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 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

The Correct Option is C

Rate constant of first order reaction
$\hspace10mm k=\frac{2.303}{t} \, log_{10} \frac{(A)_0}{(A)_t}$
or $\hspace10mm k=\frac{2.303}{1} \times log_{10} \, \frac{0.8}{0.2} \hspace15mm ...(i)$
(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 $A = 0.9 - 0.675 = 0.225$
$\therefore \hspace10mm k=\frac{2.303}{t_1} \, log_{10} \, \frac{0.9}{0.225} \hspace15mm ...(ii)$
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$
$\hspace25mm t_1=1 \, h$