1. Tardigrade
  2. Question
  3. Chemistry
  4. A substance 'A' decomposes in solution following the first order kinetics. Flask I contains 1 litre of 1 M solution of A and flask II contains 100 mL of 0.6 M solution of A. After 8 hours the concentration of A in flask I becomes 0.25 M. What will be the time for concentration of A in flask II to become 0.3 M ?

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

Chemical Kinetics

Solution:

For flask I
$k=\frac{2.303}{8} log \frac{1}{0.25} =\frac{2.303}{8} log 4=\frac{2.303}{8} \times0.6021$
For flask II
$k=\frac{2.303}{k} log \frac{0.6}{0.3} =\frac{2.303\times8}{2.303 \times0.6021} \times0.3010 =4\,hours$
Alternatively, From concentration $ 0.6$ to $0.3$ it means $t_1/2 $ has been asked. Then
$t_{1 2 }=\frac{0.693\times8}{2.303\times 0.6021} 4\, hours$