Question

In a second order reaction, when the concentration of both the reactants are equal, the reaction is completed $20 \%$ in $500 {s}$. How long would it take for the reaction to go to $60 \%$ completion? 

(a) $3000 {~s}$

(b) $5000 {~s}$

(c) $1000 {~s}$

(d) $2000 {~s}$

Answer 1

Correct Option : (A)

Explanation —

In first case,

Given that, $t=500 {s} ; a=100$

$a-x=80 \%$ of $100=100 \times \frac{80}{100}=80$

For second order reaction, $k=\frac{1}{t \times a} \cdot \frac{x}{a-x}$

$=\frac{1}{500 \times 100} \times \frac{20}{80}=5 \times 10^{-6}$

In second case, $a=100$

$a-x=40 \% \text { of } 100=100 \times \frac{40}{100}=40$

$\therefore \quad t=\frac{1}{k \times a} \cdot \frac{x}{a-x}=\frac{1}{5 \times 10^{-6} \times 100} \times \frac{60}{40}=3000 {s}$