Question

A first-order reaction is $40\%$ complete after $8min$. How long will it take for the reaction to be $90\%$ complete?
$\ln \frac{5}{3}=0.511ln10=2.303$

• A. $50.20min$
• B. $36.36min$
• C. $24.52min$
• D. $12.52min$

Answer 1

The correct option is B $36.36min$

For a 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 $8min$, the remaining reactant is $60\%$

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

Time when the reaction is $90\%$ complete i.e. $(a-x)=0.1a$

So
Case I: $t=8min;(a-x)=0.6a$
Case II: $t=?(a-x)=0.1a$
Since the reaction is of the first order $k$ is same in both cases so using the formula $\left(i\right)$ for both the cases and equating $k$,
$\frac{1}{8}\times ln\left(\frac{1}{0.6}\right)=\frac{1}{t}\times ln\left(\frac{1}{0.1}\right)$
$\Rightarrow t=8\times \frac{ln\left(1/0.1\right)}{ln\left(1/0.6\right)}=8\times \frac{2.302}{0.511}=36.36$
Hence, (b) is the correct option.