Question

The hydrolysis of ethyl acetate in an acidic medium is a:
                   (A) Zero order reaction
                   (B) First order reaction
                   (C) Pseudo first order reaction
                   (D) Second order reaction

Answer 1

Hint: In the hydrolysis of ethyl acetate the concentration of water does not change much with proceeding reaction and hence does not affect the rate of the reaction. So, it can be taken constant.

Complete answer:
-Order of the reaction is defined as defined as the sum of the powers of the concentration of the reactants in the rate law expression.
For a rate law: $Rate = k{[A]^x}{[B]^y}$, the order = x + y.
The order of any reaction can be 0, 1, 2, 3 or even a fraction number.
-The reaction for hydrolysis of ethyl acetate is written as:

 $C{H_3}COO{C_2}{H_5} + {H_2}O\xrightarrow{{{H^ + }}}C{H_3}COOH + {C_2}{H_5}OH$

If we hydrolyse 0.01 mole of ethyl acetate ($C{H_3}COO{C_2}{H_5}$) with 10 moles of water, the concentrations of various constituents at t=0 and time t can be written as:

 $C{H_3}COO{C_2}{H_5} + {H_2}O\xrightarrow{{{H^ + }}}C{H_3}COOH + {C_2}{H_5}OH$

Here we can see that the concentration of water does not get altered much during the progress of the reaction.
The rate equation for this reaction can be written as:

 $Rate = k'[C{H_3}COO{C_2}{H_5}][{H_2}O]$

Since the concentration of water is very high and does not get altered much during the progress of the reaction, it can be said that the reaction is independent of the change in the concentration of water. Thus the term [${H_2}O$] can be taken constant.
The new rate equation will be:
                                              $Rate = k[C{H_3}COO{C_2}{H_5}]$
Where, $k = k'[{H_2}O]$
-Now we can see from the rate law that this reaction behaves as a first order reaction. So, such reactions are known as pseudo-first order reactions.
-Another example of pseudo first order reaction is inversion of cane sugar.
  ${C_{12}}{H_{22}}{O_{11}} + {H_2}O\xrightarrow{{{H^ + }}}{C_6}{H_{12}}{O_6} + {C_6}{H_{12}}{O_6}$
Rate law equation for this will be:
                                      $Rate = k[{C_{12}}{H_{22}}{O_{11}}]$
So, the correct answer is “Option C”.

Note: Don’t confuse order of a reaction with molecularity. Order of a reaction is the sum of powers of concentration while molecularity is the number of reacting species taking part to bring about the chemical reaction. Order can be 0 or even a fraction but molecularity can never be a zero or a non integer number.
For e.g.: $RCl + {H_2}O \to ROH + HCl$, for this reaction order will be 1 and molecularity will be 2.