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Consider the following gaseous reaction where initial pressure of $A (g) is 200 torr. A (g) \to 3 B (g) + C (g)$ . After time t, the pressure of system increased to 650 torr. Calculate the value of t (in seconds) (Given : rate constant of reaction, k=4.606 x.10^-3 sec^-1, log 2=0.3)

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300

answer is B.

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Detailed Solution

$A (g) \to 3 B (g) + C (g) at t = 0, 200 torr 0 0 at t, 200 - p 3 p p 200 - p + 3 p + p = 650 p = 150 torr t = \frac{1}{K} \times 2.303 \times \log \frac{200}{200 - p} = \frac{1 \times 2.303 \times \log 4}{4.606 \times 10^{- 3}} = 300 \sec$

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