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

Entry	$[A]$ in $M$	$[B]$ in $M$	Initial rate $(M/s)$
$1$	$0.02$	$0.02$	$2 \times 10^{-2}$
$2$	$0.02$	$0.04$	$4 \times 10^{-2}$
$3$	$0.04$	$0.04$	$8 \times 10^{-2}$

$[X_{2}] \ (M)$	$[Y_{2}] \ (M)$	Rate of appearance of $XY \ (M \ sec^{-1})$
$0.1$	$0.1$	$5 \times 10^{-6}$
$0.2$	$0.1$	$10^{-5}$
$0.2$	$0.2$	$4 \times 10^{-5}$

Similar Questions

For a reaction $A + B \longrightarrow P$,the following data are provided. The rate constant for this reaction in standard units is:
Entry $[A]$ in $M$ $[B]$ in $M$ Initial rate $(M/s)$
$1$ $0.02$ $0.02$ $2 \times 10^{-2}$
$2$ $0.02$ $0.04$ $4 \times 10^{-2}$
$3$ $0.04$ $0.04$ $8 \times 10^{-2}$

The rate equation for the reaction $2 A + B \longrightarrow$ products is $\text{rate} = k[A][B]^2$. If $k$ at $T \ K$ is $5.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$,the initial rate of the reaction,when $[A] = 0.05 \ mol \ L^{-1}$ and $[B] = 0.1 \ mol \ L^{-1}$ is:

The rate of a reaction is given by $r = K[x][y] / [OH^-]$. If the concentration of $[OH^-]$ is increased,the order of the reaction will be ........

The following are the rate constants of two different reactions. What is the overall order of each reaction?
$(a)$ $2.1 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$
$(b)$ $4.5 \times 10^{-3} \ min^{-1}$

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