JEE Main 2025 Apr 4 Shift 2, Chemistry Q20: Reactions of Various Orders

JEE Main2025Apr 4, Shift 2Chemistry

Half life of zero order reaction $A \to$ product is 1 hour, when initial concentration of reaction is $2.0$ mol L $^{- 1}$ . The time required to decrease concentration of A from $0.50$ to $0.25$ mol L $^{- 1}$ is:

A
0.5 hour
B
4 hour
C
15 min
D
60 min

Solution

For a zero-order reaction, $t_{1/2} = \frac{[ A ] _{0}}{2 k}$ .

Given $t_{1/2} = 60$ min and $[A]_{0} = 2.0$ mol L $^{- 1}$ :

$60 = \frac{2}{2 k} \Rightarrow k = \frac{1}{60} mol L^{- 1} min^{- 1}$ .

The zero-order integrated rate law is $[A]_{t} = [A]_{0} - k t$ , so

$t = \frac{[ A ] _{0} - [ A ] _{t}}{k} = \frac{0.50 - 0.25}{1/60} = 0.25 \times 60 = 15$ min.

$∴$ The required time is 15 min.

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JEE Main 2025 Apr 4 Shift 2, Chemistry Q20: Reactions of Various Orders

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