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)

A_ g -> B_ g+ C_ g 1 - - 1-P P P \ P_ total =1+ P 1.5=1+ P P=0.5 K=(1 / 100) ln (1 / 0.5) =(0.693 / 100) =6.9 x 10^-3 min^-1

Decomposition of A is a first order reaction at T K and... | null (Chemistry)

JEE Mains 2026 21st 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}) \rightarrow \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 (in

\min ^{-1}

) of the reaction ? (

\log 2=0.3)

6.9 \times 10^{-1}

6.9 \times 10^{-3}

6.9 \times 10^{-2}

6.9 \times 10^{-4}

Correct Answer:

\quad \mathrm{A}_{\mathrm{g}} \rightarrow \mathrm{B}_{\mathrm{g}}+\mathrm{C}_{\mathrm{g}}\\

1 - - 1-P P P

\\ \mathrm{P}_{\text {total }}=1+\mathrm{P}\\

1.5=1+\mathrm{P}\\

\mathrm{P}=0.5\\

\mathrm{K}=\frac{1}{100} \ln \frac{1}{0.5}\\

=\frac{0.693}{100}\\

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

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