Lab Help (click on pics to enlarge)

16.1: Acid-Ionization Equilibria

Recap:

In pure water, [H₃0⁺] = [OH^-] = 1 x 10^-7 M and pH 7 is neutral.

Acidic Solution:
[H₃0⁺] is greater than [OH^-]
[H₃0⁺] is greater than 1 x 10^-7 M
[OH^-] is less than 1 x 10^-7 M
pH is less than 7

Basic Solution:
[OH^-] is greater than [H₃0⁺]
[H₃0⁺] is less than 1 x 10^-7 M
[OH^-] is greater than 1 x 10^-7 M
pH is greater than 7

Weak Acids/Bases:

Example:

A strong acid, HCl dissociates in water 100%, so a concentration of 0.0100 M HCl would yield +0.0100 M H₃O⁺, with a pH of -log(0.0100) = 2.
A weak acid dissociates less than 100%. So 0.0100 M of a weak acid would yield less than 0.0100 M H₃O⁺. For a weak acid, the concentrations of ions in solution are determined from the acid-ionization constant.

Acid-ionization constant: the equilibrium constant for the ionization of a weak acid. Denoted K_a.

Shorthand for a weak acid is HA (where H is the hydrogen and A is the rest) for a monoprotic acid,
H₂A for a diprotic acid, and H₃A for a triprotic acid. So:

HA (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + A^- (aq)




where [H₃0⁺] and [H⁺] are interchangeable.

Example:

HC₆H₄NO₂ (aq) + H₂0 (l) ⇌ H₃O⁺ (aq) + C₆H₄NO₂^- (aq)
is the same as
HA (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + A^- (aq)

[HA] = 0.012 M
pH = 3.39
What is the K_a and what is the degree of ionization?

Degree of Ionization: the fraction of molecules that react with water to give ions. Given as a percentage it is called the percent ionization.




Since we know the pH is 3.39, we can find [H₃O⁺].

[H₃O⁺] = 10^-pH
[H₃O⁺] = 4.07 x 10^-4

So, x = 4.07 x 10^-4

K_a = (4.07 x 10^-4)(4.07 x 10^-4) / (0.0120)
K_a = 1.38 x 10^-5

To obtain the degree of ionization, notice that 4.07 x 10^-4 mol out of 0.0120 mol of acid ionizes, so

degree of ionization = (4.07 x 10^-4) / (0.0120) = 0.0339
percent ionization = 0.0339 x 100 = 3.39 %

Approximation Method:

Let C_a represent the initial concentration of the weak acid.

There will be less than a 5% error with the approximation method if C_a / K_a is greater than 100.

Example:

What is the pH for 0.500 M of acetic acid? K_a = 1.7 x 10^-5




K_a = 1.7 x 10^-5 = (x²) / (0.500 - x)

To solve this, we need to use the quadratic equation. Or, we can test to see if we are able to apply the approximation method and use that instead, since it is much quicker and easier.

Is C_a / K_a greater than 100?

(0.500) / (1.7 x 10^-5) = 29411.8

Yes.

Here is the Approximation: in the term (0.500 - x), the x will be so small that subtracting it from 0.500 will be negligible, and the term is roughly the same as simply 0.500. This approximation greatly simplifies the equation:

1.7 x 10^-5 = (x²) / (0.500)
x = 0.00292
pH = -log(0.00292) = 2.54

Remember, this substitution only works if C_a / K_a is greater than 100. If it's not, then the quadratic equation must be used.

15.1-15.8: All of Chapter 15

Strong Acids and Bases:

(If it is an acid or a base and it is not in this table, then it is a weak acid or base. An indicator that something is an acid is that it will start with a hydrogen, ex: H₃PO₃, H₃PO₄).

Three Definitions:

1. Arrhenius Concept of Acids and Bases:

-Acid: a substance that, when dissolved in water, increases the concentration of hydronium ion, H₃O⁺ (aq). H⁺ (aq) is interchangeable with H₃O⁺ (aq), but remember that it is not just a lone proton, but a proton chemically bonded to water (and with other water molecules via hydrogen bonding).

-Strong Acid: a substance that completely ionizes in aqueous solution to give H₃O⁺ (aq) and an anion.

Example:

HCl (aq) + H₂O (l) → H₃O⁺ (aq) + Cl^- (aq)
Hydrochloric acid donates a hydrogen to the water molecule to yield a hydronium ion. Hydrochloric acid thus increases the hydronium ion concentration of the solution and is considered an Arrhenius acid.

-Base: a substance that, when dissolved in water, increases the concentration of hydroxide ion, OH^- (aq).

-Strong Base: a substance that completely ionizes in aqueous solution to give OH^- and a cation.

-Weak Acids and Bases: are not completely ionized in solution and exist in reversible reaction with the corresponding ions.

Example:

NaOH (aq) → Na⁺ (aq) + OH^- (aq)
Sodium hydroxide dissociates in water and increases the concentration of hydroxide ion. It is thus considered an Arrhenius base.

2. BrØnsted-Lowery Concept of Acids and Bases:

Independently, these two guys came up with a more flexible definition of acids and bases.

Acid: the species donating a proton in a proton-transfer reaction.

Base: the species accepting the proton in a proton-transfer reaction.

Example:

HCl (aq) + H₂O (l) → H₃O⁺ (aq) + Cl^- (aq)
Hydrochloric acid donates a proton to water to yield H₃O⁺, and water accepts a proton to become H₃O⁺. So here, Hydrochloric acid is the acid and water is the base.

Example:

NH₃ (aq) + H₂O (l) → NH₄⁺ (aq) + OH^- (aq)
Ammonia accepts a proton from water to yield an ammonium ion and a hydroxide ion. So here, water is the acid, and ammonia is the base.

Conjugate Acid-Base Pair: consists of two species in an acid-base reaction, one acid and one base, that differ by the loss or gain of a proton. The acid of the pair is called the conjugate acid of the base, and the base is called the conjugate base of the acid. (ex: NH₃ and NH₄⁺, remember when you add the hydrogen to add one positive charge)


-Amphiprotic/Amphoteric Species: a species that can act as either an acid or a base (it can gain or lose a proton) depending on the other reaction. (ex: OH^– ← H₂O → H₃O⁺).

3. Lewis Concept of Acids and Bases:

Lewis Acid: a species that can form a covalent bond by accepting an electron pair from another species.

Lewis Base: a species that can form a covalent bond by donating an electron pair to another species.

Example:



Relative Strengths of Acids and Bases:


As seen in the above table, the strongest acids have the weakest conjugate bases, and the strongest bases have the weakest conjugate acids. This table can be used to predict the direction of an acid-base reaction: the normal direction of reaction is from the stronger acid and base to the weaker acid and base.

Molecular Structure and Acid Strength:

-The strength of an acid depends on how easily the proton, H⁺, is lost or removed from an H-X bond in the acid species. (ex: H-X = H-Cl, H-X = H-Br, etc.)

1) In going down a column of elements of the periodic table, the size of atom X increases (have more and more electrons), the H-X bond strength decreases, thus increases the strength of the acid.

Example: Group VIIA elements: HF, HCl, HBr, HI = HI > HBr > HCl > HF

2) Going across a row of elements of the periodic table, the electronegativity increases, the H-X bond polarity increases, and the acid strength increases.

3) An oxoacid has the structure H-O-Y- where the acidic H atom is always attached to an O atom, which, in turn, is attached to an atom Y. Other groups may be attached to Y.

-For a series of oxoacids of the same structure, differing only in the atom Y, the acid strength increases with the electronegativity of Y.

Example: HClO, HBrO, and HIO. Consider the electronegativities of Cl, Br, and I.
HClO > HBrO > HIO

4) For a series of oxoacids with the structure H-O-Y-O_n, with the same Y, more O's = higher acid strength.



5) The acid strength of a polyprotic acid and its anions decreases with increasing negative charge.

Example: H₃PO₄ > H₂PO₄^- > HPO₄^2- (the negative charge acts on the proton to hold it in tighter and make it less acidic).


Self Ionization of Water:

Self-Ionization/Autoionization: a reaction in which two like molecules react to give ions.

H₂O (l) + H₂O (l) ⇌ H₃O⁺ (aq) + OH^– (aq)

Ion-Product Constant for Water: (a different version of K_c)

K_w = [H₃O⁺][OH^-] = 1.00 x 10^-14 (at 25 degrees Celsius)
-Just like K_c, no units of measure
-Can be rearranged to find hydronium and hydroxide ion concentrations

[H₃O⁺] = 1.00 x 10^-14 / [OH^-]
[OH^-] = 1.00 x 10^-14 / [H₃O⁺]

Example:

What is the [OH^-] for 0.150 M HCl in the following reaction?

HCl (aq) + H₂O (l) → H₃O⁺ (aq) + Cl^- (aq)



We know the concentration of water is constant because it is a pure liquid, and we know that HCl dissociates completely (to 0M) because it is a strong acid.

So now we know the [H₃O⁺] is 0.150 M, so:

[OH^-] = 1.00 x 10^-14 / 0.150M = 6.67 x 10^-14 M

-In a neutral solution, [H₃O⁺] = [OH^-]
-In an acidic solution [H₃O⁺] > [OH^-]
-In a basic solution [H₃O⁺] < [OH^-]

The pH of a Solution:

-Defined as the negative logarithm of the molar hydronium ion concentration. (pH = percent hydrogen).

pH = -log [H₃O⁺]
or
[H₃O+] = 10^-pH

(Note: it is possible for pH to be larger than 14 and less than 0, it's just very uncommon)

-The lower the pH, the more acidic; the higher the pH, the more basic. Can also be used with [OH^-].

pOH = -log [OH^-]
or
[OH^-] = 10^-pOH

Also:

pH + pOH = 14

14.4-14.9: Qualitatively Interpreting the Equilibrium Constant-Effect of a Catalyst

Qualitatively Interpreting the Equilibrium Constant:

-If K_c is a large number (example: 4.1x10⁸) the products are favored at equilibrium, and there will be more product than reactant. To obtain K_c, the reactants are in the denominator, and when you divide by a smaller and smaller number (as the reactants dwindle and the products emerge), the value of K_c will get bigger and bigger.

-If K_c is a very small number (example: 4.6x10^-31) the reactants are favored at equilibrium, and there will be more reactant than product. Again, to obtain K_c, the products are in the numerator, and when you divide a small number (products) by a larger number (reactants), the value of K_c will be very small.

-If K_c is within an order of magnitude near 1 (example: 7.2), there is an even distribution of reactants and products.

-If K_c = 0, there has been no reaction.

Predicting the Direction of Reaction:

Q_c = reaction quotient
-Obtained the same exact way used to obtain K_c. Plug in the concentrations of the products raised to the power of their coefficients, divided by the concentrations of the reactants raised to the power of their coefficients.



-The only difference is K_c is at equilibrium and Q_c is not. So we can use Q_c to see how to get to K_c.

-If Q_c > K_c, numerator is too big, need more reactant, reaction will shift to the left
-If Q_c < K_c, denominator is too big, need more product, reaction will shift to the right -If Q_c = K_c, already at equilibrium

Calculating Equilibrium Concentrations:

1. Set up a table of concentrations (Initial, Change, and Equilibrium expressions in x)
2. Substitute expressions in x for equilibrium concentrations into the equilibrium-constant equation.
3. Solve the equilibrium-constant equation for the values of the equilibrium concentrations.

Note: If you cannot get rid of a squared x term easily by simply taking the square root of everything (which will happen if your initial concentrations are the same), you will have to use the quadratic equation.

This will yield two answers, only one answer will be correct. If one is positive and the other is negative, the positive answer is the correct answer. If both are positive, plug the answers into your x expressions to see which one is realistically plausible (doesn't yield a negative number for any concentrations).

Changing Reaction Conditions:

-Le Chatelier's Principle: when a system in chemical equilibrium is disturbed by a change in temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change variable.

-Three ways to alter equilibrium concentration of a gaseous reaction mixture:
1. Changing concentrations
2. Changing the partial pressures of the products and reactants by changing the volume
3. Changing the temperature

-Concentration:
-When more reactant is added or product removed from the reaction vessel, the reaction will shift to the right to compensate for this gain/loss.
-When more product is added or reactant removed, the reaction will shift to the left to compensate. More reactants will be produced and amount of product will be subsequently decreased.

-Pressures/Volume:
-Pressure is directly proportional to moles of gas, so a decrease in the volume of a reaction vessel will cause the reaction to move in the direction that has a smaller number of moles of gas.
-An increase in volume will cause the reaction to move in the direction that has a larger number of moles of gas.
-When there are equal numbers of moles of gas on either side of the reaction (example: 2 total moles of reactant yields 2 total moles of product), then a change in pressure causes no shift in the reaction.

-Temperature:
-Must know the change in enthalpy of a reaction (ΔH) to know how temperature will shift the reaction.
-If ΔH is negative, the reaction is exothermic. Heat can be "imagined" to be a product of the reaction, and when you increase a product, the reaction will shift to the left to compensate. So adding heat to an exothermic reaction will cause it to shift to the left. And a temperature decrease will cause the reaction to shift to the right.
-If ΔH is positive, the reaction is endothermic. Heat can now be imagined to be a reactant. When you increase a reactant, the reaction will shift to the right to compensate. So adding heat to an endothermic reaction will cause it to shift to the right. And a temperature decrease will cause the reaction to shift to the left.

Effect of a Catalyst:

K_c is unaffected by a catalyst. The rate of reaction increases, but the final composition at equilibrium will be the same!

14.1-14.2: Chemical Equilibrium-The Equlibrium Constant

Chemical Equilibrium:

At dynamic equilibrium, Rate_f = Rate_r, the forward rate equals the reverse rate. Both sides are always reacting at a microscopic level at the same rate.

Chemical Equilibrium: the state reached by a reaction mixture when the rates of forward and reverse reactions have become equal.

In a Moles vs Time graph, notice the amounts of substances become constant at equilibrium:

In a Rate vs Time graph, notice that the forward rate is large at first but steadily decreases, while the reverse rate starts at zero and steadily increases, until both rates are equal at equilibrium:


Molar Composition:

Example:

For the reaction

CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)

When 1.000 mol CO and 3.000 mol H₂ are placed in a 10.00 L vessel at constant temperature and allowed to come to equilibrium, the mixture is found to contain 0.387 mol H₂O. How many moles of each substance are present in the equilibrium mixture?

Think about the starting amounts in moles, which are known. The change in the amount of moles as the reaction proceeds is an unknown, so we can label it "x". Each reactant will decrease by a factor of x (the factor is indicated by the coefficient of that particular reactant) and each product will increase by a factor of x (also found by the coefficient). The resulting amounts can be found by combining our known information into equations and solving for x.

It is easiest to do this by making a table. I, C, and E indicate Initial, Change, and Equilibrium.


If the initial amounts of products isn't stated, it is assumed to be 0. Also, if we know each product (with a coefficient of 1) increases by x, and we know the resulting amount of that product, then we know x. So here, x = 0.387 mol.

If the problem were to ask for concentrations, divide the results by the volume given, in this case 10.00L.

The Equilibrium Constant:

General Form of an Equation:
aA + bB ⇌ cC + dD

Where capital letters are reactants and products and lowercase letters are coefficients of the balanced chemical equation.

-The equilibrium constant can be obtained by multiplying the concentrations of products, dividing by the product of the concentrations of reactants, and raising each concentration term to a power equal to the coefficient in the chemical equation. This is much easier to see when it's written out:



-K_c can also be denoted as K_eq
-K_c has no units
-Concentrations must be entered in mols/Liter (M), otherwise convert to these units
-Must have an equilibrium to find K_c
-Ignore concentrations of pure liquids and solids, they are constant. Another way of writing concentrations is density (same dimensions of mass over volume) which will remain the same no matter how much of a solid or liquid you have.
-Just use the concentrations of gaseous and aqueous reactants and products.

Example:

CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)

K_c = [CH₄][H₂O] / [CO][H₂]³

Example:

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

K_c = [NH₃]² / [NH₂][H₂]³

Example:

2A (g) + B(s) + 7C (aq) ⇌ 14D (l) + 3/2E (g)

K_c = [E]^3/2 / [A]²[C]⁷

-Manipulating the balanced chemical equation will change K_c:

-If you multiply the coefficients of a reaction through by a number, A, then K_c = K_c^A
-Suppose you multiply a reaction through by 2, then you must square K_c
-If you flip a reaction around so the reactants are now the products and the products are now the reactants, then you get the reciprocal, K_c = 1 / K_c

The Pressure Equilibrium Constant, K_p:

-Won't use this until later when we study Thermodynamics, but it is relevant now.
-K_p is another version of K_c
-The values plugged into K_p are partial pressures converted to atm instead of concentrations.
-Related to K_c via the ideal gas law (PV = nRT → n/V = P/RT)
-In other words, the molar concentration of a gas equals its partial pressure divided by RT, which is constant at a given temperature.


Example:

For the reaction
CH₄ (g) + 2O₂ (g) ⇌ CO₂ (g) + 2H₂O (g)

K_c = [CO₂][H₂O]² / [CH₄][O₂]²

K_p = P_CO2 · P_H2O² / P_CH4 · P_O2²

-Sometimes K_c = K_p, but not always, it depends on the coefficients of the chemical equation.

K_c = K_p when the sum of reactant coefficients equals the sum of the product coefficients.

Example:

For the same reaction
CH₄ (g) + 2O₂ (g) ⇌ CO₂ (g) + 2H₂O (g)

1 + 2 = 1 + 2

K_c = K_p

Equation to relate K_p to K_c:

K_p = K_c (RT)^Δn

R = the ideal gas constant (0.0821 atm.L/mol.K)
T = Temperature in degrees Kelvin
Δn = sum of the coefficients of the products – the sum of the coefficients of the reactants.

Note: when the sum of the reactant coefficients equals the sum of product coefficients, Δn will be zero. Anything raised to the zero power is 1, so RT^Δn will equal 1, and K_p will equal K_c times 1, or just K_c.

Equilibrium Constant for the Sum of Reactions:

-If a given chemical equation can be obtained by taking the sum of other equations, the equilibrium constant for the given equation equals the product of the equilibrium constants of the other equations.
-Note: use Hess's Law

Example:

Equation 1:
CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)
Equation 2:
CH₄ (g) + 2H₂S (g) ⇌ CS₂ (g) + 4H₂ (g)
Hess's Law:
CO + 3H₂ + CH₄ + 2H₂S ⇌ CH₄ + H₂O + CS₂ + 4H₂
Equation 3:
CO (g) + 2H₂S (g) ⇌ H₂O (g) + CS₂ (g) + H₂ (g)

K_c for Equation 1 = K₁
K₁ = [CH₄][H₂O] / [CO][H₂]³

K_c for Equation 2 = K₂
K₂ = [CS₂][H₂]⁴ / [CH₄][H₂S]²

K_c for Equation 3 = K₁ · K₂
K₃ = [CS₂][H₂O][H₂] / [CO][H₂S]²
(K₃ can be verified by looking at Equation 3 and writing K_c directly from it)

13.7-13.9: Mechanisms-Catalysis

Mechanisms:

Reaction Mechanism: The set of elementary reactions whose overall effect is given by the net chemical equation. An elementary reaction is a single molecular event, such as a collision of molecules, resulting in a reaction.

Example:

Consider the reaction of nitrogen dioxide with carbon monoxide:

NO₂ (g) + CO (g) → NO (g) + CO₂ (g) (net chemical equation)
NO₂ + NO₂ → NO₃ + NO (elementary reaction)
NO₃ + CO → NO₂ + CO₂ (elementary reaction)

Note: Elementary reactions don't have phase labels. Also, NO₃ is a molecule, not the nitrate ion, and is highly unstable. It is both created and destroyed by the reaction and is called a reaction intermediate. In the last step, the CO molecule needs something desperately energetic to smack into in order to react, and the NO₃ provides that.

Following Hess's Law, obtain the overall equation by adding the elementary reactions and canceling species that occur on both sides:

NO₂ + NO₂ → NO₃ + NO (elementary reaction)
NO₃ + CO → NO₂ + CO₂ (elementary reaction)
-------------------------------------
NO₂ +NO₂ + NO₃ + CO → NO₃ + NO + NO₂ + CO₂ (overall)

Molecularity:

Molecularity: the number of molecules on the reactant side of an elementary reaction. Products are not considered in molecularity, only reactants of the elementary reaction.

Example:

Unimolecular: an elementary reaction that involves one reactant molecule.
O₂ → 2O

Bimolecular: an elementary reaction that involves two reactant molecules.
NO₂ + CO → NO + CO₂

Termolecular: an elementary reaction that involves three reactant molecules.
Br₂ +Cl₂ + C → CBr₂Cl₂ (generally exceptionally slow because highly unlikely statistically, remember Civil War colliding bullets example).

Determining the Rate Law from an Elementary Reaction:

Example:

What is the rate law for an overall reaction? In order to do this, slow step and fast step must be labeled. Need to know the slow step because it moderates the reaction; the reaction can't proceed faster than the slowest step. So the rate law for the slow step is the rate law for the overall reaction.
Note: These are both elementary reactions, the (slow step) and (fast step) imply that. Also, k is unique for each step. Once the slow step is identified, use the general form to assume the rate equation for this step:

Rate = k₁[NO₂][F₂]

Note: The components in the rate law must be part of the overall reaction. Intermediates cannot be part of the rate law.

This is pretty straightforward for reactions with an initial slow step. It's more complicated for reactions with an initial fast, equilibrium step.

Example:

Mechanism:


When we build the rate law from our slow step, we have an intermediate which needs to be eliminated:
Rate = k₂[NO₂][NO₃]

Use the fast step to find something plausible to replace [NO₃] with:

Rate_f = k₁[N₂O₅]
Rate_r = k_-1[NO₂][NO₃]

Where Rate_f is the forward rate and Rate_r is reverse rate (k₁ and k_-1 also mean forward and reverse rate). Also, at equilibrium, by definition, the forward and reverse rates are equal to each other. They are at dynamic equilibrium, where the forward reaction and the reverse reaction are happening at the same time, all the time.

So:
k₁[N₂O₅] = k_-1[NO₂][NO₃]
Solve for the component you want to eliminate, the intermediate [NO₃]:
[NO₃] = k₁[N₂O₅] / k_-1[NO₂]
Substitute this into the rate law from the slow step for [NO₃]:
Rate = k₂[NO₂] x (k₁[N₂O₅] / k_-1[NO₂])
([NO₂] will cancel)
Rate = (k₂k₁ / k_-1) [N₂O₅]
Where all k represent a constant and can be combined into a single k to get:
Rate = k[N₂O₅]

Catalysis:

Catalysts: cause a reaction to proceed faster without being consumed by the reaction. Catalysts do this by lowering the activation energy, E_a.



The solid line represents the reaction without a catalyst. It requires more energy to reach the activated state and thus has a higher E_a. The dotted line represents the reaction with the addition of a catalyst. It requires much less energy and thus has a much lower E_a.

Homogeneous Catalysis: the use of a catalyst in the same phase as the reacting species. An example would be a gas phase catalyst and gas phase reacting species; they are all in the same phase (gas).

Heterogeneous Catalysis: the use of a catalyst that exists in a different phase from the reacting species, usually a solid catalyst in contact with a gaseous or liquid solution of reactants.

13.5-13.6: Potential Energy Diagrams-Arrhenius Equation

Potential Energy Diagrams:

Example:

*Endothermic Reaction

E_a = Activation Energy, the minimum energy of collision required for two molecules to react.
Activated Complex = ǂ
Reaction Pathway = progress of reaction
ΔH = Heat of Reaction, the difference between the energy of the products and the energy of reactants. In this example, the energy of the products is higher than the energy of the reactants, which means energy was put in, and the reaction is endothermic.

Simply: The potential energy of the reactants increases as the reaction progresses until it increases to the value for the activated complex. The activated complex is unstable and breaks down into products, the products go to lower potential energy.

*Exothermic Reaction


In this example, the energy of the reactants is higher than that of the products, so heat energy is released when the reaction progresses, and the reaction is exothermic.

Arrhenius Equation:

epic mustache man - Svante August Arrhenius

Arrhenius Equation allows us to find values for k at different temperatures (because k is temperature dependent).

ln (k₂ / k₁) = (E_a / R) [(1 / T₁) – (1 / T₂)]

Here, brackets don't relate to any concentration, they just denote a second pair of parentheses for algebraic purposes.
ln = natural log
k₁ = rate constant at T₁
k₂ = rate constant at T₂
(units for k defined by reaction order, should cancel each other in Arrhenius Equation, should be given)
E_a = Activation Energy in J / mol
R = The Ideal Gas Constant converted to different units of measure. Equal to 8.31 J / (K . mol)
T₁ = Temperature 1 in degrees K (temp. associated with k₁)
T₂ = Temperature 2 in degrees K (temp. associated with k₂)
Degrees K = Degrees Celsius + 273.15

Example:

*The rate constant for the formation of hydrogen iodide from the elements Hydrogen and Iodine is 2.7x10^-4 L / (mol . s) at 600 K and 3.5x10^-3 L / (mol . s) at 650 K. Find the activation energy.

Establish which values are given and which value is unknown:
(It doesn't matter which k becomes k₁ or k₂, as long as k₁ matches T₁ and k₂ matches T₂. This is also the part where anything not already in the proper units should be converted to the units of the Arrhenius Equation. When everything is in the appropriate units, the units don't have to be carried through the calculations, units will cancel to what needs to be left over for the unknown)
k₂ = 3.5x10^-3 L / (mol . s)
k₁ = 2.7x10^-4 L / (mol . s)
E_a = unknown
R = 8.31 J / (K . mol)
T₁ = 600 K
T₂ = 650 K

ln ( 3.5x10^-3 / 2.7x10^-4) = (E_a / 8.31) [(1 / 600) – (1 / 650)]

Perform operations, E_a = 1.66x10⁵ J / mol

*Find the rate constant at 700 K.

To establish the values for the new equation, choose either previous value for k to be the new k₁, remember its corresponding temperature is the new T₁.
k₂ = unknown
k₁ = 2.7x10^-4 L / (mol . s)
E_a = 1.66x10⁵ J / mol
R = 8.31 J / (K . mol)
T₁ = 600 K
T₂ = 700 K

ln (k₂ / 2.7x10^-4) = (1.66x10⁵ / 8.31) [(1 / 600) – (1 / 700)]

Perform operations, k₂ = 3.2x10^-2 L / (mol . s)

Note: to “undo” the ln on the left hand side, apply the inverse. The inverse of the natural log is e. Take e to the power of both sides. e^ln (k₂ / 2.7x10^-4) = k₂ / 2.7x10^-4

13.4-13.5 (Half Life-Transition State Theory)

First Order Half Life:

Half Life: time it takes for the reactant concentration to decrease to one half its initial value (t_½) .
Value of “t” when [A]_t = (½) [A]₀
So ln (½) = -kt → 0.693 = kt → t_½ = 0.693 / k

Simply: t_½ = 0.693 / k

The half life of a first order reaction is independent of initial concentration:

Example:

*SO₂Cl₂ decomposes in a first order reaction to produce SO₂ and Cl₂. What is the half life if the rate constant at a given temperature is 2.20 x 10^-5 s^-1 ?

t_½ = 0.693 / k
t_½ = 0.693 / 2.20 x 10^-5 s^-1
t_½ = 3.15 x 10⁴ s

*How long would it take for 75% of the SO₂Cl₂ to decompose?

75% is half of the initial concentration plus half of the concentration left over after the first half life, so it is two times the half life.

Second Order Half Life:

t_½ = 1 / k[A]₀

This time, the half life depends on initial concentration and each subsequent half life becomes larger as time progresses.



Zeroth Order Half Life:

t_½ = [A]₀ / 2k

Again, the half life is dependent upon initial concentration of the reactant, but instead of becoming larger as the reaction progresses, as a zeroth order reaction proceeds, each half-life gets shorter.




Graphing of Kinetic Data:

It is possible to determine the order of a reaction by graphical plotting of the data for a particular experiment.

If you plot the data for ln[A]_t on the vertical axis against time “t” on the horizontal axis, a first order reaction will give you a straight line.

If you plot the data for 1 / [A]_t on the vertical axis against time “t” on the horizontal axis, a second order reaction will give you a straight line.

If a curved line results for either graph, the reaction order is not the one that matches that graph.

An organized explanation of the key points in the chapter thus far is found in Table 13.2, pg 544.


Collision Theory: (3 Stooges Universe)



Three things must happen between objects before there is a reaction:
1. Objects must “hit” (Collision).
2. Must collide with enough energy to break bonds (Activation Energy = E_a).
3. Objects must smack into each other the right way (Orientation).

Example:

O=N + Cl ̶ Cl → [O=N ̶ Cl···Cl]ǂ
This ǂ symbol denotes a highly unstable, activated state that quickly falls apart to
O=N ̶ Cl + Cl
Notice the orientation of N to Cl. If O smacked into Cl:

N=O + Cl ̶ Cl → No Reaction

Nitrogen must collide with Chlorine, not Oxygen in order for a reaction to occur here.

Enzymes function the same way:

13.1-13.4 (Reaction Rates-Integrated Rate Laws)

Rates of Reaction (rxn):

Reaction Rate: the increase in molar concentration of product of a reaction per unit time or the decrease in molar concentration of reactant per unit time.

In units of mol/L . s = moles per liter per second. Since molarity is moles per liter, this unit is also molars per second, or M/s.

Rate = Δconc/Δtime


Four variables affect reaction rate:
1. Concentration of reactant
2. Concentration of catalyst (speeds up reaction without being consumed)
3. Temperature (Generally, as the temp. goes up, the rate goes up and vice versa)
4. Surface area of reactant and catalyst (greater surface area means faster reaction)

Example:

2N₂O₅ (g) → 4NO₂ (g) + O₂ (g)

Rate of O₂ formation: Δ[O₂] / Δt (M/s)

where Δ[O₂] = change in O₂ concentration
Δt = change in time


Rate of NO₂ formation: Δ[NO₂ ] / Δt (M/s)
Rate of N₂O₅ decomposition: Δ[N₂O₅] / Δt (M/s)

*Relate the rates of reaction for O₂ and NO₂:

(1/1) Δ[O₂] / Δt = (¼) Δ[NO₂] / Δt

Note that Δ[O₂] = ([O₂]_t – [O₂]₀) / (t_f – t_i)
Which is read as: the change in oxygen concentration is understood to be the concentration of oxygen at time “t” minus the initial concentration of oxygen, all divided by the final time minus the initial time.

Since [O₂] is a product, it is greater than zero and the sign is positive. As a reaction proceeds, we lose reactant to gain product. If [O₂] were a reactant, it would be less than zero and the sign would be negative.

*Relate the rates of reaction for O₂ and N₂O₅:

- (½) Δ[N₂O₅] / Δt = (1/1) Δ[O₂] / Δt

Note that to relate a product to a reactant, you must remember to add the negative sign to the reactant. With a reactant to a reactant, the negative signs will cancel. You can also reduce fractions if there is a least common denominator.

Example done in class not included.


Rate Laws:

Rate Law: an equation that relates the rate of a reaction to the concentrations of reactants (and catalyst) raised to various powers.

Example:

2NO₂ (g) + F₂ (g) → 2NO₂F (g)

Simply: A Rate Law determines the rate of a rxn as a function of concentrations of each reactant.
Rate = k[A]^a[B]^b[C]^c...

Where exponents are integers (for this class).
Uppercase letters in brackets are concentrations, units M.
k is a predefined proportionality constant. Has a fixed value at any given temperature.

Rate = k[NO₂][F₂] (both are to the first power)

The reaction order with respect to NO₂ is first order
See table in book, page 534, to see how reaction order affects reaction rate.

Note that [A]⁰ = 1 so it doesn't affect the Rate Law and can be disregarded.
Overall reaction order is the sum of the reaction orders in the Rate Law.
So the above reaction is first order with respect to NO₂, first order with respect to F₂ and second order overall.

Note also the you CAN NOT write the Rate Law directly from the balanced reaction. The mechanism from which it is derived is more complex, and it is only by accident that it sometimes happens to match up this way.


Determining Rate Law Constant (k) Units:

First Order: k = s^-1
Second Order: k = L/mol . s
Zeroth Order: k = mol/L . s

Initial Rates Method:

PO₃ (g) + H₂ (g) → H₂PO₃ (g)

Finding Rate Law, know general form: Rate = k[A]^a[B]^b

Example:

Rate = k[PO₃]^a[H₂]^b
*Find a, b, and k (the actual numbers) with given experimental data:
Always need one more experiment than we have exponents. With two exponents, we need three experiments.






Knowing the general form of the rate law, assume the following:

*Experiment 1:
1.50 (mol/L . s) = k (0.100 M)^a(0.100 M)^b

*Experiment 2:
3.00 (mol/L . s) = k(0.200 M)^a(0.100 M)^b

*Experiment 3:
6.00 (mol/L . s) = k(0.100 M)^a(0.200 M)^b

Divide Experiment 2 by Experiment 1:
3.00/1.50 = k(0.200 M)^a(0.100 M)^b / k(0.100 M)^a(0.100 M)^b

2 = (0.200)^a/ (0.100)^a

2 = 2^a

a = 1

Repeat the process dividing Experiment 3 by Experiment 1 to get b = 2. Substitute known values to get k.

Analyzing this data, the pattern emerges.
-If you double the concentration and the rate goes down by ¼, the order is -2.
-If you double the concentration and the rate goes down by ½, the order is -1.
-If you double the concentration and the rate doesn't change, the order is 0.
-If you double the concentration and the rate doubles, the order is 1.
-If you double the concentration and the rate quadruples, the order is 2
-If you double the concentration and the rate goes up by 8x, the order is 3.


First Order Integrated Rate Law:

Remember that the average rate over a limited amount of time for component A:

Δ[A] / Δt

Example of first order reaction: Cl₂ (g) → 2Cl (g)
(Anything involving only one molecule will be decomposition, and always first order)

The integrated rate law that will tell us the concentration for all t for a first order reaction is:

ln ([A]_t / [A]₀) = -kt


When given values for initial concentration, concentration at time t, k, or t, plug them in and find the missing variable to solve. Note that the units for k are s^-1.

Second Order Integrated Rate Law:

When two molecules are colliding:

(1 / [A]_t) = kt + (1 / [A]₀)

Note that units for k are m^-1s^-1.

Zeroth Order Integrated Rate Law:

An example is the use of Pt as a catalyst, where there is a reaction in a monolayer at the surface of the metal:

[A]_t = -kt + [A]₀

Note that units for k are M/s.