Question

A substance ${}^{\prime }{A}^{\prime }$ decomposes in solution following the first order kinetics. Flask $I$ contains $1$ litre of $1M$ solution of $A$ and flask II contains $100mL$ of $0.6M$ solution of $A$. After $8$ hours the concentration of $A$ in flask I becomes $0.25M$. What will be the time for concentration of $A$ in flask II to become $0.3M$ ?

• A. $0.4hours$
• B. $2.4hours$
• C. $4.0hours$
• D. Unpredictable as rate constant is not given.

Answer 1

Answer: (C)

For flask I
$k=\frac{2.303}{8}log\frac{1}{0.25}=\frac{2.303}{8}log4=\frac{2.303}{8}\times 0.6021$
For flask II
$k=\frac{2.303}{k}log\frac{0.6}{0.3}=\frac{2.303\times 8}{2.303\times 0.6021}\times 0.3010=4hours$
Alternatively, From concentration $0.6$ to $0.3$ it means $t_1/2$ has been asked. Then
$t_{12}=\frac{0.693\times 8}{2.303\times 0.6021}4hours$