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.

The rate equation is the mathematical expression which explains the the 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 is obtained by summing up the three steps, cancelling the intermediates and combining the formulas shown as below:

The overall reaction: \({\bf{C}}{{\bf{l}}_{\bf{2}}}\left( g \right) + {\bf{CO}}\left( g \right) \to {\bf{COC}}{{\bf{l}}_{\bf{2}}}\left( g \right)\)

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.

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:

\(\begin{align}{l}rat{e_{forward}} = rat{e_{backward}}\\{k_1}({\bf{C}}{{\bf{l}}_{\bf{2}}}) = {k_{ - 1}}{({\bf{Cl}})^2}\end{align}\)

For second elementary reaction the rate law is:

\(rate = {k_2}(CO)(Cl)\)

For third elementary reaction the rate law is:

\(rate = {k_3}(COCl)(Cl)\)

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:

\(\begin{align}rate &= {k_2}(CO)(Cl)\\rate &= {k_2}(CO){\left( {\frac{{{k_1}}}{{{k_{ - 1}}}}} \right)^{\frac{1}{2}}}{\left( {{\bf{C}}{{\bf{l}}_{\bf{2}}}} \right)^{\frac{1}{2}}}\\rate &= {k_2}{\left( {\frac{{{k_1}}}{{{k_{ - 1}}}}} \right)^{\frac{1}{2}}}(CO){\left( {{\bf{C}}{{\bf{l}}_{\bf{2}}}} \right)^{\frac{1}{2}}}\end{align}\)

Thus, overall reaction, reaction intermediates, rate law for each elementary step and overall rate law expression are determined.