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 \times60} sec^{-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 $1 -\frac{x}{a} = 0.96$
$\frac{x}{a} = 0.04 = 4\%$