Question

The half-life for the reaction, $N_{2}O_{5}$⇌$2NO_{2} + \frac{1}{2}O_{2}$ in 24 hrs. at $30^{o}C$. Starting with $10g$ of $N_{2}O_{5}$ how many grams of $N_{2}O_{5}$ will remain after a period of 96 hours

• A. $1.25g$
• B. $0.63g$
• C. $1.77g$
• D. $0.5g$

Answer 1

Answer: (B)

$0.63g$

Answer 2

$k = \frac{0.693}{t_{1/2}} = \frac{0.69}{24} = \frac{2.303}{96}\log_{10}\frac{1}{(a - x)}$

Or $\log\frac{10}{(a - x)} = 1.2036$ or $1 - \log(a - x) = 1.2036$

Or $\log(a - x) = - 0.2036;$ $(a - x) = 0.6258g$