Question

If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately: (log 4 = 0.60, log 5 = 0.69)

• A. 40 minutes
• B. 50 minutes
• C. 45 minutes
• D. 60 minutes

Answer 1

The correct option is C 45 minutes

Rate of a general $1^{st}$ order reaction.

$\frac{-d\left[A\right]}{dt}=k\left[A\right]$

$\Rightarrow {\int }_{{\left[A\right]}_0}^{{\left[A\right]}_t}\frac{d\left[A\right]}{\left[A\right]}=-{\int }_0^{t}kdt$

$ln\frac{{\left[A\right]}_t}{{\left[A\right]}_0}=-kt$

$\Rightarrow 2.303log\frac{{\left[A\right]}_t}{{\left[A\right]}_0}=-kt⟶\left(1\right)$

Given : 60% of reaction completes in 60 minutes.

Applying (1).

$2.303log\frac{0.4{\left[A\right]}_0}{\left[A_0\right]}=-60k$

$2.303log0.4=-60k$

$\Rightarrow$$k=\frac{-2.303log0.4}{60}$

Using the value of k

$2.303log\frac{0.5{\left[A\right]}_0}{{\left[A\right]}_0}=-xk$

$2.303log0.5=-x\left(\frac{-2.303log0.4}{60}\right)$

$\Rightarrow \frac{60log0.5}{log0.4}=x$

$\Rightarrow x=\frac{60(log5-1)}{(log4-1)}$

$\Rightarrow x=\frac{60(0.69-1)}{0.60-1}$

$\Rightarrow x=\frac{60\times 0.31}{0.4}$

$\therefore x\simeq 45min.$