Question

For a second order reaction in which both the reactants have equal initial concentration, the time taken for 20% completion of reaction is 500 second. What will be the time taken for 60% of the reaction?

• A. 500 sec
• B. 1000 sec
• C. 300 sec
• D. 1500 sec

Answer 1

Answer: (C)

We know that $k=\frac{1}{t}\times \frac{x}{a\left(a-x\right)}$
[For a second order reaction]
$Sincex=\frac{20}{100}=0.2,a=1$
$k=\frac{1\times 0.2}{500\times 1\left(1-0.2\right)}=\frac{1}{500\times 4}$
$Now,k=\frac{1}{t}\times \frac{0.6}{1\left(1-0.6\right)}$
$\therefore \frac{1}{500\times 4}=\frac{1}{t}\times \frac{0.6}{0.4}\Rightarrow t=3000seconds$