JEE MAIN - Chemistry (2020 - 6th September Evening Slot - No. 7)
The rate of a reaction decreased by 3.555
times when the temperature was changed
from 40oC to 30oC. The activation energy
(in kJ mol–1) of the reaction is _______.
Take; R = 8.314 J mol–1 K–1 ln 3.555 = 1.268
Take; R = 8.314 J mol–1 K–1 ln 3.555 = 1.268
Answer
100
Explanation
k = A$${e^{ - {{{E_a}} \over {RT}}}}$$
$$ \therefore $$ $$\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
$$ \Rightarrow $$ ln (3.555) = $${{{E_a}} \over {8.314}}\left( {{1 \over {303}} - {1 \over {313}}} \right)$$
$$ \Rightarrow $$ Ea = $${{1.268 \times 8.314 \times 3.3 \times 313} \over {10}}$$
= 99980.7 = 99.98 kJ/mol $$ \simeq $$ 100 kJ/mol
$$ \therefore $$ $$\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
$$ \Rightarrow $$ ln (3.555) = $${{{E_a}} \over {8.314}}\left( {{1 \over {303}} - {1 \over {313}}} \right)$$
$$ \Rightarrow $$ Ea = $${{1.268 \times 8.314 \times 3.3 \times 313} \over {10}}$$
= 99980.7 = 99.98 kJ/mol $$ \simeq $$ 100 kJ/mol
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