Question
The rate of a reaction doubles when its temperature changes from $$300 K$$ to $$310 K.$$ Activation energy of such a reaction will be : $$\left( {R = 8.314\,J{K^{ - 1}}mo{l^{ - 1}}\,{\text{and}}\,\log 2 = 0.301} \right)$$
A.
$$53.6\,kJ\,mo{l^{ - 1}}$$
B.
$$48.6\,kJ\,mo{l^{ - 1}}$$
C.
$$58.5\,kJ\,mo{l^{ - 1}}$$
D.
$$60.5\,kJ\,mo{l^{ - 1}}$$
Answer :
$$53.6\,kJ\,mo{l^{ - 1}}$$
Solution :
Activation energy can be calculated from the equation
$$\eqalign{
& \frac{{\log {k_2}}}{{\log {k_1}}} = \frac{{{E_a}}}{{2.303R}}\left( {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right) \cr
& {\text{given}}\frac{{{k_2}}}{{{k_1}}} = 2\,\,\,\,{T_2} = 310\,K\,\,\,\,{T_1} = 300K \cr
& = \log 2 = \frac{{ - {E_a}}}{{2.303 \times 8.314}}\left( {\frac{1}{{310}} - \frac{1}{{300}}} \right) \cr
& {E_a} = 53598.6\,J/mol = 53.6kJ/mol. \cr} $$