Question

The rate constant for forward and backward reactions of hydrolysis of an ester are $1.1\times {{10}^{-2}}$ and $1.5\times {{10}^{-3}}$ per minute, respectively. The reaction's equilibrium constant is
$CH_3COOC_2H_5 \ + \ H_2O\rightleftharpoons CH_3COOH \ + C_2H_5OH$
(A) 4.33
(B) 5.33
(C) 6.33
(D) 7.33

Answer 1

Hint: The ratio of the forward and reverse rate constants, which are both constant values at a given temperature, is equal to the equilibrium constant (K). Both the forward reaction rate and the backward reaction rate are equal at equilibrium.

Formula Used: If for any equilibrium reaction, the rate of forward reaction is ${{k}_{f}}$ and the rate of backward reaction is ${{k}_{b}}$, then the equilibrium constant is $K=\frac{{{k}_{f}}}{{{k}_{b}}}$.

Complete Step by Step Answer:
The given reaction is:
$CH_3COOC_2H_5 \ + \ H_2O\rightleftharpoons CH_3COOH \ + C_2H_5OH$
The rate constant for forward reaction, ${{k}_{f}}=1.1\times {{10}^{-2}}$
And the rate constant for backward reaction,${{k}_{b}}=1.5\times {{10}^{-3}}$
The equilibrium constant for this reaction is $K=\frac{{{k}_{f}}}{{{k}_{b}}}$
$K=\frac{1.1\times {{10}^{-2}}}{1.5\times {{10}^{-3}}}$
$K=\frac{0.011}{0.0015}$
$K=7.33$
Thus, the value of the equilibrium constant (K) is 7.33.
Correct Option: (D) 7.33.

Additional Information: Chemical equilibrium is a state in which the rates of both the forward and backward reactions are equal and the concentration of both the reactants and products is constant. At equilibrium, there is no net change in the number of moles, although conversion from reactants to products or products to reactants is still occurring.

Note: If the K value is large ,i.e.,$K>>1$ , the equilibrium lies to the right and the reaction mixture contains mostly products, If the K value is very small ,i.e., $K<<1$ , the equilibrium lies to the left and the reaction mixture contains mostly reactants, If the K value is close to 1, the mixture contains appreciable amounts of both reactants and products.