Half life of zero order reaction $\mathrm{A} \rightarrow$ product is 1 hour, when initial concentration of reactant is $2.0 \mathrm{~mol} \mathrm{~L}{ }^{-1}$. The time required to decrease concentration of A from 0.50 to $0.25 \mathrm{~mol} \mathrm{~L}^{-1}$ is :
Solution
<p>For zero order reaction</p>
<p>$$\begin{aligned}
& \text { Half life }=\frac{\mathrm{A}_{\mathrm{o}}}{2 \mathrm{k}} \\
& 60 \min =\frac{2}{2 \mathrm{k}} \\
& \mathrm{k}=\frac{1}{60} \mathrm{M} / \min
\end{aligned}$$</p>
<p>Now</p>
<p>$$\begin{aligned}
& A_t=A_o-k t \\
& t=\frac{A_o-A_t}{k} \\
&=\frac{0.5-0.25}{1 / 60} \\
& 0.25 \times 60 \\
& t=15 \mathrm{~min}
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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