I know what a rate determining step is, and also have knowledge about zero order and first order reactions. But, in one of my tests, I was unable to solve the following question:

A hypothetical reaction A2 + B2 --> 2AB has a rate law, Rate = k[A2]1 [B2]0 and mechanism of reaction is

Step-I: B2 --> 2B

Step-II: A2 --> 2A

Step-III: 2A + 2B --> 2AB

The correct statement is

A) Step-I of mechanism is rate determining step of the reaction

B) Step-II of mechanism is rate determining step of the reaction

C) Step-III of mechanism is rate determining step of the reaction if step I and II are reversible

D) Step-I and Step-III both are rate determining step of the reaction if step II are reversible

The correct answer is option B, but how would you approach the problem and solve it?

EDIT: What if rate was given as

1) Rate = k[A2]1 [B2]1

2) Rate = k[A2]2 [B2]1


The reaction is 1st-order in A2, so for starters we look for a step with A2 on the left, with a coefficient of 1. In fact, only step II has A2 on the left at all, so really we don't have to look any further.

Comments which may (or may not) make it clearer:

The rate equation says that greater the A2 concentration, the faster the reaction goes, in a linear relationship, and the rate doesn't depend on [B2] at all, since [B2] has an exponent of 0, and anything to the zeroth power is 1. (I suspect that was just put there to make it hard.)

If the whole reaction can be speeded up by adding A2 alone, [A2] must be rate-determining; i.e., [A2] determines the rate, and A2 reacts only in step II.


The answer is dependent on the assumption that the reactions are elementary reactions. In other words that the rate equation has reaction orders equal to the stoichiometric coefficients for each reactant. In chemical kinetics that doesn't necessarily have to be true.

So for the irreversible reactions only the forward rate, $r_f$, needs to be considered.

Step-I: $\quad\ce{B2 ->[r_f] 2B}$

$\ce{r_f = k_1[B2]}$

Step-II: $\quad\ce{A2 ->[r_f] 2A}$

$\ce{r_f = k_2[A_2]}$

Step-III: $\quad\ce{2A + 2B ->[r_f] 2AB}$

$\ce{r_f = k_3[A]^2[B]^2}$

For the reversible reactions both the forward rate, $r_f$, and the backwards rate, $r_b$ need to be considered to get the overall rate, $r$.

$r = r_f - r_b$

Step-I: $\quad\ce{B2 <=>[r_f][r_b] 2B}$

$\ce{r= r_f - r_b = (k_{1f}[B2]) - (k_{1b}[B]^2)}$

Step-II: $\quad\ce{A2 ->[r_f] 2A}$

$\ce{r= r_f - r_b = (k_{2f}[A2]) - (k_{2b}[A]^2)}$

Step-III: $\quad\ce{2A + 2B ->[r_f] 2AB}$

$\ce{r= r_f - r_b = (k_{3f}[A]^2[B]^2) - (k_{3b}[AB]^2)}$

So for the hypothetical reaction $\ce{A2 + B2 -> 2AB}$ with a rate law, $\ce{Rate = k[A2]^1 [B2]^0} = k[A2]$ and we just match to the various steps to see that Step II is the rate determining step.


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