JEE MAIN - Chemistry (2020 - 4th September Evening Slot - No. 9)
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)
(Take ln 5 = 1.6094; R = 8.314 J mol–1 K–1)
Answer
84297.47to84297.48
Explanation
$$ \because $$ k = A$${e^{ - {{{E_a}} \over {RT}}}}$$
T1 = 300K, T2 = 315K
As per question KT2 = 5KT2 as molecules activated are increased five times so K will increases five time.
$$\ln \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right)$$ = $${{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
$$ \Rightarrow $$ ln 5 = $${{{E_a}} \over R}\left( {{{15} \over {300 \times 315}}} \right)$$
$$ \Rightarrow $$ Ea = $${{1.6094 \times 8.314 \times 300 \times 315} \over {15}}$$
= 84297.47 Joules/mole
T1 = 300K, T2 = 315K
As per question KT2 = 5KT2 as molecules activated are increased five times so K will increases five time.
$$\ln \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right)$$ = $${{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
$$ \Rightarrow $$ ln 5 = $${{{E_a}} \over R}\left( {{{15} \over {300 \times 315}}} \right)$$
$$ \Rightarrow $$ Ea = $${{1.6094 \times 8.314 \times 300 \times 315} \over {15}}$$
= 84297.47 Joules/mole
Comments (0)
