Question

In a first order reaction, the concentration of the reactant decrease from $ 0.6 \,M $ to $ 0.3\, M $ in $ 15 $ min . The time taken for the concentration to change from $ 0.1 \,M $ to $ 0.025 \,M $ in minutes is

• A. 1.2
• B. 12
• C. 30
• D. 3

Answer 1

Answer: (C)

Since, concentration of the reactant decreases from $0.6\, M$ to $0\, M$ (i.e., halved) in $15 $ minutes, therefore half time for this reaction be $15 $ min.

i.e.,$t_{1 / 2}=15$ min

Now for a first order reaction

$k =\frac{0.693}{t_{1 / 2}}=\frac{0.693}{15} min ^{-1} $

Again, $k =\frac{2.303}{t} \log \frac{[ A ]_{0}}{[ A ]} $

$\therefore \frac{0.693}{15} =\frac{2.303}{t} \log \frac{0.1}{0.025}$

$t =\frac{2.303 \times 15}{0.693} \times \log 4 $

$ =\frac{2.303 \times 15}{0.693} \times 0.6020=30$ min

Alternate method

$\because$ Reaction is of first order with $t_{1 / 2}=15$ min

$\therefore 0.1 \,M \ce{->[{15\text{ min} }]} \frac{0.1}{2} =0.05\, M \ce{->[{15 \text{min} }]}\frac{0.05}{2} $

$=0.025\, M$

$\therefore $ Total time $=15+15=30$ min