JEE Main 2026 — Jan 21, Shift 2 — Chemistry solution

JEE Main2026Jan 21, Shift 2Chemistry

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

A
$6.9 \times 1 0^{- 1}$
B
$6.9 \times 1 0^{- 3}$
C
$6.9 \times 1 0^{- 2}$
D
$6.9 \times 1 0^{- 4}$

Solution

Let $p$ = pressure of A that has decomposed. Then $P_{A} = 1 - p$ , $P_{B} = P_{C} = p$ . Total: $P_{t o t} = 1 + p$ .

Given $P_{t o t} = 1.5$ bar ⇒ $p = 0.5$ bar. So fraction of A remaining $= \frac{1 - 0.5}{1} = 0.5$ .

First-order integrated rate law: $k = \frac{1}{t} ln \frac{P _{0}}{P _{A}} = \frac{1}{100} ln \frac{1}{0.5} = \frac{ln 2}{100} = \frac{0.693}{100} = 6.9 \times 1 0^{- 3}$ min $^{- 1}$ .

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