g Cu x (1 mol Cu / 63.5g Cu) x (2 mol e– / 1 mol Cu) x (96485 C / 1 mol e–) = 20512.6 C
To get the time, convert 6.50h to seconds:
6.50h x (3600 s / 1 h) = 23400s
Divide Coulombs by Seconds to get Amperes:
Amperes = 20512.6 C / 23400s = 0.876 Amps
Downs Cell: a commercial electrochemical cell used to obtain sodium metal by the electrolysis of molten sodium chloride.
The half-reactions are:
Na+ (l) + e– → Na (l)
Cl– (l) → (½)Cl2 (g) + e–
Where the reduction of Na+ to Na occurs at the cathode and the oxidation of Cl– to Cl2 occurs at the anode. No water is used.
CaCl2 is added to reduce the melting point of NaCl from 801oC to 580oC. And the electrodes do not need to be separated because the non-spontaneous reaction does not want to proceed anyway. This is the only way to commercially produce Na(s).
The cell reaction is:
Na+ (l) + Cl– (l) → Na (l) + (½)Cl2 (g)
Overvoltage:
“The presence of water in a solution of sodium chloride must be examined in respect to its reduction and oxidation in both electrodes. Usually, water is electrolysed as mentioned in electrolysis of water yielding gaseous oxygen in the anode and gaseous hydrogen in the cathode. On the other hand, sodium chloride in water dissociates in Na+ and Cl– ions, cation, which is the positive ion, will be attracted to the cathode (+), thus reducing the sodium ion. The anion will then be attracted to the anode (–) oxidizing chloride ion.
The following half reactions describes the process mentioned:
1. Cathode: Na+(aq) + e– → Na(s) (E° = –2.71 V)
2. Anode: 2 Cl–(aq) → Cl2(g) + 2 e– (E° = +1.36 V)
3. Cathode: 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq) (E° = –0.83 V)
4. Anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e– → (E° = +1.23 V)
Reaction 1 is discarded as it has the most negative value on standard reduction potential thus making it less thermodynamically favorable in the process.
When comparing the reduction potentials in reactions 2 and 4, the reduction of chloride ion is favored. Thus, if the Cl– ion is favored for reduction, then the water reaction is favored for oxidation producing gaseous oxygen, however experiments show gaseous chlorine is produced and not oxygen.
Although the initial analysis is correct, there is another effect that can happen, known as the overvoltage effect. Additional voltage is sometimes required, beyond the voltage predicted by the E°cell. This may be due to kinetic rather than thermodynamic considerations. In fact, it has been proven that the activation energy for the chloride ion is very low, hence favorable in kinetic terms. In other words, although the voltage applied is thermodynamically sufficient to drive electrolysis, the rate is so slow that to make the process proceed in a reasonable time frame, the voltage of the external source has to be increased (hence, overvoltage.)
Finally, reaction 3 is favorable because it describes the proliferation of OH– ions thus letting a probable reduction of H+ ions less favorable an option.
The overall reaction for the process according to the analysis would be the following:
Anode (oxidation): 2 Cl–(aq) → Cl2(g) + 2 e–
Cathode (reduction): 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
Overall reaction: 2 H2O + 2 Cl–(aq) → H2(g) + Cl2(g) + 2 OH–(aq)
As the overall reaction indicates, the concentration of chloride ions is reduced in comparison to OH– ions (whose concentration increases). The reaction also shows the production of gaseous hydrogen, chlorine and aqueous sodium hydroxide.”
Source
Electroplating of Metals:
Copper (II) ion leaves the anode and plates out on the cathode. Unreactive substances (such as gold) collect as anode mud under the anode (very valuable).
Stoichiometry of Electrolysis:
Ampere: Denoted A, is the SI base unit for current. The coulomb ( C ) is the SI unit of electric charge and is equivalent to an ampere-second.
Coulombs = Amperes x Seconds
(Electric Charge = Electric Current x Time Lapse)
and
Amperes = Coulombs / Seconds
One faraday (96,485 Coulombs) is equivalent to the charge on one mole of electrons.
Example:
System: Cu2+ (aq) + 2e– → Cu (s)
A constant current passes through this system for 6.50 hours and 6.75g Cu (s) was deposits. What was the current?
Amperes = Coulombs / Seconds
To get the electric current, convert Cu from grams to moles, use stoichiometry to determine moles of e– to moles of Cu, then use moles of e– to convert to Coulombs:
6.75g Cu x (1 mol Cu / 63.5g Cu) x (2 mol e– / 1 mol Cu) x (96485 C / 1 mol e–) = 20512.6 C
To get the time, convert 6.50h to seconds:
6.50h x (3600 s / 1 h) = 23400s
Divide Coulombs by Seconds to get Amperes:
Amperes = 20512.6 C / 23400s = 0.876 Amps
Sunday, November 14, 2010
19.4-19.7: Cell Potential-Dependence of Cell Potential on Concentration
Wmax = ∆Go = –nFEcell
Example:
What is the Wmax that can be obtained from 351g of zinc metal in the galvanic cell with the overall cell reaction of
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
Wmax = ∆Go = –nFEcell
n = moles of e– transferred = 2
F = 96485 C / mol e–
to find the Wmax we need Ecell.
To find Ecell, reference Table 19.1: Standard Electrode (Reduction) Potentials(Eo), page 786.
First, write the two half-reactions from the overall cell reaction:
Zn (s) → Zn2+ (aq) + 2e– (oxidation half-reaction)
Cu2+ (aq) + 2e– → Cu (s) (reduction half-reaction)
This table gives Eo values only for the reduction half-reactions. In order to get the Eo value for an oxidation half-reaction, reverse the order to find it as it is written in the table, and take the negative of the given Eo number given for that reaction.
Zn (s) → Zn2+ (aq) + 2e– is not a reaction that is in Table 19.1 because it is an oxidation reaction. Reverse the reaction to find it as it is written in the table:
Zn2+ (aq) + 2e– → Zn (s)
The given Eo value is – 0.76 V. Since the reaction we are working with is the reverse of this one, take the negative of this value. So:
Zn (s) → Zn2+ (aq) + 2e– (Eo = + 0.76 V)
Cu2+ (aq) + 2e– → Cu (s) (Eo = + 0.34 V)
To get the value for Ecell for the overall reaction, add these two Eo values together.
0.76 V + 0.34 V = 1.10 V = Ecell
Now back to the original question:
Wmax = ∆Go = –nFEcell
Wmax = –(2)(96485)(1.10) = –212 kJ per mole of Zn
Remember that the problem stated that we had 351g of Zn, not one mole of Zn, so:
(351g Zn) x (1 mol Zn / 65.4g Zn) x (–212 kJ / 1 mol Zn) = 1.14 x 103kJ
Standard Cell Potential: Denoted Eocell, it is the cell potential of a voltaic cell operating under standard-state conditions (solute concentrations are each 1 M, gas pressures are each 1 atm, and the temperature has a specified value, usually 25oC.) The degree sign signifies standard-state conditions.
Standard Electrode Potential: Denoted Eo, it is the electrode potential under standard-state conditions. The degree sign signifies standard-state conditions.
Standard electrode potentials help determine strengths of oxidizing and reducing agents.
-The strength of reducing agents increases going up the table.
-Example: Lithium is the strongest reducing agent.
-The strength of oxidizing agents increases going down the table.
-Example: F2 is the strongest oxidizing agent.
Example:
Which is the stronger reducing agent? Li (s) or Li+?
Li, it has more e– to give.
To determine the direction of spontaneity with this information as well:
-The stronger oxidizing agent will be on the reactant side when the equation is written as a spontaneous reaction.
-The stronger reducing agent will be on the reactant side of the spontaneous reaction.
Equilibrium Constants from Cell Potentials:
If Wmax = ∆Go
and
∆Go = –nFEocell
and
∆Go = –RtlnK
then
–nFEocell = –RtlnK
nFEocell = RtlnK
Since we're at standard conditions, we can rearrange the equation and plug in known values to get:
Eocell = (0.0592 / n)logK
Example:
Find K for the overall reaction:
2H3+ (aq) + 3Fe (s) → 2Al (s) + 3 Fe2+ (aq)
Eocell = (0.0592 / n)logK
To find K, need n and Eocell. So start, like always, with writing the half-reactions:
2Al3+ (aq) + 6e– → 2Al (s) (Eo = –1.66 V)
3Fe (s) → 3Fe2+ (aq) + 6e– (Eo = +0.41 V)
Eocell = –1.66 V + 0.41 V = –1.25 V
n = 6
(–1.25 V) = (0.0592 / 6)(logK)
logK = –127
K = 10–127
Nernst Equation:
What if we're at non-standard conditions (assuming temperature remains at 298K)? Luckily, the monopoly guy derived an equation that relates cell potential to standard cell potential.
Walther Hermann Nernst
Ecell = Eocell – (0.0592 / n)(logQ)
where Q is the reaction quotient (remember this is the same as K except not at equilibrium).
Example:
Zn (s) │Zn2+ (0.155M) ║ Ag+ (0.0200M) │ Ag (s)
Find Ecell. (Must use Nernst equation because it is at non-standard conditions.)
Ecell = Eocell – (0.0592 / n)(logQ)
First convert notation into half-reactions and find Eocell:
Zn (s) → Zn2+ (aq) + 2e– (Eo = +0.76 V)
2Ag+ (aq) + 2e– → 2Ag (s) (Eo = +0.80 V)
Eocell = +1.56 V
n = 2
Add the half-reactions to get the overall reaction:
Zn (s) + 2Ag+ (aq) → Zn2+ (aq) + 2Ag (s)
Write Q using the same format used to write K:
Q = [Zn2+] / [Ag+]2
Q = (0.155) / (0.0200)2
Q = 387.5
Ecell = Eocell – (0.0592 / n)(logQ)
Ecell = (1.56) – (0.0592 / 2)(log 387.5)
Ecell = 1.48 V
Determination of pH:
A pH meter actually measures Ecell and then backtracks to pH.
pH = (0.76 – Ecell) / (0.0592)
Example:
If the pH = 4.3, what is the Ecell of the test system?
4.3 = (0.76 – Ecell) / (0.0592)
Ecell = 0.505 V
Example:
What is the Wmax that can be obtained from 351g of zinc metal in the galvanic cell with the overall cell reaction of
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
Wmax = ∆Go = –nFEcell
n = moles of e– transferred = 2
F = 96485 C / mol e–
to find the Wmax we need Ecell.
To find Ecell, reference Table 19.1: Standard Electrode (Reduction) Potentials(Eo), page 786.
First, write the two half-reactions from the overall cell reaction:
Zn (s) → Zn2+ (aq) + 2e– (oxidation half-reaction)
Cu2+ (aq) + 2e– → Cu (s) (reduction half-reaction)
This table gives Eo values only for the reduction half-reactions. In order to get the Eo value for an oxidation half-reaction, reverse the order to find it as it is written in the table, and take the negative of the given Eo number given for that reaction.
Zn (s) → Zn2+ (aq) + 2e– is not a reaction that is in Table 19.1 because it is an oxidation reaction. Reverse the reaction to find it as it is written in the table:
Zn2+ (aq) + 2e– → Zn (s)
The given Eo value is – 0.76 V. Since the reaction we are working with is the reverse of this one, take the negative of this value. So:
Zn (s) → Zn2+ (aq) + 2e– (Eo = + 0.76 V)
Cu2+ (aq) + 2e– → Cu (s) (Eo = + 0.34 V)
To get the value for Ecell for the overall reaction, add these two Eo values together.
0.76 V + 0.34 V = 1.10 V = Ecell
Now back to the original question:
Wmax = ∆Go = –nFEcell
Wmax = –(2)(96485)(1.10) = –212 kJ per mole of Zn
Remember that the problem stated that we had 351g of Zn, not one mole of Zn, so:
(351g Zn) x (1 mol Zn / 65.4g Zn) x (–212 kJ / 1 mol Zn) = 1.14 x 103kJ
Standard Cell Potential: Denoted Eocell, it is the cell potential of a voltaic cell operating under standard-state conditions (solute concentrations are each 1 M, gas pressures are each 1 atm, and the temperature has a specified value, usually 25oC.) The degree sign signifies standard-state conditions.
Standard Electrode Potential: Denoted Eo, it is the electrode potential under standard-state conditions. The degree sign signifies standard-state conditions.
Standard electrode potentials help determine strengths of oxidizing and reducing agents.
-The strength of reducing agents increases going up the table.
-Example: Lithium is the strongest reducing agent.
-The strength of oxidizing agents increases going down the table.
-Example: F2 is the strongest oxidizing agent.
Example:
Which is the stronger reducing agent? Li (s) or Li+?
Li, it has more e– to give.
To determine the direction of spontaneity with this information as well:
-The stronger oxidizing agent will be on the reactant side when the equation is written as a spontaneous reaction.
-The stronger reducing agent will be on the reactant side of the spontaneous reaction.
Equilibrium Constants from Cell Potentials:
If Wmax = ∆Go
and
∆Go = –nFEocell
and
∆Go = –RtlnK
then
–nFEocell = –RtlnK
nFEocell = RtlnK
Since we're at standard conditions, we can rearrange the equation and plug in known values to get:
Eocell = (0.0592 / n)logK
Example:
Find K for the overall reaction:
2H3+ (aq) + 3Fe (s) → 2Al (s) + 3 Fe2+ (aq)
Eocell = (0.0592 / n)logK
To find K, need n and Eocell. So start, like always, with writing the half-reactions:
2Al3+ (aq) + 6e– → 2Al (s) (Eo = –1.66 V)
3Fe (s) → 3Fe2+ (aq) + 6e– (Eo = +0.41 V)
Eocell = –1.66 V + 0.41 V = –1.25 V
n = 6
(–1.25 V) = (0.0592 / 6)(logK)
logK = –127
K = 10–127
Nernst Equation:
What if we're at non-standard conditions (assuming temperature remains at 298K)? Luckily, the monopoly guy derived an equation that relates cell potential to standard cell potential.
Walther Hermann Nernst
Ecell = Eocell – (0.0592 / n)(logQ)
where Q is the reaction quotient (remember this is the same as K except not at equilibrium).
Example:
Zn (s) │Zn2+ (0.155M) ║ Ag+ (0.0200M) │ Ag (s)
Find Ecell. (Must use Nernst equation because it is at non-standard conditions.)
Ecell = Eocell – (0.0592 / n)(logQ)
First convert notation into half-reactions and find Eocell:
Zn (s) → Zn2+ (aq) + 2e– (Eo = +0.76 V)
2Ag+ (aq) + 2e– → 2Ag (s) (Eo = +0.80 V)
Eocell = +1.56 V
n = 2
Add the half-reactions to get the overall reaction:
Zn (s) + 2Ag+ (aq) → Zn2+ (aq) + 2Ag (s)
Write Q using the same format used to write K:
Q = [Zn2+] / [Ag+]2
Q = (0.155) / (0.0200)2
Q = 387.5
Ecell = Eocell – (0.0592 / n)(logQ)
Ecell = (1.56) – (0.0592 / 2)(log 387.5)
Ecell = 1.48 V
Determination of pH:
A pH meter actually measures Ecell and then backtracks to pH.
pH = (0.76 – Ecell) / (0.0592)
Example:
If the pH = 4.3, what is the Ecell of the test system?
4.3 = (0.76 – Ecell) / (0.0592)
Ecell = 0.505 V
Saturday, November 13, 2010
19.2-19.4: Construction of Voltaic Cells-Cell Potential
Construction of Voltaic Cells:
Terms:
Electrochemical Cell: a system consisting of electrodes that dip into an electrolyte and in which a chemical reaction either uses or generates an electric current.
Voltaic/Galvanic Cell: an electrochemical cell in which a spontaneous reaction generates an electric current.
Electrolytic Cell: an electrochemical cell in which an electric current drives an otherwise non-spontaneous reaction.
Galvanic/Voltaic Cell:
A voltaic cell consists of two half-cells.
Half-cell: the portion of an electrochemical cell in which a half reaction takes place, electrically connected to the other half-cell.
Here, one is a zinc/zinc ion half-cell where a strip of zinc metal is in a solution of a zinc salt. The other is a copper metal strip in a solution of a copper salt. These are called a zinc electrode (the anode) and copper electrode (the cathode), respectively.
An external circuit must connect the two zinc and copper half-cells two allow electron flow between the metal electrodes, represented in the diagram by the placement of the voltmeter.
The zinc electrode is dissolved as it's oxidized (as the zinc metal atom loses two e–). These e– flow through the zinc electrode to the external circuit. From the external circuit, they reach the copper electrode and the other half-cell. This gives the copper electrode a negative charge, which attracts the positively charged copper cations in solution. Copper is then deposited on the copper electrode as copper cations become reduced to copper metal.
One metal must be able to reduce the cation of the other without allowing the two solutions to mix. The charges in the solution need to remain in contact without allowing the metals to react. The solution to this problem is a salt bridge, the peach-colored tube filled with KCl in the diagram.
Salt Bridge: a tube of an electrolyte in a gel that is connected to the two half-cells of a voltaic cell; the salt bridge allows the flow of ions but prevents the mixing of the different solutions that would allow direct reaction of the cell reactants.
The two half-cell reactions are:
Zn (s) → Zn2+ (aq) + 2e–
(oxidation half-reaction, a species loses electrons)
Cu2+ (aq) + 2e– → Cu (s)
(reduction half-reaction, a species gains electrons)
Anode: the electrode at which oxidation occurs. Electrons are given up by the anode and flow to the cathode. Represented with a negative sign to show that electrons flow from it.
Cathode: the electrode at which reduction occurs. Electrons flow from the anode to the cathode. Represented with a positive sign.
Cell Reaction: the sum of the two half-reactions, the net reaction that occurs in the voltaic cell.
Cell Reaction:
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
Note: electrons are never aqueous, don't give them phase labels.
To sketch and label a voltaic cell, it helps to remember the key information:
1. Remember the basic setup of two half-cells and their electrodes connected by an external current and an internal salt bridge.
2. Recognize any information in the problem that will help you figure out which half-cell reaction is the reduction and which is the oxidation. If one half-cell is known to be reduced, then we know that it is the cathode and the other is the anode.
3. Remember the electrons flow away from the anode to the cathode.
Notation:
Example:
For the voltaic cell in the above diagram, it is written as:
Zn (s) │ Zn2+ (aq) ║ Cu2+ (aq) │ Cu (s)
-The anode (oxidation half-cell) is written on the left, the cathode (reduction half-cell) is written on the right.
-The cell terminals are written on the far right and far left of the notation. The salt solutions are written on the inner right and inner left of the notation.
-The single vertical line represents a phase boundary, for example between the solid and the solution.
-The double vertical line represents the salt bridge.
-When the half reaction involves a gas, an inert material such as platinum may be used as a surface for the reaction. In this case, Pt will be added to the outside end of that electrode half-reaction.
-Coefficients are not included.
Example:
Write the following reaction in cell notation:
2Tl (s) + Sn2+ (aq) → 2Tl+ (aq) + Sn (s)
First write the half-reactions and find out what is reduced and what is oxidized.
2Tl (s) → 2Tl+ (aq) + 2e– (oxidation)
2e– + Sn2+ (aq) → Sn (s) (reduction)
Apply notation:
Tl (s) │ Tl+ (aq) ║ Sn2+ (aq) │ Sn (s)
Example:
Write the following reaction in cell notation:
Zn (s) + 2Fe3+ (aq) → Zn2+ (aq) + 2 Fe2+ (aq)
(with platinum used as a catalyst).
First write the half-reactions and find out what is reduced and what is oxidized.
Zn (s) → Zn2+ (aq) + 2e–
Fe3+ (aq) + e– → Fe2+ (aq)
Apply notation:
Zn (s) │ Zn2+ (aq) ║ Fe3+ (aq) , Fe2+ (aq) │ Pt
Note: There is a comma between Fe3+ and Fe2+ on the right side because they are both in the same phase, and platinum appears on the end.
To write a cell reaction from the cell notation, write each half-reaction from the notation and add them together.
Example:
Tl (s) │ Tl+ (aq) ║ Sn2+ (aq) │ Sn (s)
Write half-reactions:
2Tl (s) → 2Tl+ (aq) + 2e–
2e– + Sn2+ (aq) → Sn (s)
Take the sum of these reactions:
2Tl (s) + 2e– + Sn2+ (aq) → 2Tl+ (aq) + 2e– + Sn (s)
(Simplify)
2Tl (s) + Sn2+ (aq) → 2Tl+ (aq) + Sn (s)
Note: These cell notations show products and reactants in standard state:
aq → 1.00 M
gas → 1 atm
When not in standard state, the information for concentration or pressure goes in parentheses just to the right for each species.
Example:
Zn (s) │ Zn2+ (aq) ║ H+ (aq) (0.200 M) │ H2 (g) │ Pt
Cell Potential:
Potential Difference: the difference in electric potential (electrical pressure) between two points. Measured in volts (V).
The electrical work expended on a charge moving through an electric potential:
Electrical work = charge x potential difference
Using corresponding units:
Joules = coulombs x volts → J = C x V
Faraday Constant: Denoted F, the magnitude of the charge on one mole of electrons; equal to 96,485 C/mol e–. The faraday is a unit of charge equal to 96,485 C.
Cell Potential: the maximum potential difference between the electrodes of a voltaic cell. Denoted Ecell. Ecell must be positive for a spontaneous process. Larger values for Ecell are desirable. The larger the Ecell, the greater the energy we get out of the system.
Maximum work obtainable form a voltaic cell:
Wmax = ∆Go = –nFEcell
where:
n = number of moles of electrons transferred in the overall cell equation.
F = the Faraday constant
Ecell = the cell potential
Note: ∆Go will be less than zero for a spontaneous process.
Terms:
Electrochemical Cell: a system consisting of electrodes that dip into an electrolyte and in which a chemical reaction either uses or generates an electric current.
Voltaic/Galvanic Cell: an electrochemical cell in which a spontaneous reaction generates an electric current.
Electrolytic Cell: an electrochemical cell in which an electric current drives an otherwise non-spontaneous reaction.
Galvanic/Voltaic Cell:
A voltaic cell consists of two half-cells.
Half-cell: the portion of an electrochemical cell in which a half reaction takes place, electrically connected to the other half-cell.
Here, one is a zinc/zinc ion half-cell where a strip of zinc metal is in a solution of a zinc salt. The other is a copper metal strip in a solution of a copper salt. These are called a zinc electrode (the anode) and copper electrode (the cathode), respectively.
An external circuit must connect the two zinc and copper half-cells two allow electron flow between the metal electrodes, represented in the diagram by the placement of the voltmeter.
The zinc electrode is dissolved as it's oxidized (as the zinc metal atom loses two e–). These e– flow through the zinc electrode to the external circuit. From the external circuit, they reach the copper electrode and the other half-cell. This gives the copper electrode a negative charge, which attracts the positively charged copper cations in solution. Copper is then deposited on the copper electrode as copper cations become reduced to copper metal.
One metal must be able to reduce the cation of the other without allowing the two solutions to mix. The charges in the solution need to remain in contact without allowing the metals to react. The solution to this problem is a salt bridge, the peach-colored tube filled with KCl in the diagram.
Salt Bridge: a tube of an electrolyte in a gel that is connected to the two half-cells of a voltaic cell; the salt bridge allows the flow of ions but prevents the mixing of the different solutions that would allow direct reaction of the cell reactants.
The two half-cell reactions are:
Zn (s) → Zn2+ (aq) + 2e–
(oxidation half-reaction, a species loses electrons)
Cu2+ (aq) + 2e– → Cu (s)
(reduction half-reaction, a species gains electrons)
Anode: the electrode at which oxidation occurs. Electrons are given up by the anode and flow to the cathode. Represented with a negative sign to show that electrons flow from it.
Cathode: the electrode at which reduction occurs. Electrons flow from the anode to the cathode. Represented with a positive sign.
Cell Reaction: the sum of the two half-reactions, the net reaction that occurs in the voltaic cell.
Cell Reaction:
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
Note: electrons are never aqueous, don't give them phase labels.
To sketch and label a voltaic cell, it helps to remember the key information:
1. Remember the basic setup of two half-cells and their electrodes connected by an external current and an internal salt bridge.
2. Recognize any information in the problem that will help you figure out which half-cell reaction is the reduction and which is the oxidation. If one half-cell is known to be reduced, then we know that it is the cathode and the other is the anode.
3. Remember the electrons flow away from the anode to the cathode.
Notation:
Example:
For the voltaic cell in the above diagram, it is written as:
Zn (s) │ Zn2+ (aq) ║ Cu2+ (aq) │ Cu (s)
-The anode (oxidation half-cell) is written on the left, the cathode (reduction half-cell) is written on the right.
-The cell terminals are written on the far right and far left of the notation. The salt solutions are written on the inner right and inner left of the notation.
-The single vertical line represents a phase boundary, for example between the solid and the solution.
-The double vertical line represents the salt bridge.
-When the half reaction involves a gas, an inert material such as platinum may be used as a surface for the reaction. In this case, Pt will be added to the outside end of that electrode half-reaction.
-Coefficients are not included.
Example:
Write the following reaction in cell notation:
2Tl (s) + Sn2+ (aq) → 2Tl+ (aq) + Sn (s)
First write the half-reactions and find out what is reduced and what is oxidized.
2Tl (s) → 2Tl+ (aq) + 2e– (oxidation)
2e– + Sn2+ (aq) → Sn (s) (reduction)
Apply notation:
Tl (s) │ Tl+ (aq) ║ Sn2+ (aq) │ Sn (s)
Example:
Write the following reaction in cell notation:
Zn (s) + 2Fe3+ (aq) → Zn2+ (aq) + 2 Fe2+ (aq)
(with platinum used as a catalyst).
First write the half-reactions and find out what is reduced and what is oxidized.
Zn (s) → Zn2+ (aq) + 2e–
Fe3+ (aq) + e– → Fe2+ (aq)
Apply notation:
Zn (s) │ Zn2+ (aq) ║ Fe3+ (aq) , Fe2+ (aq) │ Pt
Note: There is a comma between Fe3+ and Fe2+ on the right side because they are both in the same phase, and platinum appears on the end.
To write a cell reaction from the cell notation, write each half-reaction from the notation and add them together.
Example:
Tl (s) │ Tl+ (aq) ║ Sn2+ (aq) │ Sn (s)
Write half-reactions:
2Tl (s) → 2Tl+ (aq) + 2e–
2e– + Sn2+ (aq) → Sn (s)
Take the sum of these reactions:
2Tl (s) +
(Simplify)
2Tl (s) + Sn2+ (aq) → 2Tl+ (aq) + Sn (s)
Note: These cell notations show products and reactants in standard state:
aq → 1.00 M
gas → 1 atm
When not in standard state, the information for concentration or pressure goes in parentheses just to the right for each species.
Example:
Zn (s) │ Zn2+ (aq) ║ H+ (aq) (0.200 M) │ H2 (g) │ Pt
Cell Potential:
Potential Difference: the difference in electric potential (electrical pressure) between two points. Measured in volts (V).
The electrical work expended on a charge moving through an electric potential:
Electrical work = charge x potential difference
Using corresponding units:
Joules = coulombs x volts → J = C x V
Faraday Constant: Denoted F, the magnitude of the charge on one mole of electrons; equal to 96,485 C/mol e–. The faraday is a unit of charge equal to 96,485 C.
Cell Potential: the maximum potential difference between the electrodes of a voltaic cell. Denoted Ecell. Ecell must be positive for a spontaneous process. Larger values for Ecell are desirable. The larger the Ecell, the greater the energy we get out of the system.
Maximum work obtainable form a voltaic cell:
Wmax = ∆Go = –nFEcell
where:
n = number of moles of electrons transferred in the overall cell equation.
F = the Faraday constant
Ecell = the cell potential
Note: ∆Go will be less than zero for a spontaneous process.
Sunday, November 7, 2010
19.1: Balancing Oxidation-Reduction Reactions in Acidic/Basic Solutions
Electrochemistry
Balancing Oxidation-Reduction Reactions in Acidic and Basic Solutions with Half Reaction Method:
If acidic, reaction takes place in solution where H2O, H+, and e– are available in solution.
Step 1: Split the equation up into a reduction half reaction and an oxidation half reaction. In a redox reaction, there is a reduction and an oxidation. Ignoring hydrogen and oxygen, be sure to match pairs with a common element, in other words, don't create elements out of thin air.
Step 2: Balance all elements that are not H or O
Step 3: Balance O atoms by adding H2O's to one side of the equation.
Step 4: Balance H atoms by adding H+ ions to one side of the equation.
Step 5: Balance the electric charge by adding e– to the more positive side. Note that you are not trying to get both sides equal to zero, just equal to each other. At this point, the reduction reaction has e– as a reactant and the oxidation reaction has e– as a product.
Step 6: Make sure the number of e– that were added to the total reduction reaction is the same as the number of e– that were added to the total oxidation reaction. If not, multiply either one or both total reactions by an integer that will make them equal.
Step 7: Combine the two half reactions and cancel species that appear on both sides and reduce the coefficients to the smallest whole numbers. No e– should appear in the final equation.
Example:
Balance using the half reaction method. (acidic)
MnO2 (aq) + HNO2 (aq) → Mn2+ (aq) + NO3– (aq)
Step 1:
MnO2 → Mn2+
HNO2 → NO3–
Step 2:
Mn is already balanced.
N is already balanced.
Step 3:
MnO2 → Mn2+ + 2H2O
HNO2 + H2O → NO3–
Step 4:
MnO2 + 4H+ → Mn2+ + 2H2O
HNO2 + H2O → NO3– + 3H+
Step 5:
MnO2 + 4H+ + 2e– → Mn2+ + 2H2O
HNO2 + H2O → NO3– + 3H+ + 2e–
Step 6:
2e– = 2e–
Step 7:
MnO2 +4H+ + 2e– + HNO2 + H2O → Mn2+ + 2H2O + NO3– + 3H+ + 2e–
=
H+ + MnO2 + HNO2 → Mn2+ + H2O + NO3–
Example:
Balance using the half reaction method. (acidic)
Cr(OH)4– + H2O2 → CrO42– + H2O
Step1:
Cr(OH)4– → CrO42–
H2O2 → H2O
Step2:
Cr is already balanced.
Step3:
Cr(OH)4– → CrO42–
H2O2 → H2O + H2O
Step4:
Cr(OH)4– → CrO42– + 4H+
H2O2 + 2H+ → 2H2O
Step5:
Cr(OH)4– → CrO42– + 4H+ + 3e–
H2O2 + 2H+ + 2e– → 2H2O
Step6:
(Cr(OH)4– → CrO42– + 4H+ + 3e–) x 2
(H2O2 + 2H+ + 2e– → 2H2O) x 3
=
2Cr(OH)4– → 2CrO42– + 8H+ + 6e–
3H2O2 + 6H+ + 6e– → 6H2O
Step7:
2Cr(OH)4– + 3H2O2 +6H+ + 6e– → 2CrO42– + 82H+ + 6e– + 6H2O
=
2Cr(OH)4– + 3H2O2 → 2CrO42– + 2H+ + 6H2O
If basic, reaction takes place in solution where H2O, OH–, and e– are available in solution. So when the reaction is in basic solution, the process of balancing by half reactions is exactly the same, except that two more steps are added on to the end.
Step 8: Take the number of H+ ions in the finished equation. Add that many OH– ions to both sides of the equation.
Step 9: Every H+ will react with a OH– to produce that many water molecules. Cancel repeating water molecules and reduce.
Example:
Balance using the half reaction method. (basic)
Bi(OH)3 + Sn(OH)3– → Sn(OH)62– + Bi
Step 1:
Bi(OH)3 → Bi
Sn(OH)3– → Sn(OH)62–
Step 2:
Bi and Sn are balanced.
Step 3:
Bi(OH)3 → Bi + 3H2O
Sn(OH)3– + 3H2O → Sn(OH)62–
Step 4:
Bi(OH)3 + 3H+ → Bi + 3H2O
Sn(OH)3– + 3H2O → Sn(OH)62– + 3 H+
Step 5:
Bi(OH)3 + 3H+ + 3e– → Bi + 3H2O
Sn(OH)3– + 3H2O → Sn(OH)62– + 3 H+ + 2e–
Step 6:
(Bi(OH)3 + 3H+ + 3e– → Bi + 3H2O) x 2
(Sn(OH)3– + 3H2O → Sn(OH)62– + 3 H+ + 2e–) x 3
=
2Bi(OH)3 + 6H+ + 6e– → 2Bi + 6H2O
3Sn(OH)3– + 9H2O → 3Sn(OH)62– + 9H+ + 6e–
Step 7:
2Bi(OH)3 +6H+ + 6e– + 3Sn(OH)3– + 93H2O → 2Bi + 6H2O + 3Sn(OH)62– + 93H+ + 6e–
=
2Bi(OH)3 + 3Sn(OH)3– + 3H2O → 2Bi + 3Sn(OH)62– + 3H+
Step 8:
2Bi(OH)3 + 3Sn(OH)3– + 3H2O + 3OH– → 2Bi + 3Sn(OH)62– + 3H+ + 3OH–
Step 9:
2Bi(OH)3 + 3Sn(OH)3– +3H2O + 3OH– → 2Bi + 3Sn(OH)62– + 3H2O
=
2Bi(OH)3 + 3Sn(OH)3– + 3OH– → 2Bi + 3Sn(OH)62–
Balancing Oxidation-Reduction Reactions in Acidic and Basic Solutions with Half Reaction Method:
If acidic, reaction takes place in solution where H2O, H+, and e– are available in solution.
Step 1: Split the equation up into a reduction half reaction and an oxidation half reaction. In a redox reaction, there is a reduction and an oxidation. Ignoring hydrogen and oxygen, be sure to match pairs with a common element, in other words, don't create elements out of thin air.
Step 2: Balance all elements that are not H or O
Step 3: Balance O atoms by adding H2O's to one side of the equation.
Step 4: Balance H atoms by adding H+ ions to one side of the equation.
Step 5: Balance the electric charge by adding e– to the more positive side. Note that you are not trying to get both sides equal to zero, just equal to each other. At this point, the reduction reaction has e– as a reactant and the oxidation reaction has e– as a product.
Step 6: Make sure the number of e– that were added to the total reduction reaction is the same as the number of e– that were added to the total oxidation reaction. If not, multiply either one or both total reactions by an integer that will make them equal.
Step 7: Combine the two half reactions and cancel species that appear on both sides and reduce the coefficients to the smallest whole numbers. No e– should appear in the final equation.
Example:
Balance using the half reaction method. (acidic)
MnO2 (aq) + HNO2 (aq) → Mn2+ (aq) + NO3– (aq)
Step 1:
MnO2 → Mn2+
HNO2 → NO3–
Step 2:
Mn is already balanced.
N is already balanced.
Step 3:
MnO2 → Mn2+ + 2H2O
HNO2 + H2O → NO3–
Step 4:
MnO2 + 4H+ → Mn2+ + 2H2O
HNO2 + H2O → NO3– + 3H+
Step 5:
MnO2 + 4H+ + 2e– → Mn2+ + 2H2O
HNO2 + H2O → NO3– + 3H+ + 2e–
Step 6:
2e– = 2e–
Step 7:
MnO2 +
=
H+ + MnO2 + HNO2 → Mn2+ + H2O + NO3–
Example:
Balance using the half reaction method. (acidic)
Cr(OH)4– + H2O2 → CrO42– + H2O
Step1:
Cr(OH)4– → CrO42–
H2O2 → H2O
Step2:
Cr is already balanced.
Step3:
Cr(OH)4– → CrO42–
H2O2 → H2O + H2O
Step4:
Cr(OH)4– → CrO42– + 4H+
H2O2 + 2H+ → 2H2O
Step5:
Cr(OH)4– → CrO42– + 4H+ + 3e–
H2O2 + 2H+ + 2e– → 2H2O
Step6:
(Cr(OH)4– → CrO42– + 4H+ + 3e–) x 2
(H2O2 + 2H+ + 2e– → 2H2O) x 3
=
2Cr(OH)4– → 2CrO42– + 8H+ + 6e–
3H2O2 + 6H+ + 6e– → 6H2O
Step7:
2Cr(OH)4– + 3H2O2 +
=
2Cr(OH)4– + 3H2O2 → 2CrO42– + 2H+ + 6H2O
If basic, reaction takes place in solution where H2O, OH–, and e– are available in solution. So when the reaction is in basic solution, the process of balancing by half reactions is exactly the same, except that two more steps are added on to the end.
Step 8: Take the number of H+ ions in the finished equation. Add that many OH– ions to both sides of the equation.
Step 9: Every H+ will react with a OH– to produce that many water molecules. Cancel repeating water molecules and reduce.
Example:
Balance using the half reaction method. (basic)
Bi(OH)3 + Sn(OH)3– → Sn(OH)62– + Bi
Step 1:
Bi(OH)3 → Bi
Sn(OH)3– → Sn(OH)62–
Step 2:
Bi and Sn are balanced.
Step 3:
Bi(OH)3 → Bi + 3H2O
Sn(OH)3– + 3H2O → Sn(OH)62–
Step 4:
Bi(OH)3 + 3H+ → Bi + 3H2O
Sn(OH)3– + 3H2O → Sn(OH)62– + 3 H+
Step 5:
Bi(OH)3 + 3H+ + 3e– → Bi + 3H2O
Sn(OH)3– + 3H2O → Sn(OH)62– + 3 H+ + 2e–
Step 6:
(Bi(OH)3 + 3H+ + 3e– → Bi + 3H2O) x 2
(Sn(OH)3– + 3H2O → Sn(OH)62– + 3 H+ + 2e–) x 3
=
2Bi(OH)3 + 6H+ + 6e– → 2Bi + 6H2O
3Sn(OH)3– + 9H2O → 3Sn(OH)62– + 9H+ + 6e–
Step 7:
2Bi(OH)3 +
=
2Bi(OH)3 + 3Sn(OH)3– + 3H2O → 2Bi + 3Sn(OH)62– + 3H+
Step 8:
2Bi(OH)3 + 3Sn(OH)3– + 3H2O + 3OH– → 2Bi + 3Sn(OH)62– + 3H+ + 3OH–
Step 9:
2Bi(OH)3 + 3Sn(OH)3– +
=
2Bi(OH)3 + 3Sn(OH)3– + 3OH– → 2Bi + 3Sn(OH)62–
All of Chapter 18: 18.1-18.7: Enthalpy-Change of Free Energy with Temperature
Thermodynamics
(If anything starts to look too unfamiliar, review Thermochemistry, Chapter 6)
Thermodynamics is a “dead” science; it's a complete science. Because of this, we can test every new hypothesis against it, because it is true. It is the only true test.
Thermodynamics works even if matter is continuous (if matter can be halved and halved and halved forever).
Terms:
System: Whatever we're studying, things that produce the energy
Surroundings: Everything else in the universe
There are the 1st, 2nd, 3rd, and 0th Laws.
0th Law of Thermodynamics: Same as the Reflexive Property of Equality for mathematics:
a = a; 3 = 3.
Imagine three closed systems in thermal equilibrium (A, B, and C). If the temperature of A equals the temperature of B, and if the temperature of B equals the temperature of C, then the temperature of A equals the temperature of C.
In other words: Thermometers work!
First Law of Thermodynamics: Law of Conservation of Energy
Total energy is constant. Energy is neither created nor destroyed. “You never get more than you pay for.” Put a certain amount in, you'll never get more out.
Internal Energy: Denoted U, is the sum of the kinetic and potential energies of the particles making up the system. Kinetic energy = energy of motion of electrons, nuclei and molecules. Potential energy = bonds in the molecules and attractions between them.
U = internal energy = PE + KE
U is relative (there is no true zero of positional energy)
ΔU: the change is what's important
Change in Internal Energy equals heat plus work → ΔU = q + w
Thermo = Heat = q
Dynamics = Work = w
Heat: energy that moves into or out of the system because of a temperature difference between the system and its surroundings. In this case, due to random motion. q units are kJ (kilojoules). Heat flows in one direction: hot to cold
heat enters the system → +q
heat leaves the system → – q
Remember from Chapter 6:
q = s x m x Δt
(s = specific heat, m = mass, Δt = change in temperature)
qp = heat at constant pressure
qp = ΔH → heat of reaction at constant pressure is the same as change in enthalpy
Work: the energy exchange that results when a force F moves an object through a distance d.
w = F x d
– w ; work done on the surroundings by the system
+ w; work done on the system by the surroundings.
Example: Piston-cylinder system:
Zn + 2HCl → ZnCl2 + H2
The constant pressure of the atmosphere is replaced by the piston and weight. As hydrogen gas evolves and pushes up against the piston, work is done by the system on the surroundings. This is represented as – w. Or:
w = – F x Δh ← (distance is represented by the change in height the piston moves in the cylinder)
Pressure = Force / Surface Area
*Solve this expression for Force:
Force = Pressure x Surface Area = P x A
*Replace this expression for Force in the first equation:
w = – P x A x Δh
*Area times height is the same as volume. Replace A x Δh with ΔV for change in volume.
w = – PΔV → “PV work”
w = – P(vf – vi)
In other words: you can calculate the work done by a chemical reaction carried out in an open vessel by multiplying the atmospheric pressure by the change in volume of the chemical system.
Review:
ΔU = q + w
qp = heat at constant pressure
w = – PΔV
ΔU = qp – PΔV
qp = ΔH
ΔU = ΔH – PΔV
Example:
qp for a reaction is – 511 kJ (heat went out). Pressure = 2.05 x 105 Pa. Initial Volume = 5.05 m3. Final Volume = 6.15 m3.
A note on units:
Pa x m3 = J
ΔU = J
w = J
– P = Pa
101,325 Pa = 1 atm
ΔV = m3
What is the internal energy?
ΔU = qp – PΔV
ΔU = – 511 kJ – (2.05 x 105 Pa)(6.15 m3 – 5.05 m3)
ΔU = – 511 kJ – 225500J → convert J to kJ by dividing by 1000
ΔU = – 511 kJ – 226 kJ = – 737 kJ
Example:
For a reaction in a piston/cylinder system, the heat at constant pressure was – 815 kJ, and the volume contracted from 3.15 m3 to 2.50 m3. Pressure = 1.56 x 105 Pa. What is the ΔU for this process?
ΔU = – 815 kJ – (1.56 x 105 Pa)(2.50 m3 – 3.15 m3)
ΔU = – 815 kJ – (– 101400 J)
ΔU = – 815 kJ + 101 kJ = – 714 kJ
From ΔU = ΔH – PΔV, rearrange to get:
ΔH = ΔU + PΔV
*If there is no change in volume (or a negligible change) then
ΔH = ΔU (because PΔV would be zero)
For example, a bomb calorimeter does not allow a change in volume. At constant volume, ΔH = ΔU.
qwhatever means constant whatever:
qv = heat at constant volume = ΔU
qp = heat at constant pressure = ΔH
State Functions: Do not depend on the history of the system. All we care about is initial and final values.
ΔH = Hf – Hi → state function
Example: Route up a hill, the change in altitude you experience from the bottom to the top of the hill is the same no matter what route you take to get to the top of the hill.
ΔH, ΔU, ΔT, ΔV, ΔS, ΔG → all state functions
q and w are NOT state functions.
But qp, qv, and (q + w) are state functions
Example of the First Law of Thermodynamics:
Perpetual Motion DEVICES:
Are possible because devices don't do work.
Perpetual Motion MACHINES:
Are not possible because machines do work. The machine stops doing work because it eventually runs out of energy. Without an input of energy, because of the first law, you can't get more than you pay for.
Second Law of Thermodynamics:
“You get cheated every time.” Energy gets lost as random motion (frictional heat). The total entropy of a system and its surroundings always increases for a spontaneous process.
Spontaneous vs. Non-Spontaneous Processes:
Spontaneous Process: a physical or chemical change that occurs by itself, without any outside force to help it along.
Non-Spontaneous Process: the opposite of a spontaneous process. A process that does not naturally occur all by itself.
-A ball rolling down a hill = spontaneous.
-A ball rolling up a hill = non-spontaneous (not that it's impossible, it's just highly improbable). The ball could be rolled up the hill, but it would require work.
Most processes are “possible” , just so very unlikely that they'll just about never happen.
Example: You scatter a deck of cards randomly, face down on a table. The odds of you picking them up in perfect order is 1 in 52! (factorial), or 1 in 8.07 x 1067. It is not impossible, it is just highly unlikely.
Dependent on an idea called entropy.
Entropy: Denoted S, is randomness. A thermodynamic quantity representing the amount of energy in a system that is no longer available for doing mechanical work; "entropy increases as matter and energy in the universe degrade to an ultimate state of inert uniformity." Entropy represents “energy dispersal”, not specifically energy.
Because you want one possibility out of many, you're fighting randomness. (This is why we age.)
The entropy of the universe increases with every spontaneous process.
Entropy = S = in units J / K (Joules / Kelvin)
Suniverse = Ssurroundings + Ssystem
ΔS = Sf – Si
ΔS = entropy created + (q / T)
The term “entropy created” equals the movement of molecules, breaking of bonds, etc. It is very hard to quantify, so it is ignored, and we use:
ΔS = (q / T)
q = random motion in units J
T = temperature in degrees Kelvin
The entropy change associated with a flow of heat, q, at an absolute temperature, T.
At constant pressure,
ΔS is greater than qp / T (which equals ΔH / T)
but at equilibrium (at phase change, going between solid and liquid, solid and gas, or liquid and gas)
ΔS = ΔH / T
but ΔS = ΔH / T exactly only when the phase change takes forever.
But as long as the phase change happens relatively slowly, this is a great approximation.
At equilibrium and at constant pressure:
ΔHfus = heat of fusion (liquid to solid; melting)
ΔHvap = heat of vaporization (liquid to gas)
ΔHcond = condensation (gas to liquid)
Example:
ΔHvap = +42.5 kJ / mol for SeF4.
If 1 mol of liquid SeF4 at 25oC has an entropy of 185 J / K, what is the entropy of 1 mol of SeF4 gas at equilibrium with the liquid?
*Everything with a temperature greater than 0 Kelvin has a positive entropy within itself (that is inherent).
ΔSgas = ΔSliquid + (ΔHvap / T)
As ΔS increases, the randomness increases.
ΔSliquid = 185 J / K
ΔHvap = 42.5 x 103 J
T = 298 K
ΔSgas = 185 J / K + (42.5 x 103 J / 298 K)
ΔSgas = 328 J / K
Example:
ΔHcond = – 58.5 kJ / mol for TeF4.
If 1 mol of gaseous TeF4 at 25oC has an entropy of 453 J / K, what is the entropy of 1 mol of TeF4 liquid at equilibrium with the gas?
ΔSliquid = ΔSgas – (ΔHcond / T)
ΔSliquid = 453 J / K – (58.5 x 103 J / 298 K)
ΔSliquid = 257 J / K
For processes at constant temperature and pressure:
ΔH – TΔS is less than zero when the process is spontaneous.
ΔH – TΔS is greater than zero when the process is non-spontaneous.
This process will be spontaneous in the opposite direction.
ΔH – TΔS is equal to zero when the reaction is at equilibrium.
Third Law of Thermodynamics:
A substance that is a perfect crystal (pure, no breaks or occlusions) at zero degrees Kelvin has 0 entropy. No randomness.
S0 K = 0 for any substance
ΔSo = So (standard conditions) = absolute entropy
This is the entropy value for the standard state, which is the pure substance at 1 atm pressure, or the substance in solution at a concentration of 1 M.
Qualitative Info From Equations on Entropy:
There is an increase in entropy when:
1. A reaction in which a molecule is broken into two smaller molecules.
2. A reaction in which there is an increase in moles of gas (moles of gas of product is greater than moles of gas of reactants).
3. A process in which a solid changes to liquid or gas, or a liquid changes to a gas.
Entropy rises gradually as the temperature increases but jumps sharply at each phase transition.
Example:
Is ΔSo positive or negative?
CaCO3 (s) → CaO (s) + CO2 (g)
ΔSo is positive, see Rule 1.
CS2 (g) → CS2 (l)
ΔSo is negative, see Rule 3.
2Hg (l) + O2 (g) → 2HgO (s)
ΔSo is negative, see Rule 1.
Note: gas formation wins out increasing in entropy.
Remember from chapter 6:
ΔHorxn = ΣmΔHof(products) – ΣnΔHof(reactants)
Example:
CaCO3 (s) → CaO (s) + CO2 (g)
ΔHof CaCO3 = 92.9 J / K • mol
ΔHof CaO = 38.21 J / K • mol
ΔHof CO2 = 213.7 J / K • mol
Plug values into the ΔHorxn equation:
[(1mol)(38.21 J / K • mol) + (1 mol)(213.7 J / K • mol)] – [(1 mol)(92.9 J / K • mol)] = 159 J / K
The standard change of entropy, ΔSo, can be found for a reaction in a similar way.
ΔSorxn = ΣmΔSo(products) – ΣnΔSo(reactants)
ΔG = free energy
If ΔG is less than zero, reaction is spontaneous.
If ΔG is greater than zero, reaction is non-spontaneous and doesn't happen.
Thus:
ΔG = wmax ← maximum work that any process can produce (but you never actually get there).
When ΔGo is a large negative number, more negative than – 10 kJ, then the reaction is spontaneous (mostly products).
When ΔGo is a large positive number, greater than 10 kJ, then the reaction is non-spontaneous (mostly reactants).
When ΔGo is between – 10 kJ and 10 kJ, there will be a mixture of reactants and products.
ΔG = – 0.0100 kJ / mol is spontaneous
ΔG = + 0.0100 kJ / mol is non-spontaneous
ΔGorxn = ΣmΔGof(products) – ΣnΔGof(reactants)
and
ΔG = ΔH – TΔS
Example:
CH4 (g) + 2O2 (g) ⇌ CO2 (g) + 2H2O (l)
**Note: ΔHof values in Table 6.2 on page 246, Sof values in Table 18.1 on page 743, and ΔGof values in Table 18.2 on page 747. Values are also together in Appendix C.**
Find ΔSo and ΔHorxn at 298 K.
Then use these values to find ΔG.
ΔHorxn = [(1 mol)(– 393.5 KJ / mol) + (2 mol)(– 285.8 KJ / mol)] – [(1 mol)(– 74.87 KJ / mol) + (2 mol)(0)] = – 890 KJ
ΔSo = [(1 mol)(213.7 J / K • mol) + (2 mol)(69.95 J / K • mol)] – [(1 mol)(186.1 J / K • mol) + (2 mols)(205.0 J / K • mol) = – 242.5 J / K
There are two methods to solve for ΔG:
ΔG = ΔH – TΔS
ΔG = (– 890 KJ) – (298 K)(– 242 J / K)
ΔG = (– 890 KJ) – (72.27 KJ) = – 817.7KJ
or plug values into the previously stated equation for ΔG:
ΔGorxn = ΣmΔGof(products) – ΣnΔGof(reactants)
When you solve for ΔG using these two methods, your answer will not be exactly the same, but very close.
Relationship Between Thermodynamics and Kinetics:
ΔG ↔ K (thermodynamic equilibrium constant)
ΔGorxn = – RTlnK
Where:
R = 8.31 J / K • mol (Ideal Gas Constant in J / K • mol)
T = Temperature in degrees Kelvin
K = Kc or Ka or Kb or Ksp or Kf or Kd, etc.
Example:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Find K.
ΔGorxn = [(1 mol)(77.1 kJ) + (1 mol)(– 133.3 kJ)] – [(1 mol)(–109.8 kJ)] = 55.6 kJ = 55.6 x 103 J
55.6 x 103 J = – (8.31 J / K • mol)(298 K)(ln K)
(55.6 x 103) / [(– 8.31)(298)] = ln K
– 22.45 = ln K
e– 22.45 = K
K = 1.78 x 10–10
Example:
Ka = 1.7 x 10–5
Find ΔG.
ΔG = (– 8.31 J / K • mol)(298 K)(ln 1.7 x 10–5) = 27196.84 J = +27.2 kJ / mol
Consider the equation ΔG = ΔH – TΔS
How is ΔG affected by temperature?
ΔG = 0 is the point where reaction goes from spontaneous to non-spontaneous.
(If anything starts to look too unfamiliar, review Thermochemistry, Chapter 6)
Thermodynamics is a “dead” science; it's a complete science. Because of this, we can test every new hypothesis against it, because it is true. It is the only true test.
Thermodynamics works even if matter is continuous (if matter can be halved and halved and halved forever).
Terms:
System: Whatever we're studying, things that produce the energy
Surroundings: Everything else in the universe
There are the 1st, 2nd, 3rd, and 0th Laws.
0th Law of Thermodynamics: Same as the Reflexive Property of Equality for mathematics:
a = a; 3 = 3.
Imagine three closed systems in thermal equilibrium (A, B, and C). If the temperature of A equals the temperature of B, and if the temperature of B equals the temperature of C, then the temperature of A equals the temperature of C.
In other words: Thermometers work!
First Law of Thermodynamics: Law of Conservation of Energy
Total energy is constant. Energy is neither created nor destroyed. “You never get more than you pay for.” Put a certain amount in, you'll never get more out.
Internal Energy: Denoted U, is the sum of the kinetic and potential energies of the particles making up the system. Kinetic energy = energy of motion of electrons, nuclei and molecules. Potential energy = bonds in the molecules and attractions between them.
U = internal energy = PE + KE
U is relative (there is no true zero of positional energy)
ΔU: the change is what's important
Change in Internal Energy equals heat plus work → ΔU = q + w
Thermo = Heat = q
Dynamics = Work = w
Heat: energy that moves into or out of the system because of a temperature difference between the system and its surroundings. In this case, due to random motion. q units are kJ (kilojoules). Heat flows in one direction: hot to cold
heat enters the system → +q
heat leaves the system → – q
Remember from Chapter 6:
q = s x m x Δt
(s = specific heat, m = mass, Δt = change in temperature)
qp = heat at constant pressure
qp = ΔH → heat of reaction at constant pressure is the same as change in enthalpy
Work: the energy exchange that results when a force F moves an object through a distance d.
w = F x d
– w ; work done on the surroundings by the system
+ w; work done on the system by the surroundings.
Example: Piston-cylinder system:
Zn + 2HCl → ZnCl2 + H2
The constant pressure of the atmosphere is replaced by the piston and weight. As hydrogen gas evolves and pushes up against the piston, work is done by the system on the surroundings. This is represented as – w. Or:
w = – F x Δh ← (distance is represented by the change in height the piston moves in the cylinder)
Pressure = Force / Surface Area
*Solve this expression for Force:
Force = Pressure x Surface Area = P x A
*Replace this expression for Force in the first equation:
w = – P x A x Δh
*Area times height is the same as volume. Replace A x Δh with ΔV for change in volume.
w = – PΔV → “PV work”
w = – P(vf – vi)
In other words: you can calculate the work done by a chemical reaction carried out in an open vessel by multiplying the atmospheric pressure by the change in volume of the chemical system.
Review:
ΔU = q + w
qp = heat at constant pressure
w = – PΔV
ΔU = qp – PΔV
qp = ΔH
ΔU = ΔH – PΔV
Example:
qp for a reaction is – 511 kJ (heat went out). Pressure = 2.05 x 105 Pa. Initial Volume = 5.05 m3. Final Volume = 6.15 m3.
A note on units:
Pa x m3 = J
ΔU = J
w = J
– P = Pa
101,325 Pa = 1 atm
ΔV = m3
What is the internal energy?
ΔU = qp – PΔV
ΔU = – 511 kJ – (2.05 x 105 Pa)(6.15 m3 – 5.05 m3)
ΔU = – 511 kJ – 225500J → convert J to kJ by dividing by 1000
ΔU = – 511 kJ – 226 kJ = – 737 kJ
Example:
For a reaction in a piston/cylinder system, the heat at constant pressure was – 815 kJ, and the volume contracted from 3.15 m3 to 2.50 m3. Pressure = 1.56 x 105 Pa. What is the ΔU for this process?
ΔU = – 815 kJ – (1.56 x 105 Pa)(2.50 m3 – 3.15 m3)
ΔU = – 815 kJ – (– 101400 J)
ΔU = – 815 kJ + 101 kJ = – 714 kJ
From ΔU = ΔH – PΔV, rearrange to get:
ΔH = ΔU + PΔV
*If there is no change in volume (or a negligible change) then
ΔH = ΔU (because PΔV would be zero)
For example, a bomb calorimeter does not allow a change in volume. At constant volume, ΔH = ΔU.
qwhatever means constant whatever:
qv = heat at constant volume = ΔU
qp = heat at constant pressure = ΔH
State Functions: Do not depend on the history of the system. All we care about is initial and final values.
ΔH = Hf – Hi → state function
Example: Route up a hill, the change in altitude you experience from the bottom to the top of the hill is the same no matter what route you take to get to the top of the hill.
ΔH, ΔU, ΔT, ΔV, ΔS, ΔG → all state functions
q and w are NOT state functions.
But qp, qv, and (q + w) are state functions
Example of the First Law of Thermodynamics:
Perpetual Motion DEVICES:
Are possible because devices don't do work.
Perpetual Motion MACHINES:
Are not possible because machines do work. The machine stops doing work because it eventually runs out of energy. Without an input of energy, because of the first law, you can't get more than you pay for.
Second Law of Thermodynamics:
“You get cheated every time.” Energy gets lost as random motion (frictional heat). The total entropy of a system and its surroundings always increases for a spontaneous process.
Spontaneous vs. Non-Spontaneous Processes:
Spontaneous Process: a physical or chemical change that occurs by itself, without any outside force to help it along.
Non-Spontaneous Process: the opposite of a spontaneous process. A process that does not naturally occur all by itself.
-A ball rolling down a hill = spontaneous.
-A ball rolling up a hill = non-spontaneous (not that it's impossible, it's just highly improbable). The ball could be rolled up the hill, but it would require work.
Most processes are “possible” , just so very unlikely that they'll just about never happen.
Example: You scatter a deck of cards randomly, face down on a table. The odds of you picking them up in perfect order is 1 in 52! (factorial), or 1 in 8.07 x 1067. It is not impossible, it is just highly unlikely.
Dependent on an idea called entropy.
Entropy: Denoted S, is randomness. A thermodynamic quantity representing the amount of energy in a system that is no longer available for doing mechanical work; "entropy increases as matter and energy in the universe degrade to an ultimate state of inert uniformity." Entropy represents “energy dispersal”, not specifically energy.
Because you want one possibility out of many, you're fighting randomness. (This is why we age.)
The entropy of the universe increases with every spontaneous process.
Entropy = S = in units J / K (Joules / Kelvin)
Suniverse = Ssurroundings + Ssystem
ΔS = Sf – Si
ΔS = entropy created + (q / T)
The term “entropy created” equals the movement of molecules, breaking of bonds, etc. It is very hard to quantify, so it is ignored, and we use:
ΔS = (q / T)
q = random motion in units J
T = temperature in degrees Kelvin
The entropy change associated with a flow of heat, q, at an absolute temperature, T.
At constant pressure,
ΔS is greater than qp / T (which equals ΔH / T)
but at equilibrium (at phase change, going between solid and liquid, solid and gas, or liquid and gas)
ΔS = ΔH / T
but ΔS = ΔH / T exactly only when the phase change takes forever.
But as long as the phase change happens relatively slowly, this is a great approximation.
At equilibrium and at constant pressure:
ΔHfus = heat of fusion (liquid to solid; melting)
ΔHvap = heat of vaporization (liquid to gas)
ΔHcond = condensation (gas to liquid)
Example:
ΔHvap = +42.5 kJ / mol for SeF4.
If 1 mol of liquid SeF4 at 25oC has an entropy of 185 J / K, what is the entropy of 1 mol of SeF4 gas at equilibrium with the liquid?
*Everything with a temperature greater than 0 Kelvin has a positive entropy within itself (that is inherent).
ΔSgas = ΔSliquid + (ΔHvap / T)
As ΔS increases, the randomness increases.
ΔSliquid = 185 J / K
ΔHvap = 42.5 x 103 J
T = 298 K
ΔSgas = 185 J / K + (42.5 x 103 J / 298 K)
ΔSgas = 328 J / K
Example:
ΔHcond = – 58.5 kJ / mol for TeF4.
If 1 mol of gaseous TeF4 at 25oC has an entropy of 453 J / K, what is the entropy of 1 mol of TeF4 liquid at equilibrium with the gas?
ΔSliquid = ΔSgas – (ΔHcond / T)
ΔSliquid = 453 J / K – (58.5 x 103 J / 298 K)
ΔSliquid = 257 J / K
For processes at constant temperature and pressure:
ΔH – TΔS is less than zero when the process is spontaneous.
ΔH – TΔS is greater than zero when the process is non-spontaneous.
This process will be spontaneous in the opposite direction.
ΔH – TΔS is equal to zero when the reaction is at equilibrium.
Third Law of Thermodynamics:
A substance that is a perfect crystal (pure, no breaks or occlusions) at zero degrees Kelvin has 0 entropy. No randomness.
S0 K = 0 for any substance
ΔSo = So (standard conditions) = absolute entropy
This is the entropy value for the standard state, which is the pure substance at 1 atm pressure, or the substance in solution at a concentration of 1 M.
Qualitative Info From Equations on Entropy:
There is an increase in entropy when:
1. A reaction in which a molecule is broken into two smaller molecules.
2. A reaction in which there is an increase in moles of gas (moles of gas of product is greater than moles of gas of reactants).
3. A process in which a solid changes to liquid or gas, or a liquid changes to a gas.
Entropy rises gradually as the temperature increases but jumps sharply at each phase transition.
Example:
Is ΔSo positive or negative?
CaCO3 (s) → CaO (s) + CO2 (g)
ΔSo is positive, see Rule 1.
CS2 (g) → CS2 (l)
ΔSo is negative, see Rule 3.
2Hg (l) + O2 (g) → 2HgO (s)
ΔSo is negative, see Rule 1.
Note: gas formation wins out increasing in entropy.
Remember from chapter 6:
ΔHorxn = ΣmΔHof(products) – ΣnΔHof(reactants)
Example:
CaCO3 (s) → CaO (s) + CO2 (g)
ΔHof CaCO3 = 92.9 J / K • mol
ΔHof CaO = 38.21 J / K • mol
ΔHof CO2 = 213.7 J / K • mol
Plug values into the ΔHorxn equation:
[(1mol)(38.21 J / K • mol) + (1 mol)(213.7 J / K • mol)] – [(1 mol)(92.9 J / K • mol)] = 159 J / K
The standard change of entropy, ΔSo, can be found for a reaction in a similar way.
ΔSorxn = ΣmΔSo(products) – ΣnΔSo(reactants)
ΔG = free energy
If ΔG is less than zero, reaction is spontaneous.
If ΔG is greater than zero, reaction is non-spontaneous and doesn't happen.
Thus:
ΔG = wmax ← maximum work that any process can produce (but you never actually get there).
When ΔGo is a large negative number, more negative than – 10 kJ, then the reaction is spontaneous (mostly products).
When ΔGo is a large positive number, greater than 10 kJ, then the reaction is non-spontaneous (mostly reactants).
When ΔGo is between – 10 kJ and 10 kJ, there will be a mixture of reactants and products.
ΔG = – 0.0100 kJ / mol is spontaneous
ΔG = + 0.0100 kJ / mol is non-spontaneous
ΔGorxn = ΣmΔGof(products) – ΣnΔGof(reactants)
and
ΔG = ΔH – TΔS
Example:
CH4 (g) + 2O2 (g) ⇌ CO2 (g) + 2H2O (l)
**Note: ΔHof values in Table 6.2 on page 246, Sof values in Table 18.1 on page 743, and ΔGof values in Table 18.2 on page 747. Values are also together in Appendix C.**
Find ΔSo and ΔHorxn at 298 K.
Then use these values to find ΔG.
ΔHorxn = [(1 mol)(– 393.5 KJ / mol) + (2 mol)(– 285.8 KJ / mol)] – [(1 mol)(– 74.87 KJ / mol) + (2 mol)(0)] = – 890 KJ
ΔSo = [(1 mol)(213.7 J / K • mol) + (2 mol)(69.95 J / K • mol)] – [(1 mol)(186.1 J / K • mol) + (2 mols)(205.0 J / K • mol) = – 242.5 J / K
There are two methods to solve for ΔG:
ΔG = ΔH – TΔS
ΔG = (– 890 KJ) – (298 K)(– 242 J / K)
ΔG = (– 890 KJ) – (72.27 KJ) = – 817.7KJ
or plug values into the previously stated equation for ΔG:
ΔGorxn = ΣmΔGof(products) – ΣnΔGof(reactants)
When you solve for ΔG using these two methods, your answer will not be exactly the same, but very close.
Relationship Between Thermodynamics and Kinetics:
ΔG ↔ K (thermodynamic equilibrium constant)
ΔGorxn = – RTlnK
Where:
R = 8.31 J / K • mol (Ideal Gas Constant in J / K • mol)
T = Temperature in degrees Kelvin
K = Kc or Ka or Kb or Ksp or Kf or Kd, etc.
Example:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Find K.
ΔGorxn = [(1 mol)(77.1 kJ) + (1 mol)(– 133.3 kJ)] – [(1 mol)(–109.8 kJ)] = 55.6 kJ = 55.6 x 103 J
55.6 x 103 J = – (8.31 J / K • mol)(298 K)(ln K)
(55.6 x 103) / [(– 8.31)(298)] = ln K
– 22.45 = ln K
e– 22.45 = K
K = 1.78 x 10–10
Example:
Ka = 1.7 x 10–5
Find ΔG.
ΔG = (– 8.31 J / K • mol)(298 K)(ln 1.7 x 10–5) = 27196.84 J = +27.2 kJ / mol
Consider the equation ΔG = ΔH – TΔS
How is ΔG affected by temperature?
ΔG = 0 is the point where reaction goes from spontaneous to non-spontaneous.
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