Question

In the first order reaction, half of the reaction is completed in $100$ second. The time for $99\%$ reaction to occur will be

• A. 664.64 sec
• B. 646.6 sec
• C. 660.9 sec
• D. 654.5 sec

Answer 1

Answer: (A)

Half of the reaction is completed in $100$ second
$\therefore t_{1/2}=100sec.$
$\therefore K=\frac{0.693}{100}se{c}^{-1}$
For a first order reaction
$K=\frac{2.303}{t}\log \frac{a}{a-x}$
$a=100,x=99,t_{99\%}=?$
$t_{99\%}=\frac{2.303}{K}\log \frac{100}{100-99}$
$=\frac{2.303\times 100}{0.693}\times \log 100sec$
$=\frac{2.303\times 100\times 2}{0.693}$
$=664.64sec$