Question

A first-order reaction is $40$% complete after $8$ min. How long will it take before it is $90$% complete?

• A. 40.20 min
• B. 36.36 min
• C. 24.52 min
• D. None of these

Answer 1

The correct option is B 36.36 min

For first order reaction we have:

$K=\frac{1}{t}$ $ln\frac{a}{a-x}$ $\left(i\right)$

where, $K$= rate constant of reaction

$t$= time

$a$= initial concentration of the reactant

$x$= amount of reactant reacted in time $t$

Since, the reactant has reacted $40$% in $8$ min, the remaining reactant is $60$%

$\Rightarrow$ At $t=8$ min , $(a-x)=0.6a$

Now, we have to find time when the reaction is $90$% complete i.e. $(a-x)=0.1a$

So Case I: $t=8$ min; $(a-x)=0.6a$

Case II: $t=?$ ; $(a-x)=0.1a$

Since the reaction is of first order $K$ is same in both cases so using the formula $\left(i\right)$ :-

$\frac{1}{2}\times ln\left(\frac{1}{0.6}\right)=\frac{1}{t}\times ln\left(\frac{1}{0.1}\right)$

$\Rightarrow$ $t=8\times \frac{ln\left(1/0.1\right)}{ln\left(1/0.6\right)}=8\times \frac{2.302}{0.511}=36.36$ min

Answer: (B) $36.36$ min