Question

Consider the mechanism. Step 1: 2A↽−−⇀B+C equilibrium Step 2: B+D⟶E slow Overall: 2A+D⟶C+E Determine the rate law for the overall reaction, where the overall rate constant is represented as ????.

Answer 1

Answer:

It should be D

Explanation:

It’s on the quiz

Answer 2

Answer : The rate law for the overall reaction is, $R=\frac{K[A]^2[D]}{[C]}$

Explanation :

As we are given the mechanism for the reaction :

Step 1 : $2A\rightleftharpoons B+C$    (equilibrium)

Step 2 : $B+D\rightarrow E$     (slow)

Overall reaction : $2A+D\rightarrow C+E$

First we have to determine the equilibrium constant from step 1.

The expression for equilibrium constant will be,

$K'=\frac{[B][C]}{[A]^2}$

Form this, the value of [B] is,

$[B]=\frac{K'[A]^2}{[C]}$       ............(1)

Now we have to determine the rate law from the slow step 2.

The expression for law will be,

$Rate=K''[B][D]$       .............(2)

Now put equation 1 in 2, we get:

$Rate=K''\frac{K'[A]^2}{[C]}[D]$

$Rate=\frac{K[A]^2[D]}{[C]}$

Therefore, the rate law for the overall reaction is, $R=\frac{K[A]^2[D]}{[C]}$