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
Answer (integer)
100
Solution
k = A${e^{ - {{{E_a}} \over {RT}}}}$
<br><br>$\therefore$ $$\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
<br><br>$\Rightarrow$ ln (3.555) = ${{{E_a}} \over {8.314}}\left( {{1 \over {303}} - {1 \over {313}}} \right)$
<br><br>$\Rightarrow$ E<sub>a</sub> = ${{1.268 \times 8.314 \times 3.3 \times 313} \over {10}}$
<br><br>= 99980.7 = 99.98 kJ/mol $\simeq$ 100 kJ/mol
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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