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Chemistry · Chemical Kinetics

JEE Main 2026 — 23 January, Evening Shift — Question 54

Observe the following reactions at T(K)\mathrm{T}(\mathrm{K}) I. A → products. II. 5Br−(aq)+BrO3−(aq)+6H+(aq)→3Br2(aq)+3H2O(l)5 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_{3}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{Br}_{2}(\mathrm{aq})+ 3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})

Both the reactions are started at 10.00 am . The rates of these reactions at 10.10 am are same. The value of −Δ[Br−]Δt-\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}} at 10.10 am is 2×10−4 mol L−1Min−12 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{Min}^{-1}. The concentration of A at 10.10 am is 10−2 mol L−110^{-2} \mathrm{~mol} \mathrm{~L}^{-1}. What is the first order rate constant (in min−1\mathrm{min}^{-1} ) of reaction I?

  1. Option A:

    2×10−32 \times 10^{-3}

  2. Option B:

    10−310^{-3}

  3. Option C:

    10−210^{-2}

  4. Option D:

    4×10−34 \times 10^{-3}

    Correct

Answer: D

Step-by-step solution

At t=10\mathrm{t}=10 minutes Rate of reaction =−15Δ[Br−]Δt=15×(2×10−4)=4×10−5=-\frac{1}{5} \frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}=\frac{1}{5} \times\left(2 \times 10^{-4}\right) =4 \times 10^{-5}

For reaction A→P\mathrm{A} \rightarrow \mathrm{P} at t=10\mathrm{t}=10 minutes Rate of reaction =4×10−5=k[A]=4 \times 10^{-5}=\mathrm{k}[\mathrm{A}] k=4×10−3 min−1\mathrm{k}=4 \times 10^{-3} \mathrm{~min}^{-1}.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
23 January, Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics
Topic
Rate Laws and Rate Constant
Observe the following reactions at T ( K ) I. A → products. II.… | JEE Main 2026 PYQ with Solution · DhiX AI