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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2026 · 21 January · Evening Shift · Chemistry · Chemical Kinetics

Decomposition of A is a first order reaction at T(K) and is given...

Decomposition of A is a first order reaction at T(K)\mathrm{T}(\mathrm{K}) and is given by A(g)⟶B(g)+C(g)\mathrm{A}(\mathrm{g}) \longrightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g}). In a closed 1 L vessel, 1 bar A(g)\mathrm{A}(\mathrm{g}) is allowed to decompose at T(K)\mathrm{T}(\mathrm{K}). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant \left(\mathrm{in} \mathrm{min}^{-1}\right)ofthereaction?of the reaction ?(\log 2=0.3)

Options

  1. A6.9×10−36.9 \times 10^{-3}
  2. B6.9×10−46.9 \times 10^{-4}
  3. C6.9×10−16.9 \times 10^{-1}
  4. D6.9×10−26.9 \times 10^{-2}

Correct answer

A. 6.9×10−36.9 \times 10^{-3}

Step-by-step solution

For the reaction A(g)→B(g)+C(g)A(g) \rightarrow B(g) + C(g), initially pressure of A is 1 bar.
Initially: total pressure = 1 bar.
After 100 min: total pressure = 1.5 bar.
Using initial and final pressures, if x bar of A decomposes, then A becomes (1-x), B = x, C = x.
Total pressure = 1-x+x+x = 1+x = 1.5, so x = 0.5 bar.
For a first-order reaction: k=2.303tlog⁡[A]0[A]k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}.
Using pressures instead: k=2.303100log⁡10.5k = \frac{2.303}{100} \log \frac{1}{0.5}
=2.303100×log⁡2=2.303×0.3100=0.691100=\frac{2.303}{100} \times \log 2 = \frac{2.303 \times 0.3}{100} = \frac{0.691}{100}
=6.91×10−3= 6.91 \times 10^{-3} min−1^{-1} ≈ 6.9×10−36.9 \times 10^{-3} min−1^{-1}
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