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First Order Reaction

The half life...

Question

The half life for the reaction, $N_{2} O_{5} ⇌ 2 {N O}_{2} + \frac{1}{2} O_{2}$ in 24 hour at $30^{0} C$ starting with 10g 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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Solution

For the first order reaction,
$t_{1}$ $_{2}$ = 0.693/ k
K = 0.693 / $t_{1}$ $_{2}$ = $\frac{0.693}{24}$ = 0.0289 $h r^{- 1}$
kt = 2.303 * log $\frac{a}{a - x}$
0.0289 * 98 = 2.3038 * log $\frac{10}{10 - x}$
log $\frac{10}{10 - x}$ = 1.2
$\frac{10}{10 - x}$ = antilog 1.2
$\frac{10}{10 - x}$ = 15.85
10 = 158.5 - 15.85 x
15.85 x = 148. 5
x = 9.36
a-x =10 - 9.36 = 0.63 g

Therefore the answer is 0.63 g

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