Q26 / 160 · AP EAMCET 2015 · Chemistry

The time required for a first order reaction to complete $90 \%$ is ' $t$ '. What is the time required to complete $99 \%$ of the same reaction?

  • Correct answer
    A
    $2t$
  • B
    $3t$
  • C
    $t$
  • D
    $4t$

Solution

For the first order reaction, $t=\frac{2303}{k} \log \frac{a}{a-x}$ Now putting the values $\begin{aligned} & t_{90 \%}=t=\frac{2303}{k} \log \left(\frac{100}{100-90}\right) \\ & t_{99 \%}=\frac{2303}{k} \log \left(\frac{100}{100-99}\right) \\ & \frac{t}{t_{99 \%}}=\frac{\frac{2303}{k} \log \left(\frac{100}{10}\right)}{\frac{2303}{k} \log \left(\frac{100}{1}\right)}=\frac{1}{2} \\ & t_{99 \%}=2 t \end{aligned}$

Hence, the correct answer is option: 'A'.

Relevant Questions