JEE MAIN - Chemistry (2013 (Offline) - No. 14)
The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such
a reaction will be:(R = 8.314 JK–1 mol–1 and log 2 = 0.301)
48.6 kJ mol–1
58.5 kJ mol–1
60.5 kJ mol–1
53.6 kJ mol–1
Explanation
Activation energy can be calculated from the equation
$${{\log \,{k_2}} \over {\log \,{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
given $${{{k_2}} \over {{k_1}}} = 2\,\,{T_2} = 310\,K\,\,{T_1} = 300\,K$$
$$ = \log 2 = {{ - {E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {310}} - {1 \over {300}}} \right)$$
$${E_a} = 53598.6J/mol$$
$$ = 53.6kJ/mol.$$
$${{\log \,{k_2}} \over {\log \,{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
given $${{{k_2}} \over {{k_1}}} = 2\,\,{T_2} = 310\,K\,\,{T_1} = 300\,K$$
$$ = \log 2 = {{ - {E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {310}} - {1 \over {300}}} \right)$$
$${E_a} = 53598.6J/mol$$
$$ = 53.6kJ/mol.$$
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