Activation energy (E$$a$$) and rate constants (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2) are related by
A
$$\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
B
$$\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)$$
C
$$\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} + {1 \over {{T_1}}}} \right)$$
D
$$\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
Explanation
Let $${k_1} = A{e^{ - {{{E_a}} \over {R{T_1}}}}}$$
$$\ln {k_1} = \ln A - {{{E_a}} \over {R{T_1}}}$$ ......(1)
$${k_2} = A{e^{ - {{{E_a}} \over {R{T_2}}}}}$$
$$\ln {k_2} = \ln A - {{{E_a}} \over {R{T_2}}}$$ ....(2)
From eq.(1) and (2), we have
$$\ln {k_2} - \ln {k_1} = \ln A - {{{E_a}} \over {R{T_2}}} - \ln A + {{{E_a}} \over {R{T_1}}}$$
$$ \Rightarrow $$ $$\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
$$ \Rightarrow $$ $$\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)$$
