Kinetic data for 2A + B to text Products : Exp I: [A]=0.10, [B]=0.20, rate=6.93times10^ -3 ; Exp II: [A]=0.10, [B]=0.25, rate=6.93times10^ -3 ; Exp III: [A]=0.20, [B]=0.30, rate=1.386times10^ -2 text mol L ^ -1 text min ^ -1 . Find the half-life (in minutes) for reactant A.

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JEE MAIN Chemistry QUESTION #5529

Question 1

Kinetic data for $2A + B \to \text{Products}$: Exp I: [A]=0.10, [B]=0.20, rate=$6.93\times10^{-3}$; Exp II: [A]=0.10, [B]=0.25, rate=$6.93\times10^{-3}$; Exp III: [A]=0.20, [B]=0.30, rate=$1.386\times10^{-2}\ \text{mol L}^{-1}\text{min}^{-1}$. Find the half-life (in minutes) for reactant A.

5
10
1
100✔️

Correct Answer Explanation

Comparing Exp I and II: [B] changes but rate is constant → order w.r.t. B = 0. Comparing Exp I and III: [A] doubles (0.10→0.20) and rate doubles → order w.r.t. A = 1. Rate law: $r = k[A]$. From Exp I: $k = \frac{6.93\times10^{-3}}{0.10} = 6.93\times10^{-2}\ \text{min}^{-1}$. Half-life: $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{6.93\times10^{-2}} = 10\ \text{min}$.

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