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  4. 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?

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Q. 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?

Chemical Kinetics

Solution:

We know that $k=\frac{1}{t} \times\frac{x}{a\left(a-x\right)}$
[For a second order reaction]
$Since\, x=\frac{20}{100}=0.2, a=1$
$k=\frac{1\times0.2}{500\times1\left(1-0.2\right)}=\frac{1}{500\times4}$
$Now, k=\frac{1}{t}\times\frac{0.6}{1\left(1-0.6\right)}$
$\therefore \, \frac{1}{500\times4}=\frac{1}{t}\times\frac{0.6}{0.4} \Rightarrow \, t=3000 seconds $