Question
The activation energy for a reaction which doubles the rate when the temperature is raised from $$298\,K$$ to $$308\,K$$ is
A.
$$59.2\,kJ\,mo{l^{ - 1}}$$
B.
$$39.2\,kJ\,mo{l^{ - 1}}$$
C.
$$52.9\,kJ\,mo{l^{ - 1}}$$
D.
$$29.5\,kJ\,mo{l^{ - 1}}$$
Answer :
$$52.9\,kJ\,mo{l^{ - 1}}$$
Solution :
Activation energy can be calculated from the equation
$$\eqalign{
& \frac{{{\text{log}}\,{K_2}}}{{{\text{log}}\,{K_1}}} = \frac{{ - {E_a}}}{{2.303\,R}}\left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right) \cr
& {\text{Given}}\,\frac{{{\text{log}}\,{K_2}}}{{{\text{log}}\,{K_1}}} = 2\,{T_2} = 308\,K;\,{T_1} = 298\,K \cr
& \therefore \,\,{\text{log}}\,2 = \frac{{ - {E_a}}}{{2.303 \times 8.314}}\left( {\frac{1}{{308}} - \frac{1}{{298}}} \right) \cr
& {E_a} = 52.9\,kJ\,mo{l^{ - 1}} \cr} $$