Question

$t_{1/2}$ for a first order reaction is $14.26\, min$. Calculate the time when $5\%$ of the reactant is left.

• A. $62\, min$
• B. $42\, min$
• C. $26\, min$
• D. $53\, min$

Answer 1

Answer: (A)

$62\, min$

Answer 2

$x=100\%-5\%=95\%$ $k=\frac{0.693}{t_{1 /2}}=\frac{0.693}{14.26}=0.0486$ $k=\frac{2.303}{t}log \frac{a}{a-x}$ $t=\frac{2.303}{0.0486}log \frac{100}{100-95}$ $t=61.65\,min\, \approx 62\,min$