# Chemical Kinetics - Rate Laws and Mechanisms

1. Dec 15, 2007

### kazimmerman

1. The problem statement, all variables and given/known data
This is straight off an old AP Exam, but I can't seem to find it online so I'm hoping you all can help me out.

2 NO(g) + Br2(g) --> 2 NOBr(g)
The following results were obtained in experiments designed to study the rate of the reaction above.
Experiment Initial Concentration (mol/L) Initial Rate of Appearance of NOBr(M/sec)
[NO] [Br2]
1 0.02 0.02 9.6e-2
2 0.04 0.02 3.8e-1
3 0.02 0.04 1.9e-1

(I'm not sure how to correctly display tables, so hopefully you can understand what everything is.

a) Write the rate law for the reaction.
b) Calculate the value of the rate constant, k, for the reaction. Include units.
c) In experiment 2, what was the concentration of NO remaining when half of the original amount of Br2 was consumed?
d) Which of the following reaction mechanisms is consistent with the rate law established in (a)? Explain your choice.
I. NO + NO --> N2O2 (fast)
N2O2 + Br2 --> 2 NOBr (slow)
II. Br2 --> Br + Br (slow)
2 (NO + Br --> NOBr) (fast)

2. Relevant equations

Rate = k[A]^n^m

3. The attempt at a solution

a) I've figured this out and I know I am correct with the following rate law:
Rate = k[NO]^2[Br2]
b) I also have this one confirmed correct:
k = 1.2e4 L^2 mol^-2 s-1
c) This is where my problems begin. I am sure I should probably be splitting the rate law and using integrated rate laws and/or half-life equations, but I'm not completely sure where to begin. I know for second-order rates, half-life is:
t = (k[A](initial))^-1
and for a first-order reaction:
t = ln(2) k^-1
d) I understand mechanisms a bit, but I guess I don't know how to differentiate between which should occur.

Thanks ahead of time for any help. ;)

2. Dec 15, 2007

### Gokul43201

Staff Emeritus
(a) and (b) look good.

For (c), you do not need to use any of the kinetics - only the stoichiometry of the balanced equation. Translating the equation into words: for every mole of Br2 consumed, there are 2 moles of NO consumed. This should lead you to the answer for (c).

(d) As far as mechanisms are concerned, a key point is that the overall rate is largely determined by the slowest step among a series of sequential steps that constitute the overall reaction. So, the essentials of the overall rate law should be extractable from the slow step.

3. Dec 15, 2007

### kazimmerman

So, for (c), wouldn't it also be half of the initial concentration of NO, .02?

And, as far as (d) goes, I know that choice I is correct, but I just don't understand why.

4. Dec 15, 2007