Question

If $75\%$ of a first order reaction was completed in 90 minutes, $60\%$ of the same reaction would be completed in approximately:

• A. $59.4min$
• B. $22.1min$
• C. $76.9min$
• D. $32.8min$

Answer 1

The correct option is A $59.4min$

For a first order reaction:
$A\to \text{Products}$
$\left[A\right]=[A{]}_0{e}^{-kt}...eqn(i)$
Here
$[A{]}_0$ is the intial concentration
$\left[A\right]$ is the concentration at time t
$k$ is the first order rate constant.

When $75\%$ of a first order reaction is complete:
$\left[A\right]=0.25[A{]}_0$
$t=90min$
Put in $eqn\left(i\right)$
$0.25[A{]}_0=[A{]}_0{e}^{-k\times 90}$
$\ln 4=k\times 90k=\frac{\ln 4}{90}mi{n}^{-1}$
When$60\%$ of the reactionis complete:
$\left[A\right]=0.4[A{]}_0$
$t=?$
$k=\frac{\ln 4}{90}mi{n}^{-1}$
Put in $eqn\left(i\right)$
$0.4[A{]}_0=[A{]}_0{e}^{-\frac{\ln 4}{90}\times t}$

$\ln 0.4=-\frac{\ln 4}{90}\times t$

$\frac{\left(\ln \right(10)-2\ln 2)\times 90}{2\ln 2}=t$

$t=59.49min$