Question

The reaction of \({\bf{CO}}\) with \({\bf{C}}{{\bf{l}}_{\bf{2}}}\) gives phosgene \(\left( {{\bf{COC}}{{\bf{l}}_{\bf{2}}}} \right)\), a nerve gas that was used in World War I. Use the mechanism shown here to complete the following exercises:(fast, \({{\bf{k}}_{\bf{1}}}\) represents the forward rate constant, \({k_{ - {\bf{1}}}}\)the reverse rate constant)\({\bf{CO}}\left( g \right){\rm{ }} + {\rm{ }}{\bf{Cl}}\left( g \right) \to {\bf{COCl}}\left( g \right)\)(slow, \({k_{\bf{2}}}\) the rate constant)\({\bf{COCl}}\left( g \right){\rm{ }} + {\rm{ }}{\bf{Cl}}\left( g \right) \to {\bf{COC}}{{\bf{l}}_{\bf{2}}}\left( g \right)\)(fast,\({k_{\bf{3}}}\)the rate constant)(a) Write the overall reaction.(b) Identify all intermediates.(c) Write the rate law for each elementary reaction.(d) Write the overall rate law expression.

Short answer

(a) The overall reaction will be \({\bf{C}}{{\bf{l}}_{\bf{2}}}\left( g \right) + {\bf{CO}}\left( g \right) \to {\bf{COC}}{{\bf{l}}_{\bf{2}}}\left( g \right)\)(b) The intermediates are \((COCl)\) and \((Cl)\).(c) The rate law expression for each elementary reaction. \(\begin{aligned}{l}{k_1}({\bf{C}}{{\bf{l}}_{\bf{2}}}) = {k_{ - 1}}{({\bf{Cl}})^2}\\rate = {k_2}(CO)(Cl)\\rate = {k_3}(COCl)(Cl)\end{aligned}\) (d) The overall rate law expression: \(rate = {k_2}{\left( {\frac{{{k_1}}}{{{k_{ - 1}}}}} \right)^{\frac{1}{2}}}(CO){\left( {{\bf{C}}{{\bf{l}}_{\bf{2}}}} \right)^{\frac{1}{2}}}\)

Step-by-step solution

Definition of Rate Equation or Rate Law, Intermediates

The rate equation is the mathematical expression which explains thethe relationship between the rate of a chemical reaction and the concentration of its reactants.\({\rm{rate = }}k{{\rm{(A)}}^x}{{\rm{(B)}}^y}{{\rm{(C)}}^z}.....\)Where,(A), (B), and (C) denotes the molar concentrations of reactants.kis the rate constant.Exponents m, n, and pare generally positive integers.

The species which are produced in one of the step or elementary reaction are used in the successive step or elementary reaction.

The Overall Reaction 

The overall reaction is obtained by summing up the three steps, cancelling the intermediates and combining the formulas shown as below: The overall reaction: \({\bf{C}}{{\bf{l}}_{\bf{2}}}\left( g \right) + {\bf{CO}}\left( g \right) \to {\bf{COC}}{{\bf{l}}_{\bf{2}}}\left( g \right)\)

Identification of All Intermediates  

By the derivation of overall reaction with the help of individual elementary reactions, the intermediates are found to be\((COCl)\)and\((Cl)\).These species are produced in one of the step or elementary reaction are used in the successive step or elementary reaction.

Rate Law for each Elementary Reaction 

Let us write the rate law expression for every elementary reaction as these elementary reactions are part of the mechanism. For first elementary reaction the rate law is:\(\begin{aligned}{l}rat{e_{forward}} = rat{e_{backward}}\\{k_1}({\bf{C}}{{\bf{l}}_{\bf{2}}}) = {k_{ - 1}}{({\bf{Cl}})^2}\end{aligned}\) For second elementary reaction the rate law is:\(rate = {k_2}(CO)(Cl)\) For third elementary reaction the rate law is:\(rate = {k_3}(COCl)(Cl)\)

Overall Rate Law Expression  

The step 2 is the slow step, which is the nothing but the rate determining step. Therefore, the overall rate law can be written as\(rate = {k_2}(CO)(Cl)\). As the intermediates are\((COCl)\)and\((Cl)\). Algebraic expression is used to represent\((Cl)\).Using elementary reaction 1,\((Cl) = {\left( {\frac{{{k_1}}}{{{k_{ - 1}}}}} \right)^{\frac{1}{2}}}{\left( {{\bf{C}}{{\bf{l}}_{\bf{2}}}} \right)^{\frac{1}{2}}}\)Now, let us substitute the algebraic expression in the overall rate law, substituting gives us the overall rate law expression:\(\begin{aligned}{l}rate = {k_2}(CO)(Cl)\\rate = {k_2}(CO){\left( {\frac{{{k_1}}}{{{k_{ - 1}}}}} \right)^{\frac{1}{2}}}{\left( {{\bf{C}}{{\bf{l}}_{\bf{2}}}} \right)^{\frac{1}{2}}}\\rate = {k_2}{\left( {\frac{{{k_1}}}{{{k_{ - 1}}}}} \right)^{\frac{1}{2}}}(CO){\left( {{\bf{C}}{{\bf{l}}_{\bf{2}}}} \right)^{\frac{1}{2}}}\end{aligned}\)Thus, overall reaction, reaction intermediates, rate law for each elementary step and overall rate law expression are determined.