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The half-life of decomposition of $N_{2} O_{5}$ is a first order reaction represented by
$N_{2} O_{5} \to N_{2} O_{4} + 1 / 2 O_{2}$
After 15 minutes the volume of $O_{2}$ produced is 9 mL and at the end of the reaction 35 mL. The rate constant is equal to:

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$\frac{1}{15} \log_{e} \frac{35}{26}$

$\frac{1}{15} \log_{e} \frac{44}{26}$

$\frac{1}{15} \log_{e} \frac{35}{36}$

15log_e $\frac{26}{35}$

answer is A.

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Detailed Solution

$K \times 15 = \ln (\frac{35 - 0}{35 - 9})$

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