Question

The reaction: $CO+C{l}_2\to COC{l}_2$ has the following mechanism.

What is the rate expression ?

(i) $C{l}_2\underset{K_2}{\stackrel{K_1}{\rightleftharpoons }}2Cl$ (fast)
(ii) $Cl+CO\underset{K_2^{\prime }}{\stackrel{K_1^{\prime }}{\rightleftharpoons }}COCl$ (fast)
(iii) $COCl+C{l}_2\stackrel{K_3}{\to }COC{l}_2+Cl$ (slow)

• A. $\frac{d\left[COC{l}_2\right]}{dt}=K\cdot \left[CO\right][C{l}_2{]}^{1/2}$
• B. $\frac{d\left[COC{l}_2\right]}{dt}=K\cdot \left[CO\right][C{l}_2{]}^{3/2}$
• C. $\frac{d\left[COC{l}_2\right]}{dt}=K\cdot \left[CO\right][C{l}_2{]}^{5/2}$
• D. None of these

Answer 1

The correct option is B $\frac{d\left[COC{l}_2\right]}{dt}=K\cdot \left[CO\right][C{l}_2{]}^{3/2}$

By step (iii) $\frac{d\left[COC{l}_2\right]}{dt}=K_3\left[COCl\right]\left[C{l}_2\right]$ .....(a)

By step (i) $\frac{K_1}{K_2}=\frac{[Cl{]}^2}{\left[C{l}_2\right]}$ .....(b)

By step (ii) $\frac{K_1^{\prime }}{K_2^{\prime }}=\frac{\left[COCl\right]}{\left[CO\right]\left[Cl\right]}$ ....(c)

Thus, from Eqs. (b) and (c):

$\left[COCl\right]=\frac{K_1^{\prime }}{K_2^{\prime }}\left[CO\right]\times \sqrt{\frac{K_1}{K_2}\cdot \left[C{l}_2\right]}$ ....(d)

From Eqs. (a) and (d):

$\frac{d\left[COC{l}_2\right]}{dt}=K_3\frac{K_1^{\prime }}{K_2^{\prime }}{\left(\frac{K_1}{K_2}\right)}^{1/2}\left[CO\right][C{l}_2{]}^{1/2}\cdot [C{l}_2]$

$\frac{d\left[COC{l}_2\right]}{dt}=K\cdot \left[CO\right][C{l}_2{]}^{3/2}$

Hence, option $B$ is correct.