Question

$t_{1/2}$ for a first order reaction is $14.26$ mins. The percentage of reactant decomposed after $50$ sec. is

• A. $8\%$
• B. $4\%$
• C. $6\%$
• D. $10.2\%$

Answer 1

Answer: (B)

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{14.26\times 60}se{c}^{-1}$
$k=\frac{2.303}{t}log\frac{a}{a-x}$ (For first order reaction )
$\frac{0.693}{14.26\times 60}=\frac{2.303}{50}log\frac{a}{a-x}$
$log\frac{a}{a-x}=\frac{0.693\times 50}{14.26\times 60\times 2.303}=0.0175$
$\frac{a}{a-x}=1.041$
or $\frac{a-x}{a}=0.96$ or $1-\frac{x}{a}=0.96$
$\frac{x}{a}=0.04=4\%$