1. Integrated Rate Laws 02/13/13
AP Chemistry
Integrated Rate Laws
02/13/13

2. Integrated vs. Differential Rate Laws
Differential Rate Law: A rate law that expresses how the rate depends on the concentration of the reactants. (Often simply called “a rate law.”)
Integrated Rate Law: A rate law that expresses how the concentrations depend on time.

3. First Order Rate Laws Consider the reaction: aA  products
The rate is -∆[A]/ ∆t = k[A]n
Since it is often useful to consider the change in concentration in terms of time, the following law is derived from the previous equation by some sort of miracle.
ln[A] = -kt + ln[A]0
ln = natural log, k = rate constant, t = time, [A] is at time t, [A]0 is at time 0 (the start of the experiment).

4. ln[A] = -kt + ln[A]0
This equation shows how the concentration of A depends on time. If the initial concentration and k are known, the concentration of A at any time can be calculated.

5. ln[A] = -kt + ln[A]0
Note that the form of this equation is y = mx + b.
y = ln[A] x = t m = -k b = ln[A]0
Therefore, plotting the natural log of concentration vs. time will always produce a straight line for a first order reaction.
If the plot is a straight line, the reaction is first order; if the plot is not a straight line, the reaction is not first order.

6. ln[A] = -kt + ln[A]0
The integrated rate law can also be expressed in terms of a ratio of [A] and [A]0.
ln([A]0/A]) = kt

7. First Order Rate Laws Consider 2N2O5(g)  4 NO2(g) + O2(g)
Using this data, verify that the rate law is first order and calculate k.
[N2O5] (mol/L)
Time (s)
0.1000
0.0707 50
0.0500
100
0.0250
200
0.0125
300
400

8. -2 -4 -6
100
200
300
400

9. First Order Rate Laws
Since the reaction is first order, then the slope of the line is equal to -k = ∆y/ ∆x = ∆(ln[N2O5])/ ∆t
Since the first and last points are exactly on the line, we will use them to calculate slope.
k = -(slope) = 6.93 x 10-3 s-1

10. First Order Rate Law
Using the previously obtained data, calculate [N2O5] at 150 seconds after the start of the reaction.

11. Half Life of a First Order Reaction
Half Life: the time required for a reactant to reach half its original concentration.
ln([A]0/A]) = kt
By definition at t =½: [A] = [A]0/2
ln(2) = kt½
t½ = 0.693/k

12. Half Life for First Order Reaction
A certain first order reaction has a half-life of 20.0 minutes.
Calculate the rate constant for this reaction.
How much time is required for this reaction to be 75% complete?
ln([A]0/A]) = kt
If 75% of the reaction is completed then 25% of the reactants still remain, [A]/[A]0 = .25 and [A]0/[A] = t½ = 0.693/k k=.03465

13. Calculate the rate constant. Calculate the half life.
The decomposition of A is first order, and [A] is monitored. The following data are recorded:
Calculate the rate constant.
Calculate the half life.
Calculate [A] when t = 5 min. How long will it take for 90% of A to decompose?
T (min) 1 2 4
[A]
0.100
0.0905
0.0819
0.0670

14. NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
Reaction Rate: Concentration
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)

15. The reaction 2ClO2 (aq) + 2OH- (aq) ClO3- (aq) + ClO2- (aq) + H2O (l) was studied with the following results.
Exp [ClO2] [OH-] Rate (M/s)
(a) Determine the rate law
(b) Calculate the value of the rate law constant