Question

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

• A. 1.2
• B. 12
• C. 30
• D. 3

Answer 1

Answer: (C)

Since, concentration of the reactant decreases from $0.6M$ to $0M$ (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

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

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

$\therefore 0.1M\stackrel{\text{15}\text{ min}}{\to }\frac{0.1}{2}=0.05M\stackrel{\text{15 }\text{min}}{\to }\frac{0.05}{2}$

$=0.025M$

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