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The half-life for the reaction $N_{2} O_{5} ⇌ 2 {NO}_{2} + \frac{1}{2} O_{2}$ is 24 hr at $30^{o} C$ . Starting with 10 g of $N_{2} O_{5}$ how many grams of $N_{2} O_{5}$ will remain after a period of 96 hours?

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1.25 g

0.5 g

1.7 g

0.63 g

answer is B.

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Detailed Solution

$k = \frac{0 .693}{24} {hr}^{- 1} = \frac{2 .303}{96} \log \frac{10}{a - x}$

$\log \frac{10}{a - x} = 1 .2036$

$1 - \log (a - x) = 1 .2036$

$\log (a - x) = - 0 .2036$

$(a - x) = antilog (- 0.2036) = 0 .6258 g$

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