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Q.

If a second order reaction involving a single reactant , 25% complete in 20 minutes then what is its half life if initial concentration of the reactant is being made double at the same temperature.

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a

30 min

b

50 min

c

60 min

d

15 min

answer is B.

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Detailed Solution

$\frac{1}{C} - \frac{1}{C_{0}} = kt (for 2^{nd} order reaction)$

$\begin{array}{l} \frac{1}{\frac{3}{4} C_{0}} - \frac{1}{C_{0}} = k \times 20 \\ \frac{1}{\frac{1}{2} (2 C_{0})} - \frac{1}{2 C_{0}} = k \times t_{1 / 2} \\ \Rightarrow \frac{\frac{1}{3 C_{0}}}{\frac{1}{2 C_{0}}} = \frac{20}{t_{1 / 2}} \Rightarrow \frac{2}{3} = \frac{20}{t_{1 / 2}} \end{array}$

$\Rightarrow t_{1 / 2} = 30 minutes$

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