JEE MAIN - Chemistry (2021 - 25th February Evening Shift - No. 22)
The rate constant of a reaction increases by five times on increase in temperature from 27$$^\circ$$C to 52$$^\circ$$C. The value of activation energy in kJ mol$$-$$1 is _________. (Rounded off to the nearest integer)
[R = 8.314 J K$$-$$1mol$$-$$1]
[R = 8.314 J K$$-$$1mol$$-$$1]
Answer
52
Explanation
T1 = (273 + 27) = 300 K, T2 = (273 + 52) = 325 K
Given, temperature coefficient of the reaction,
$${\alpha _T} = {{{K_{325}}} \over {{K_{300}}}} = 5$$
$$\log {{{K_{{T_2}}}} \over {{K_{{T_1}}}}} = {{{E_a}} \over {2.303R}} \times \left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}}} \right)$$
$$\log {{{K_{325}}} \over {{K_{300}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{{325 - 300} \over {300 \times 325}}} \right)$$
$$\log 5 = {{{E_a}} \over {2.303 \times 8.319}} \times {{25} \over {300 \times 325}}$$
Ea = 52194.78 J mol$$-$$
= 52.194 kJ mol$$-$$1
$$\simeq$$ 52 kJ mol$$-$$1
Given, temperature coefficient of the reaction,
$${\alpha _T} = {{{K_{325}}} \over {{K_{300}}}} = 5$$
$$\log {{{K_{{T_2}}}} \over {{K_{{T_1}}}}} = {{{E_a}} \over {2.303R}} \times \left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}}} \right)$$
$$\log {{{K_{325}}} \over {{K_{300}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{{325 - 300} \over {300 \times 325}}} \right)$$
$$\log 5 = {{{E_a}} \over {2.303 \times 8.319}} \times {{25} \over {300 \times 325}}$$
Ea = 52194.78 J mol$$-$$
= 52.194 kJ mol$$-$$1
$$\simeq$$ 52 kJ mol$$-$$1
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