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Q.

If the half lives of first order reaction at 350 K and 300 K are 2 and 20 seconds respectively. The activation energy of the reaction in kJ/mol is

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a

20.1

b

30.2

c

60.3

d

40.2

answer is A.

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Detailed Solution

T₁=300k T₂=350 K
$\log (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{2.303 \times R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$ $\log (\frac{\frac{0.693}{2}}{\frac{0.693}{20}}) = \frac{E_{a}}{2.303 \times 8.314} [\frac{1}{300} - \frac{1}{350}] \Rightarrow E_{a} = 40.2 KJ mol$

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