Question

It takes $1\ h$ for a first order reaction to go to $50%$ completion. The to ta l time required for the same reaction to reach $87.5%$ completion will be

• A. 1.75 h
• B. 6.00 h
• C. 3.50 h
• D. 3.00 h

Answer 1

Answer: (D)

For Ist order reaction
$t_{1/ 2} =\frac{0.693}{k}$
$1=\frac{0.693}{k}$
$\therefore k=\frac{0.693}{1}$
Also, $t=\frac{2.303}{k} log \frac{a}{a-x }$
For $87.5\%$ completion
$t=\frac{2.303}{k}log \frac{a}{a-0.875}$
$=\frac{2.303}{0.693}$ log 8
$=\frac{2.303}{0.693}\times3\times0.3010=3 h$