Question

Sulfonation of benzene has the following mechanism: (1) \(2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HSO}_{4}^{-}+\mathrm{SO}_{3} \quad[\) fast \(]\) (2) \(\mathrm{SO}_{3}+\mathrm{C}_{6} \mathrm{H}_{6} \longrightarrow \mathrm{H}\left(\mathrm{C}_{6} \mathrm{H}_{5}^{+}\right) \mathrm{SO}_{3}^{-} \quad\) [slow] (3) \(\mathrm{H}\left(\mathrm{C}_{6} \mathrm{H}_{5}^{+}\right) \mathrm{SO}_{3}^{-}+\mathrm{HSO}_{4}^{-} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3}^{-}+\mathrm{H}_{2} \mathrm{SO}_{4} \quad\) [fast] (4) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3} \mathrm{H}+\mathrm{H}_{2} \mathrm{O} \quad\) [fast \(]\) (a) Write an overall equation for the reaction. (b) Write the overall rate law in terms of the initial rate of the reaction.

Short answer

Overall reaction: \( \mathrm{C}_{6}\mathrm{H}_{6} + 2 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3}\mathrm{H} + \mathrm{H}_{2}\mathrm{O} \). Rate law: \ \text{Rate} = k[\mathrm{SO}_{3}][\mathrm{C}_{6} \mathrm{H}_{6}] \.

Step-by-step solution

Understanding the Mechanism

The mechanism consists of four steps, two of which are fast and one slow. Identify each step and its role in the reaction.

Identify the Key Species and Steps

The species involved are \(\mathrm{H}_{2} \mathrm{SO}_{4}\), \(\mathrm{H}_{3}\mathrm{O}^{+}\), \(\mathrm{HSO}_{4}^{-}\), \(\mathrm{SO}_{3} \), and \(\mathrm{C}_{6} \mathrm{H}_{6} \), among others. Note that step (2) is the slow step, which will determine the rate law.

Write the Overall Chemical Equation

Combine all steps to write the complete reaction, canceling out the intermediates that appear on both sides: \[ \mathrm{C}_{6}\mathrm{H}_{6} + 2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3}\mathrm{H} + \mathrm{H}_{2}\mathrm{O} \]

Determine the Rate-Determining Step

The overall rate will be based on the slow step, which is: \( \mathrm{SO}_{3} +\mathrm{C}_{6} \mathrm{H}_{6} \longrightarrow \mathrm{H}\mathrm{(C}_{6}\mathrm{H}_{5}^{+})\mathrm{SO}_{3}^{-} \).

Write the Overall Rate Law

The rate law depends on the concentrations of \( \mathrm{SO}_{3} \) and \( \mathrm{C}_{6} \mathrm{H}_{6} \). Thus, the rate law is: \ \text{Rate} = k[\mathrm{SO}_{3}][\mathrm{C}_{6} \mathrm{H}_{6}] \.

Key concepts

reaction mechanism

The sulfonation of benzene follows a specific sequence of steps, known as the reaction mechanism. Each step involves different reactants and intermediates. By understanding the sequence, we can see how reactants turn into products.
In this mechanism, sulfuric acid (\(H_2SO_4\)) plays a vital role. It first reacts to produce hydronium ions (\(H_3O^+\)), bisulfate ions (\(HSO_4^-\)), and sulfur trioxide (\(SO_3\)). This is a fast step.
Next, \muy benzene (\(C_6H_6\)) reacts with \muy \(SO_3\)), forming an intermediate called \muy aryl sulfonium ion (\(H(C_6H_5^+)SO_3^-\)), in a slower, rate-determining step.
The intermediate then rapidly reacts with another bisulfate ion to form benzene sulfonate (\(C_6H_5SO_3^-\)) and regenerate sulfuric acid.
Finally, benzene sulfonate reacts quickly with a hydronium ion to form benzene sulfonic acid (\(C_6H_5SO_3H\)) and water. Each step, when combined, provides the detailed picture of how the reaction proceeds from start to finish.

rate-determining step

In any multi-step reaction, the rate-determining step (RDS) is crucial. It is the slowest step in the sequence and determines the overall rate of the reaction.
For the sulfonation of benzene, the rate-determining step is the second one: the reaction between sulfur trioxide (\(SO_3\)) and benzene (\(C_6H_6\)). This step forms the aryl sulfonium ion (\(H(C_6H_5^+)SO_3^-\)).
Since this step is the slowest, it acts like a bottleneck. No matter how fast the other steps are, the overall reaction cannot go faster than this crucial slow step.
Understanding which step is the rate-determining step allows chemists to predict and influence the reaction rate by changing conditions like concentration or temperature.

overall rate law

The overall rate law of a reaction describes how the rate depends on the concentration of reactants. It is derived from the rate-determining step.
For the sulfonation of benzene, the rate law is based on the slow step, which involves \(SO_3\) and benzene (\(C_6H_6\)).
Therefore, the rate of reaction can be given by:
\text{Rate} = k[\text{SO}_3][\text{C}_6\text{H}_6]\br>This tells us that the rate of the reaction is directly proportional to the concentrations of sulfur trioxide and benzene.
In practical terms, increasing the concentration of either \(SO_3\) or benzene will increase the reaction rate. Conversely, decreasing those concentrations will slow down the reaction.

chemical equation

The overall chemical equation provides a summary of the reactants and products in a reaction. For sulfonation of benzene, it accounts for all species involved and cancels out intermediates.
Combining all steps from the mechanism, the overall balanced chemical equation is:
\[ \text{C}_6\text{H}_6 + 2 \text{H}_2\text{SO}_4 \rightarrow \text{C}_6\text{H}_5\text{SO}_3\text{H} + \text{H}_2\text{O} \]
This equation shows that one molecule of benzene reacts with two molecules of sulfuric acid to produce one molecule of benzene sulfonic acid and water.
By looking at the overall chemical equation, one can understand the stoichiometry and the reactants to products relationship in the reaction, which is essential for predicting yields and scaling up the reaction.