Your AI-Powered Personal Tutor
Question
Activation energy (E$a$) and rate constants (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2) are related by
$\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$
$\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)$
$\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} + {1 \over {{T_1}}}} \right)$
$\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$

Solution

Please login to view the detailed solution steps...

Go to DASH