Decomposition of A is a first order reaction at T( K) and is given by A( g) 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…

Given: Reaction is A( g) B( g) + C( g). Initial pressure of A is P0 = 1 bar in a closed vessel. After t = 100 min, the total pressure is Pt = 1.5 bar. Find: The first-order rate constant k. For a first-order gas-phase reaction, use k = 2.303t P0PA So we first need the partial pressure of A after 100 min. Let the decrease in pressure of A be x. Then PA = P0 - x, PB = x, PC = x Hence total pressure at time t is Pt = (P0 - x) + x + x = P0 + x Given 1 + x = 1.5 Therefore, x = 0.5 bar Now the partial pressure of A after 100 min is PA = 1 - 0.5 = 0.5 bar Substitute in the first-order equation: k = 2.303100 1.00.5 Since 1.00.5 = 2 and 2 = 0.3, k = 2.303100 0.3 k = 0.6909100 6.9 10^-3 min^-1 Therefore, the rate constant is 6.9 10^-3 min^-1 and the correct option is D.

Decomposition of A is a first order reaction at T( K) and is give… — Solution | JEE Chemistry

MCQMediumJEE 2026Integrated Rate Laws

Decomposition of $A$ is a first order reaction at $T(\text{K})$ and is given by $A(\text{g}) \rightarrow B(\text{g}) + C(\text{g})$ .

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

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

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Reaction is $A(\text{g}) \rightarrow B(\text{g}) + C(\text{g})$ . Initial pressure of $A$ is $P_0 = 1 \, \text{bar}$ in a closed vessel. After $t = 100 \, \text{min}$ , the total pressure is $P_t = 1.5 \, \text{bar}$ .

Find: The first-order rate constant $k$ .

For a first-order gas-phase reaction, use

k = \frac{2.303}{t} \log \frac{P_0}{P_A}

So we first need the partial pressure of $A$ after $100 \, \text{min}$ .

Let the decrease in pressure of $A$ be $x$ . Then

P_A = P_0 - x, \quad P_B = x, \quad P_C = x

Hence total pressure at time $t$ is

P_t = (P_0 - x) + x + x = P_0 + x

Given

1 + x = 1.5

Therefore,

x = 0.5 \, \text{bar}

Rate Constant Calculation

Now the partial pressure of $A$ after $100 \, \text{min}$ is

P_A = 1 - 0.5 = 0.5 \, \text{bar}

Substitute in the first-order equation:

k = \frac{2.303}{100} \log \frac{1.0}{0.5}

Since

\frac{1.0}{0.5} = 2

and $\log 2 = 0.3$ ,

k = \frac{2.303}{100} \times 0.3

k = \frac{0.6909}{100} \approx 6.9 \times 10^{-3} \, \text{min}^{-1}

Therefore, the rate constant is $6.9 \times 10^{-3} \, \text{min}^{-1}$ and the correct option is D.

Common mistakes

Using the total pressure $1.5 \, \text{bar}$ directly in the first-order formula is incorrect because the formula requires the partial pressure of reactant $A$ , not the total pressure. First find $P_A$ from the stoichiometry of the reaction.
Assuming the total pressure remains equal to the initial pressure is wrong because one mole of $A$ forms two moles of gaseous products. The total number of moles increases, so total pressure increases during decomposition.
Taking $x = 1.5 \, \text{bar}$ instead of solving $P_t = P_0 + x$ is a setup error. Since $P_0 = 1 \, \text{bar}$ and $P_t = 1.5 \, \text{bar}$ , the correct extent is $x = 0.5 \, \text{bar}$ .

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Answer

Step-by-step solution

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