Question

A possible mechanism for the overall reaction Br2 (g) + 2NO (g) → 2NOBr (g) is NO (g) + Br2 (g) rightwards arrow with k space 1 on top leftwards arrow for k minus 1 of NOBr2 (g) (fast) NOBr2 (g) + NO (g) rightwards arrow with k subscript 2 on top 2NOBr (slow) The rate law for formation of NOBr based on this mechanism is rate = ________.

Answer 1

Answer : The rate law for formation of NOBr based on this mechanism is, $\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2$

Explanation :

The overall reaction is:

$Br_2(g)+2NO(g)\rightleftharpoons 2NOBr(g)$

Rate law = $k[Br_2][NO]^2$

The first step of the overall reaction is:

$NO(g)+Br_2(g)\overset{k_1}{\rightarrow} NOBr_2(g)$

$NO(g)+Br_2(g)\overset{k_1^-}{\leftarrow} NOBr_2(g)$

Rate law 1 = $k_1[Br_2][NO]$

Rate law 2 = $k_1^-[NOBr_2]$

The second step of the overall reaction is:

$NOBr_2(g)+NO(g)\overset{k_2}{\rightarrow} 2NOBr$

Rate law 3 = $k_2[NOBr_2][NO]$

Now rate law of overall reaction can be obtained as follows.

We are multiplying rate law 1 and rate law 3 and dividing by rate law 2, we get:

Rate law = $\frac{[k_1[Br_2][NO]]\times [k_2[NOBr_2][NO]]}{[k_1^-[NOBr_2]]}$

Rate law = $\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2$

Rate law = $k[Br_2][NO]^2$

The rate law for formation of NOBr based on this mechanism is, $\frac{k_1\times k_2}{k_1^-}[Br_2][NO]^2$