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JEE Main 202623 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2026 · 23 January · Evening Shift · Chemistry · Chemical Kinetics

Observe the following reactions at T(K) . I. A arrow products. II....

Observe the following reactions at T(K)\mathrm{T}(\mathrm{K}).
I. A→\mathrm{A} \rightarrow 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−1 min−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 II ?

Options

  1. A10−210^{-2}
  2. B10−310^{-3}
  3. C2×10−32 \times 10^{-3}
  4. D4×10−34 \times 10^{-3}

Correct answer

D. 4×10−34 \times 10^{-3}

Step-by-step solution

For reaction II:
5Br−^- + BrO3−_3^- + 6H+^+ →\rightarrow 3Br2_2 + 3H2_2O,
The overall reaction rate is:
r=−15Δ[Br−]Δt=−13Δ[Br2]Δtr = -\frac{1}{5}\frac{\Delta[Br^-]}{\Delta t} = -\frac{1}{3}\frac{\Delta[Br_2]}{\Delta t}.
Given Δ[Br−]Δt=2×10−4\frac{\Delta[Br^-]}{\Delta t} = 2 \times 10^{-4} mol L−1^{-1} min−1^{-1}
The reaction rate is r=2×10−45=4×10−5r = \frac{2 \times 10^{-4}}{5} = 4 \times 10^{-5} mol L−1^{-1} min−1^{-1}.
Since the rates of both reactions are equal at 10.10 am and reaction I is first-order:
r=k[A]r = k[A], so 4×10−5=k×10−24 \times 10^{-5} = k \times 10^{-2}, giving k=4×10−3k = 4 \times 10^{-3} min−1^{-1}.
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