Question

The reaction of formatiom of phosgene from CO and $C{l}_2$ is $CO+C{l}_2\to COC{l}_2$
The proposed mechanism is
(i) $C{l}_2\underset{K_{-1}}{\overset{K_1}{\to }}2Cl$ (fast equilibrium)
(ii) $Cl+CO\underset{K_{-2}}{\overset{K_2}{\to }}COCl$ (fast equilibrium)
(iii) $COCl+C{l}_2\stackrel{K_3}{\to }COC{l}_2+Cl$ (slow)
The above mechanism leads to the following rate law $\frac{d\left[COC{l}_2\right]}{dt}=K\left[CO\right][C{l}_2{]}^{3/2}$,
Then determine the value of K.

• A. $K=k_3\frac{k_2}{k_{-2}}{\left(\frac{k_1}{k_{-1}}\right)}^{1/2}$
• B. $K=\frac{k_1}{k_{-1}}{\left(\frac{k_2}{k_{-2}}\right)}^{1/2}$
• C. $K=\frac{k_2}{k_{-2}}{\left(\frac{k_{-1}}{k_1}\right)}^{1/2}$
• D. $K=\frac{k_{-2}}{k_2}{\left(\frac{k_1}{k_{-1}}\right)}^{1/2}$

Answer 1

Answer: (A)

$K=k_3\frac{k_2}{k_{-2}}{\left(\frac{k_1}{k_{-1}}\right)}^{1/2}$

From the slow step (rate determining step)
$rate=k_3\left[COCl\right]\left[C{l}_2\right]$......(1)
From the first fast equilibrium
$\frac{k_1}{k_{-1}}=\frac{C{l}^2}{\left[C{l}_2\right]}$
$\left[Cl\right]=\sqrt{\frac{k_1}{k_{-1}}\times \left[C{l}_2\right]}$.....(2)
From second equilibrium
$\frac{k_2}{k_{-2}}=\frac{\left[CoCl\right]}{\left[CO\right]\left[Cl\right]}$
$\left[COCl\right]=\frac{k_2}{k_{-2}}\times \left[CO\right]\left[Cl\right]$......(3)
Substitute (3) in (1)
$rate=k_3\frac{k_2}{k_{-2}}\times \left[CO\right]\left[Cl\right]\left[C{l}_2\right]$.....(4)
Substitute (2) in (4)
$rate=k_3\frac{k_2}{k_{-2}}\times \left[CO\right]\sqrt{\frac{k_1}{k_{-1}}\times \left[C{l}_2\right]}\left[C{l}_2\right]$.....(4)
$rate=k_3\frac{k_2}{k_{-2}}\times \sqrt{\frac{k_1}{k_{-1}}}\times \left[CO\right][C{l}_2{]}^{3/2}$.....(5)
But the rate equation is $rate=k\times \left[CO\right][C{l}_2{]}^{3/2}$
Hence
$k=k_3\frac{k_2}{k_{-2}}\times \sqrt{\frac{k_1}{k_{-1}}}$