Question

\[C{H_3}COO{C_2}{H_5}(aq) + {H_2}O(l) \overset{H^{+}(aq)}{\rightarrow} C{H_3}COOH(aq) + {C_2}{H_5}OH(aq)\] What type of reaction is this?
A. Unimolecular elementary
B. Pseudo first order
C. Zero order
D. Second order

Answer 1

Hint: For the reaction in question here, it may appear as if the rate will depend on the concentrations of both Ethyl acetate (\[C{H_3}COO{C_2}{H_5}\]) and water thereby making it a second-order reaction. However, since water is present in excess in this reaction, its concentration will not affect the rate of the reaction. Thus, it is a first-order reaction that only appears to be second-order.

Complete Step by Step Solution:
According to the law of mass action, the rate of a reaction is directly proportional to the product of the concentrations of reactants. This law is applicable only for elementary reactions (reactions that occur in a single step). For an elementary reaction \[A + B \to \Pr oducts\], its rate (denoted as r), as dictated by the law of mass action would be \[r = k[A][B]\]where k is the constant of proportionality which is called the rate constant.

For a general reaction of the type:
\[aA + bB + cC \to \Pr oducts\]
It has been empirically found that the rate of the reaction is given by:
\[r = k{[A]^a}{[B]^b}{[C]^c}\] … (1)
Where k = rate constant
             [A], [B], [C] = concentrations of reactants A, B and C respectively
              a, b, c = stoichiometric coefficients of reactants A, B and C respectively
Equation (1) is called the rate-law expression for the reaction. The sum of the exponents in the rate law expression is called the order of the reaction.

Proceeding similarly, the rate-law expression for the reaction given in the question can be found as follows:
\[C{H_3}COO{C_2}{H_5}(aq) + {H_2}O(l) \overset{H^{+}(aq)}{\rightarrow} C{H_3}COOH(aq) +
{C_2}{H_5}OH(aq)\] Therefore, the rate-law expression will be \[r = k[C{H_3}COO{C_2}{H_5}][{H_2}O]\] … (2)
From this expression, it appears that the order of the reaction is 2 since both concentration terms have a power of 1.

However, it has been experimentally observed that this reaction does not behave like a second-order reaction. Rather, it behaves like a first-order reaction. Reactions that appear to be of second-order from their rate-law expression but behave like a first-order reaction are called pseudo-first-order reactions. The reaction given in the question (acid-catalyzed ester hydrolysis) is one of the most common examples of pseudo-first-order reactions.
Reactions become pseudo-first-order if the concentration of a reactant remains constant. This can occur if the reactant is in large excess compared to the other reactants. In this reaction, water is present in large excess compared to the ester. Thus \[[{H_2}O]\]remains constant and equation (2) can be rewritten as:
 \[r = k[C{H_3}COO{C_2}{H_5}][{H_2}O]\]
\[ \Rightarrow r = k'[C{H_3}COO{C_2}{H_5}]\] … (3)
where \[k' = k \times [{H_2}O]\]. k’ is called the pseudo-first-order rate constant.

Thus, option B is correct

Note: The key to solving this question was the fact that water is present in large excess in acid-catalysed ester hydrolysis. Had we not known that fact, we would have been convinced that the reaction was second order. Therefore, the student must know the actual conditions of the reaction given. There is no way to theoretically predict that the given reaction is a pseudo-first-order and not a second-order reaction.