Question

The reaction: ${{CO + C}}{{{l}}_{{2}}}\xrightarrow{{}}{{COC}}{{{l}}_{{2}}}$has the following mechanism
What is the rate expression?

$\left( {{{iii}}} \right){{COCl + C}}{{{l}}_{{2}}}\xrightarrow{{{{{K}}_{{3}}}}}{{COC}}{{{l}}_{{2}}}{{ + Cl}}\left( {{{slow}}} \right)$
A. $\dfrac{{{{d}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]}}{{{{dt}}}}{{ = K}}\left[ {{{CO}}} \right]{\left[ {{{C}}{{{l}}_{{2}}}} \right]^{{{1/2}}}}$
B. $\dfrac{{{{d}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]}}{{{{dt}}}}{{ = K}}\left[ {{{CO}}} \right]{\left[ {{{C}}{{{l}}_{{2}}}} \right]^{{{3/2}}}}$
C. $\dfrac{{{{d}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]}}{{{{dt}}}}{{ = K}}\left[ {{{CO}}} \right]{\left[ {{{C}}{{{l}}_{{2}}}} \right]^{{{5/2}}}}$
D. None of these

Answer 1

Rate of the reaction is defined as the speed at which a chemical reaction proceeds i.e. time taken by a reaction to reach the equilibrium. The sum of powers to which the concentration terms of reactants are raised in a reaction is called Order of a reaction. Rate of the reaction which has the molar concentration of reactants as unity is called rate constant. The reaction is said to be in equilibrium when rate of forward reaction is equal to rate of backward reaction.

Complete step by step answer:
Equilibrium constant for ( i ) and ( ii ) step are given by
${{{K}}_{{1}}}{{ = }}{\dfrac{{\left[ {{{Cl}}} \right]}}{{\left[ {{{C}}{{{l}}_{{2}}}} \right]}}^{{2}}} \Rightarrow \left[ {{{Cl}}} \right]{{ = }}{{{K}}_{{1}}}^{{{1/2}}}{\left[ {{{C}}{{{l}}_{{2}}}} \right]^{{{1/2}}}} \to {{1}}$
${{{K}}_{{2}}}{{ = }}\dfrac{{\left[ {{{COCl}}} \right]}}{{\left[ {{{CO}}} \right]\left[ {{{Cl}}} \right]}} \Rightarrow \left[ {{{COCl}}} \right]{{ = }}{{{K}}_{{2}}}\left[ {{{CO}}} \right]\left[ {{{Cl}}} \right] \to {{2}}$
Substituting $\left[ {{{Cl}}} \right]$ from 1 in 2 we get,
$\left[ {{{COCl}}} \right]{{ = }}{{{K}}_{{1}}}^{{{1/2}}}{{{K}}_{{2}}}\left[ {{{CO}}} \right]{\left[ {{{C}}{{{l}}_{{2}}}} \right]^{{{1/2}}}} \to {{3}}$
In the above equations the reaction is fast in case of (i) and (ii) and (iii) is the slow and rate determining step
Thus the rate of formation of Phosgene [$COCl_2$] is controlled by the reaction (iii)
Rate of the reaction can be explained by concentration change with time
Thus the rate constant can be written as $K_3$
$\dfrac{{{{d}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]}}{{{{dt}}}}{{ = }}{{{k}}_{{3}}}\left[ {{{COCl}}} \right]\left[ {{{C}}{{{l}}_{{2}}}} \right] \to {{4}}$
On substituting equation 3 in 4
$\dfrac{{{{d}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]}}{{{{dt}}}}{{ = }}{{{k}}_{{3}}}{{{K}}_{{1}}}^{{{1/2}}}{{{K}}_{{2}}}\left[ {{{CO}}} \right]{\left[ {{{C}}{{{l}}_{{2}}}} \right]^{{{1/2}}}}\left[ {{{C}}{{{l}}_{{2}}}} \right]$
On adding the powers of ${[Cl_2]}^{1/2}$ and [$Cl_2$] we get ${[Cl_2]}^{3/2}$
We get
$\dfrac{{{{d}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]}}{{{{dt}}}}{{ = }}{{{k}}_{{3}}}{{K}}_{{1}}^{{{1/2}}}{{{K}}_{{2}}}\left[ {{{CO}}} \right]{\left[ {{{C}}{{{l}}_{{2}}}} \right]^{{{3/2}}}}$
In general $k_3,{K_1}^{½},K_2$ can be written as K
Thus we have the final rate equation as
$\dfrac{{{{d}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]}}{{{{dt}}}}{{ = K}}\left[ {{{CO}}} \right]{\left[ {{{C}}{{{l}}_{{2}}}} \right]^{{{3/2}}}}$

So, the correct answer is Option B.

Additional Information:
The process of formation of Phosgene is a slow step and it is the rate determining step. Similarly the decomposition of Phosgene is given by
$\dfrac{{{{ - d}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]}}{{{{dt}}}}{{ = }}{{{k}}_{{{ - 3}}}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]\left[ {{{Cl}}} \right]$
Substituting the $\left[ {{{Cl}}} \right]$ from 1 in the above equation
$\dfrac{{{{ - d}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]}}{{{{dt}}}}{{ = }}{{{k}}_{{{ - 3}}}}{{{K}}_{{1}}}^{{{1/2}}}\left[ {{{COC}}{{{l}}_{{2}}}} \right]{\left[ {{{C}}{{{l}}_{{2}}}} \right]^{{{1/2}}}}$
The negative sign indicates the decrease in the number of reactants while formation of the products

Note: The rate of a reaction can also be defined as the change in concentration of reactant or product in unit time. It can be expressed in terms of Units of rate are concentration${{tim}}{{{e}}^{{{ - 1}}}}$ if concentration is in ${{mol}}{{{L}}^{{{ - 1}}}}$ and time is in seconds than the units will be${{mol}}{{{L}}^{{{ - 1}}}}{{{s}}^{{{ - 1}}}}$.