JEE MAIN - Chemistry (2017 - 9th April Morning Slot - No. 17)
The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is :
(Assume activation energy and preexponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol−1 K−1)
(Assume activation energy and preexponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol−1 K−1)
107.2 kJ mol$$-$$1
53.6 kJ mol$$-$$1
26.8 kJ mol$$-$$1
214.4 kJ mol$$-$$1
Explanation
According to Arrhenius equation,
$$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}}} \right)$$
Substituting the given values in the equation, we get
$$\log 4 = {{{E_a}} \over {2.303 \times 8.314\,J\,{K^{ - 1}}mo{l^{ - 1}}}}\left( {{{310 - 300} \over {300 \times 310}}} \right)$$
$${E_a} = {{0.602 \times 2.303 \times 8.314 \times 300 \times 310} \over {10}}$$
= 107197.12 J/mol$$-$$1 or 107.2 kJ/mol$$-$$1
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