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.