Q7 / 120 · VITEEE 2014 · Chemistry

The rate constant for forward reaction and backward reaction of hydrolysis of ester are $1.1 \times 10^{-2}$ and $1.5 \times 10^{-3}$ per minute respectively. Equilibrium constant for the reaction is $\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}$ $+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$

  • A
    $33.7$
  • Correct answer
    B
    $7.33$
  • C
    $5.33$
  • D
    $33.3$

Solution

Given $\mathrm{k}_{\mathrm{f}}=1.1 \times 10^{-2}, \mathrm{k}_{\mathrm{b}}=1.5 \times 10^{-3}$ $\mathrm{k}_{\mathrm{c}}=\frac{\mathrm{k}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{b}}}=\frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}}=7.33$

Hence, the correct answer is option: 'B'.

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