Consider a first order gas phase reaction,
$A (g) ⟶ 2 B (g) + C (g)$
The initial pressure taking only A in a rigid vessel was found to be P atm. After a lapse of 10 minutes, the pressure of the system increases by x units and became $p_{10}$ atm. The rate constant for the reaction is given by:

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$k = \frac{2.303}{10} \log \frac{P}{P + x} \min^{- 1}$

$k = 2.303 \times 6 \log \frac{2 P}{3 P - P_{10}} {hr}^{- 1}$

$k = \frac{2.303}{10} \log \frac{P}{P - \frac{x}{2}} {hr}^{- 1}$

$k = 2.303 \times 6 \log \frac{P}{P_{10} - P} {hr}^{- 1}$

answer is D.

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Detailed Solution

$A (g) ⟶ 2 B (g) + C (g) \begin{array}{lcll} t = 0 & P & - & - \\ t = 10 & P - y & 2 y & y \end{array} \begin{array}{l} P_{10} = P + 2 Y = P + x \\ k \times 10 = \ln [\frac{P}{P - y}] \\ \Rightarrow k = 2.303 \times 6 \log [\frac{2 P}{3 P - P_{10}}] {hr}^{- 1} \end{array}$