For a gaseous first order reaction
$A (g) ⟶ 2 B (g) + 3 C (g)$
the rate constant was determined by monitoring total pressure at different time. Calculate the value of average life in sec. [Given : In 2 = 0. 7]
t 0 90 min -
Total (in mm of Hg) pressure 500 1900 2100

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answer is 2571.

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Detailed Solution

$\begin{aligned} A (g) ⟶ 2 B (g) + 3 C (g) \\ \begin{array}{ccc} 500 & - & - \\ 500 - P & 2 P & 3 P \end{array} \end{aligned}$

$\begin{array}{l} k \times 90 = \ln (\frac{2100 - 500}{2100 - 1900}) \\ k = \frac{3 \times 0.7}{90 \times 60} \sec^{- 1} \\ t_{a v} = \frac{1}{k} = \frac{30 \times 60}{0.7} = 25718 \end{array}$

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