The number of molecules with energy greater
than the threshold energy for a reaction
increases five fold by a rise of temperature
from 27oC to 42oC. Its energy of activation in
J/mol is _____.
(Take ln 5 = 1.6094; R = 8.314 J mol–1 K–1)
Answer (integer)
84297
Solution
$\because$ k = A${e^{ - {{{E_a}} \over {RT}}}}$
<br><br>T<sub>1</sub> = 300K, T<sub>2</sub> = 315K
<br><br>As per question K<sub>T<sub>2</sub></sub> = 5K<sub>T<sub>2</sub></sub> as molecules activated are increased five times so K will increases five time.
<br><br>$\ln \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right)$ = ${{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$
<br><br>$\Rightarrow$ ln 5 = ${{{E_a}} \over R}\left( {{{15} \over {300 \times 315}}} \right)$
<br><br>$\Rightarrow$ E<sub>a</sub> = ${{1.6094 \times 8.314 \times 300 \times 315} \over {15}}$
<br><br>= 84297.47 Joules/mole
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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