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 was 1.5 bar. What is the rate constant $({in min}^{- 1})$ of the reaction (log 2 = 0.3) [2026]

Question

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 was 1.5 bar. What is the rate constant $({in min}^{- 1})$ of the reaction? (log 2 = 0.3) [2026]

Smart Achievers · Accepted Answer

(4)

$\begin{matrix} A_{g} & \to & B_{g} & + & C_{g} \\ 1 & - & - \\ 1 - P & P & P \end{matrix}$

$P_{total} = 1 + P$

$1.5 = 1 + P$

$P = 0.5$

$k = \frac{1}{100} \ln \frac{1}{0.5}$

$= \frac{0.693}{100}$

$= 6.9 \times 10^{- 3} {min}^{- 1}$

Solution

1 $6.9 \times 10^{- 2}$

2 $6.9 \times 10^{- 1}$

3 $6.9 \times 10^{- 4}$

4 $6.9 \times 10^{- 3}$

Ans.
(4)

$\begin{matrix} A_{g} & \to & B_{g} & + & C_{g} \\ 1 & - & - \\ 1 - P & P & P \end{matrix}$

$P_{total} = 1 + P$

$1.5 = 1 + P$

$P = 0.5$

$k = \frac{1}{100} \ln \frac{1}{0.5}$

$= \frac{0.693}{100}$

$= 6.9 \times 10^{- 3} {min}^{- 1}$

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Solution

Q. 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 was 1.5 bar. What is the rate constant (in min-1) of the reaction (log 2 = 0.3) [2026]

1 6.9×10-2

2 6.9×10-1

3 6.9×10-4

4 6.9×10-3