The reaction $\mathrm{A} ightarrow \mathrm{B}$ follows first order kinetics. The time taken for $0.8$ mole…

The reaction $\mathrm{A} ightarrow \mathrm{B}$ follows first order kinetics. The time taken for $0.8$ mole of $A$ to produce $0.6$ mole of $B$ is 1 hour. What is the time taken for conversion of $0.9$ mole of $A$ to produce $0.675$ mole of B?
  1. 2 hours
  2. 1 hour
  3. $0.5$ hour
  4. $0.25$ hour

Solution

$\mathrm{A} ightarrow \mathrm{B}$ For a first order reaction
Given $a=0.8 \mathrm{~mol},(a-x)=0.8-0.6=0.2$
$k=\frac{2.303}{1} \log \frac{0.8}{0.2}$ or $k=2.303 \log 4$
again $a=0.9, a-x=0.9-0.675=0.225$
$k=\frac{2.303}{t} \log \frac{0.9}{0.225}$
$2.303 \log 4=\frac{2.303}{t} \log 4$
Hence $t=1$ hour ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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