In the reaction ${NH}_{4} {NO}_{2} (a q) ⟶ N_{2} (g) + 2 H_{2} O (l)$ the volume of $N_{2}$ after 20 min and after a long time is 40 ml and 70 ml respectively. The value of rate constant is :

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$(\frac{1}{20}) \log (\frac{7}{3}) min^{- 1}$

$(\frac{1}{20}) \ln (\frac{7}{4}) min^{- 1}$

$(\frac{2.303}{1200}) \log (\frac{7}{3}) \sec^{- 1}$

$(\frac{2.303}{10}) \log (\frac{11}{7}) min^{- 1}$

answer is B.

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Detailed Solution

${NH}_{\underset{α}{4}} {NO}_{2} ⟶ \underset{\underset{\underset{α}{x}}{-}}{N_{2}} + \underline{2} H_{2} O \begin{aligned} k \times 20 = \ln (\frac{70}{30}) \\ k = (\frac{2.303}{1200}) \log (\frac{7}{3}) \sec^{- 1} \end{aligned}$

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