The gaseous reaction $A (g) \to 2 B (g) + C (g)$ is found to be first order. If the reaction is started with $p_{A} = 90 mmHg$ , the total pressure after 10 min is found to be $180 mmHg$ . The rate constant of the reaction is

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$1.15 \times 10^{- 3} s^{- 1}$

$2.30 \times 10^{- 3} s^{- 1}$

$3.45 \times 10^{- 3} s^{- 1}$

$4.60 \times 10^{- 3} s^{- 1}$

answer is A.

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Detailed Solution

$\underset{p_{0} - p}{A (g)} \to \underset{2 p}{2 B (g}) + \underset{p}{C (g)}$

Total pressure $= (p_{0} - p) + (2 p) + p = p_{0} + 2 p$

$180 mmHg = 90 mmHg + 2 p$ or $p = 45 mmHg$

Now, $\log \frac{p_{0} - p}{p_{0}} = - \frac{k}{2.303} t$

$\log \frac{45}{90} = - \frac{k}{2, 303} (10 \times 60 s)$ . Thus $k = \frac{0.301 \times 2.303}{10 \times 60 s} = 1.155 \times 10^{- 3} s^{- 1}$

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