Question

The reaction, $A\to B$ follows first order kinetics. The time taken for $0.8$ mole of A to produce $0.6$ mole of $B$ is $1h$. 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)
$\frac{2.303}{1}{\log }_{10}\frac{0.8}{0.2}=\frac{2.303}{t_1}{\log }_{10}\frac{0.9}{0.225}$
$2.303{\log }_{10}4=\frac{2.303}{t_1}{\log }_{10}4$
$t_1=1h$