Question

The reaction, $A \rightarrow B$ follows first order kinetics. The time taken for $0.8$ mole of A to produce $0.6$ mole of $B$ is $1\,h$. What is the time taken for the conversion of $0.9$, mole of $A$ to $0.675$ mole of $B$?

• A. 0.25 h
• B. 2 h
• C. 1 h
• D. 0.5 h

Answer 1

Answer: (C)

Rate constant of first order reaction
$k=\frac{2.303}{t} \, \log_{10} \frac{(A)_0}{(A)_t}$
or $k=\frac{2.303}{1} \times \log_{10} \, \frac{0.8}{0.2} ...(i)$
(because $0.6$ mole of $B$ is formed)
Suppose $t_1$ hour are required for changing the concentration of $A$ from $0.9$ mole to $0.675$ mole of $B$.
Remaining mole of $A = 0.9 - 0.675 = 0.225$
$\therefore k=\frac{2.303}{t_1} \, \log_{10} \, \frac{0.9}{0.225} ...(ii)$
From Eqs. (i) and (ii)
$\frac{2.303}{1} \, \log_{10} \, \frac{0.8}{0.2}=\frac{2.303}{t_1} \log_{10} \, \frac{0.9}{0.225}$
$ 2.303 \, \log_{10} \, 4=\frac{2.303}{t_1} \, \log_{10} \, 4$
$t_1=1 \, h$