JEE MAIN - Chemistry (2025 - 4th April Evening Shift - No. 9)
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 :
0.5 hour
15 min
60 min
4 hour
Explanation
For zero order reaction
$$\begin{aligned} & \text { Half life }=\frac{\mathrm{A}_{\mathrm{o}}}{2 \mathrm{k}} \\ & 60 \min =\frac{2}{2 \mathrm{k}} \\ & \mathrm{k}=\frac{1}{60} \mathrm{M} / \min \end{aligned}$$
Now
$$\begin{aligned} & A_t=A_o-k t \\ & t=\frac{A_o-A_t}{k} \\ &=\frac{0.5-0.25}{1 / 60} \\ & 0.25 \times 60 \\ & t=15 \mathrm{~min} \end{aligned}$$
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