Question

The rate constant for the forward and backward reactions of hydrolysis of ester are $1.1\times {10}^{-2}$ and $1.5\times {10}^{-3}$ respectively. The equilibrium constant of the following reaction is:

$C{H}_{3}COO{C}_2{H}_5+H^{+}\rightleftharpoons C{H}_{3}COOH+C_2{H}_{5}OH$

• A. $6.33$
• B. $7.75$
• C. $7.33$
• D. $8.33$

Answer 1

The correct option is C $7.33$

Rate of forward reaction $=1.1\times {10}^{-2}\left[C{H}_{3}COO{C}_2{H}_5\right]\left[H^{+}\right]$
Rate of backward reaction $=1.5\times {10}^{-3}\left[C{H}_{3}COOH\right]\left[C_2{H}_{5}OH\right]$
Now at equilibrium,

Rate of forward reaction $=$ Rate of backward reaction
$1.1\times {10}^{-2}\left[C{H}_{3}COO{C}_2{H}_5\right]\left[H^{+}\right]$ $=1.5\times {10}^{-3}\left[C{H}_{3}COOH\right]\left[C_2{H}_{5}OH\right]$
$k_c=\frac{\left[C{H}_{3}COOH\right]\left[C_2{H}_{5}OH\right]}{\left[C{H}_{3}COO{C}_2{H}_5\right]\left[H^{+}\right]}$
$=\frac{1.1\times {10}^{-2}}{1.5\times {10}^{-3}}$ $=\frac{1.1\times 10}{1.5}=7.33$