JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

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 ? $(\log 2=0.3)$

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

A
$6.9 \times 10^{-3}$
B
$6.9 \times 10^{-4}$
C
$6.9 \times 10^{-1}$
D
$6.9 \times 10^{-2}$

Did you get this right?

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 $^{-1}$

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JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Official previous-year question

Held on 21 Jan 2026 · Verified 6 Jul 2026.

Options

A
$6.9 \times 10^{-3}$
B
$6.9 \times 10^{-4}$
C
$6.9 \times 10^{-1}$
D
$6.9 \times 10^{-2}$

Did you get this right?

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 $^{-1}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry