The decomposition of $N_{2} O_{5}$ in carbon tetrachloride was followed by measuring the volume of $O_{2}$ gas evolved : $2 N_{2} O_{5} (C C l_{4}) \to_{}^{} 2 N_{2} O_{4} (C C l_{4}) + O_{2} (g)$ .
The maximum volume of $O_{2}$ gas obtained was 100 $c m^{3}$ . In 500 minutes, $90 c m^{3}$ of $O_{2}$ were evolved. The first order rate constant $(i n \min^{- 1})$ for the disappearance of $N_{2} O_{5}$ is

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$\frac{2.303}{500} \log \frac{90}{100}$

$\frac{2.303}{500}$

$\frac{2.303}{500} \log \frac{100}{90}$

$\frac{100}{10 \times 500}$

answer is A.

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Detailed Solution

According to first order reaction,

$K = \frac{2.303}{500} \log \frac{100}{100 - 90} K = \frac{2.303}{500} \log \frac{100}{10} K = \frac{2.303}{500} \log 10 (\log 10 = 1) K = \frac{2.303}{500}$

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