Chemical Kinetics - Result Question 44
####44. A substance ‘A’ decomposes by a first order reaction starting initially with $[A]=2.00 m$ and after $200 min$, [A] becomes $0.15 m$. For this reaction $t _{1 / 2}$ is
(a) $53.72 min$
(b) $50.49 min$
(c) $48.45 min$
(d) $46.45 min$
[1995]
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Solution:
- (a) Given initial concentration $(a)=2.00 m$; Time taken $(t)=200 min$ and final concentration $(a-x)=0.15 m$. For a first order reaction, rate constant,
$k=\frac{2.303}{t} \log \frac{a}{a-x}=\frac{2.303}{200} \log \frac{2.00}{0.15}$ $=\frac{2.303}{200} \times(0.301+0.824)=1.29 \times 10^{-2} min^{-1}$.
Further
$(t _{1 / 2})=\frac{0.693}{k}=\frac{0.693}{1.29 \times 10^{-2}}=53.49 min$.
