Q6 / 75 · JEE Main 2026 (21 January Shift 2) · Chemistry
Decomposition of A is a first order reaction at $\mathrm{T}(\mathrm{K})$ and is given by $\mathrm{A}(\mathrm{g}) \longrightarrow \mathrm{B}(\mathrm{g})+\mathrm{C}(\mathrm{g})$. In a closed 1 L vessel, 1 bar $\mathrm{A}(\mathrm{g})$ is allowed to decompose at $\mathrm{T}(\mathrm{K})$. After 100 minutes, the total pressure was 1.5 bar. What is the rate constant $\left(\mathrm{in} \mathrm{min}^{-1}\right)$ of the reaction ? $(\log 2=0.3)$
- Correct answerA$6.9 \times 10^{-3}$
- B$6.9 \times 10^{-4}$
- C$6.9 \times 10^{-1}$
- D$6.9 \times 10^{-2}$
Solution
For the reaction $A(g) \rightarrow B(g) + C(g)$, 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: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$.
Using pressures instead: $k = \frac{2.303}{100} \log \frac{1}{0.5}$
$=\frac{2.303}{100} \times \log 2 = \frac{2.303 \times 0.3}{100} = \frac{0.691}{100}$
$= 6.91 \times 10^{-3}$ min$^{-1}$ ≈ $6.9 \times 10^{-3}$ min$^{-1}$
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: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$.
Using pressures instead: $k = \frac{2.303}{100} \log \frac{1}{0.5}$
$=\frac{2.303}{100} \times \log 2 = \frac{2.303 \times 0.3}{100} = \frac{0.691}{100}$
$= 6.91 \times 10^{-3}$ min$^{-1}$ ≈ $6.9 \times 10^{-3}$ min$^{-1}$
Hence, the correct answer is option: 'A'.