Explanation

Initial pressure of A = 1 bar

Reaction:

A(g) → B(g) + C(g)

If x bar of A decomposes, then:

A remaining = (1 − x)

B formed = x

C formed = x

Total pressure after time t:

= (1 − x) + x + x

= 1 + x

Given total pressure after 100 min = 1.5 bar

Therefore,

1 + x = 1.5

x = 0.5

Pressure of A remaining = 1 − 0.5 = 0.5 bar

For first order reaction:

\[ k = \frac{2.303}{t} \log \frac{P_0}{P_t} \]

Here,

P₀ = 1 bar

Pₜ = 0.5 bar

t = 100 min

\[ k = \frac{2.303}{100} \log \frac{1}{0.5} \]

\[ \log \frac{1}{0.5} = \log 2 \]

Given log 2 = 0.3

\[ k = \frac{2.303}{100} × 0.3 \]

\[ k ≈ \frac{0.69}{100} \]

\[ k = 6.9 × 10^{-3} \, min^{-1} \]

Correct Option: B

Related JEE Main Questions