The reaction of formation of phosgene from CO | vedclass

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Question

(A) The rate of the reaction is determined by the slow step $(iii)$:Rate $= k_5 [COCl][Cl_2]$Assuming equilibrium for step $(ii)$:$K_{eq2} = \frac{k_3

Vedclass · Accepted Answer

$r = k_5 	imes \frac{k_3}{k_4} 	imes \left( \frac{k_1}{k_2} ight)^{1/2} [CO][Cl_2]^{3/2}$

Vedclass · Answer

(A) The rate of the reaction is determined by the slow step $(iii)$:Rate $= k_5 [COCl][Cl_2]$Assuming equilibrium for step $(ii)$:$K_{eq2} = \frac{k_3}{k_4} = \frac{[COCl]}{[Cl][CO]} \Rightarrow [COCl] = \frac{k_3}{k_4} [Cl][CO]$Assuming equilibrium for step $(i)$:$K_{eq1} = \frac{k_1}{k_2} = \frac{[Cl]^2}{[Cl_2]} \Rightarrow [Cl] = \left( \frac{k_1}{k_2} ight)^{1/2} [Cl_2]^{1/2}$Substituting $[Cl]$ into the expression for $[COCl]$:$[COCl] = \frac{k_3}{k_4} 	imes \left( \frac{k_1}{k_2} ight)^{1/2} [Cl_2]^{1/2} [CO]$Substituting $[COCl]$ into the rate expression:Rate $= k_5 	imes \left( \frac{k_3}{k_4} 	imes \left( \frac{k_1}{k_2} ight)^{1/2} [Cl_2]^{1/2} [CO] ight) 	imes [Cl_2]$Rate $= k_5 	imes \frac{k_3}{k_4} 	imes \left( \frac{k_1}{k_2} ight)^{1/2} [CO][Cl_2]^{3/2}$

$Observation$	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	$Rate \ (mol \ L^{-1} \ sec^{-1})$
$1$	$0.1$	$0.1$	$2 \times 10^{-3}$
$2$	$0.2$	$0.1$	$4 \times 10^{-3}$
$3$	$0.1$	$0.2$	$8 \times 10^{-3}$

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