Q109CP - Chemistry: Molecular Nature Of Matter And Change

Question

Question:Sulfonation of benzene has the following mechanism: 

$$\begin{gathered} \mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\to }{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{+}{{\mathbf{HSO}}_{\mathbf{4}}}^{\mathbf{-}}\mathbf{+}{\mathbf{SO}}_{\mathbf{3}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\left[}\mathbf{fast}\mathbf{\right]} \\ \mathbf{\left(}\mathbf{2}\mathbf{\right)}{\mathbf{SO}}_{\mathbf{3}}\mathbf{+}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{6}}\mathbf{\to }\mathbf{H}\mathbf{\left(}{\mathbf{C}}_{\mathbf{6}}{{\mathbf{H}}_{\mathbf{5}}}^{\mathbf{+}}\mathbf{\right)}{{\mathbf{SO}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\left[}\mathbf{slow}\mathbf{\right]} \\ \mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{H}\mathbf{(}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}\mathbf{+}\mathbf{)}{{\mathbf{SO}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{+}{{\mathbf{HSO}}_{\mathbf{4}}}^{\mathbf{-}}\mathbf{\to }{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{{\mathbf{SO}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\left[}\mathbf{fast}\mathbf{\right]} \\ \mathbf{\left(}\mathbf{4}\mathbf{\right)}{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{{\mathbf{SO}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{+}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\to }{\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{SO}}_{\mathbf{3}}\mathbf{H}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{\left[}\mathbf{fast}\mathbf{\right]} \end{gathered}$$

(a) Write an overall equation for the reaction. 

(b) Write the overall rate law for the initial rate of the reaction

Step-by-Step Solution

Answer

       (a) The overall reaction is 

$$\begin{gathered} 2{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{H}}_3{\mathrm{O}}^{+}+{{\mathrm{HSO}}_4}^{-}+{\mathrm{SO}}_3 \\ {\mathrm{SO}}_3+{\mathrm{C}}_6{\mathrm{H}}_6\to \mathrm{H}\left({\mathrm{C}}_6{{\mathrm{H}}_5}^{+}\right){{\mathrm{SO}}_3}^{-} \\ \mathrm{H}({\mathrm{C}}_6{\mathrm{H}}_5+){{\mathrm{SO}}_3}^{-}+{{\mathrm{HSO}}_4}^{-}\to {\mathrm{C}}_6{\mathrm{H}}_5{{\mathrm{SO}}_3}^{-}+{\mathrm{H}}_2{\mathrm{SO}}_4 \\ {\mathrm{C}}_6{\mathrm{H}}_5{{\mathrm{SO}}_3}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}\to {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{SO}}_3\mathrm{H}+{\mathrm{H}}_2\mathrm{O} \\ \overline{{\mathrm{C}}_6{\mathrm{H}}_6+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{SO}}_3\mathrm{H}+{\mathrm{H}}_2\mathrm{O} } \end{gathered}$$

      (b) The overall rate law is ${\mathrm{rate}}_2=\mathrm{k}{\left[{\mathrm{H}}_2{\mathrm{SO}}_4\right]}^2\left[{\mathrm{C}}_6{\mathrm{H}}_6\right]$

Step 1:(a) Write an overall equation for the reaction

The overall reactioncan be obtained by adding all

$$\begin{gathered} 2{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{H}}_3{\mathrm{O}}^{+}+{{\mathrm{HSO}}_4}^{-}+{\mathrm{SO}}_3 \\ {\mathrm{SO}}_3+{\mathrm{C}}_6{\mathrm{H}}_6\to \mathrm{H}\left({\mathrm{C}}_6{{\mathrm{H}}_5}^{+}\right){{\mathrm{SO}}_3}^{-} \\ \mathrm{H}({\mathrm{C}}_6{\mathrm{H}}_5+){{\mathrm{SO}}_3}^{-}+{{\mathrm{HSO}}_4}^{-}\to {\mathrm{C}}_6{\mathrm{H}}_5{{\mathrm{SO}}_3}^{-}+{\mathrm{H}}_2{\mathrm{SO}}_4 \\ {\mathrm{C}}_6{\mathrm{H}}_5{{\mathrm{SO}}_3}^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}\to {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{SO}}_3\mathrm{H}+{\mathrm{H}}_2\mathrm{O} \\ \overline{{\mathrm{C}}_6{\mathrm{H}}_6+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{SO}}_3\mathrm{H}+{\mathrm{H}}_2\mathrm{O} } \end{gathered}$$

Step2: Rate law

Rate is expressed in terms of the product of the concentration of the reactants raised to the power of their respective stoichiometric coefficients. The actual rate law is written on the basis of the slow step:

$$\begin{gathered} {\mathrm{SO}}_3+{\mathrm{C}}_6{\mathrm{H}}_6 \stackrel{ {\mathrm{k}}_{2 }}{\to } \mathrm{H}\left({\mathrm{C}}_6{{\mathrm{H}}_5}^{+}\right){{\mathrm{SO}}_3}^{-} \\ {\mathrm{rate}}_2 ={\mathrm{k}}_2\left[{\mathrm{SO}}_3\right]\left[{\mathrm{C}}_6{\mathrm{H}}_6\right] \end{gathered}$$

Where is the rate constant for the step. Since, ${\mathrm{SO}}_3$ is an intermediate, we have to substitute its concentration. Write the rate of reaction in terms of formation of ${\mathrm{SO}}_3$ from the first reaction as follows:

$$\begin{gathered} 2{\mathrm{H}}_2{\mathrm{SO}}_4 \to {\mathrm{H}}_3{\mathrm{O}}^{+}+{{\mathrm{HSO}}_4}^{-}+{\mathrm{SO}}_3 \\ \mathrm{rate}=\frac{∆\left[{\mathrm{SO}}_3\right]}{∆\mathrm{t}} \\ {\mathrm{k}}_1{\left[{\mathrm{H}}_2{\mathrm{SO}}_4\right]}^2=\left[{\mathrm{SO}}_3\right] \end{gathered}$$

Where ${\mathrm{k}}_1$ is the rate constant for the step.

Step3:(b) Write the overall rate law for the initial rate of the reaction

Substitute the value in rate law expression to obtain the overall rate law:

$$\begin{gathered} {\mathrm{rate}}_2={\mathrm{k}}_2{\mathrm{k}}_1{\left[{\mathrm{H}}_2{\mathrm{SO}}_{ 4 }\right]}^2\left[{\mathrm{C}}_6{\mathrm{H}}_6 \right] \\ =\mathrm{k} {\left[{\mathrm{H}}_2{\mathrm{SO}}_{ 4 }\right]}^2\left[{\mathrm{C}}_6{\mathrm{H}}_6 \right] \end{gathered}$$

Where, $\mathrm{k}={\mathrm{k}}_2{\mathrm{k}}_1$

Thus, the overall rate law is ${\mathrm{rate}}_2=\mathrm{k} {\left[{\mathrm{H}}_2{\mathrm{SO}}_{ 4 }\right]}^2\left[{\mathrm{C}}_6{\mathrm{H}}_6 \right]$