JEE MAIN - Chemistry (2017 (Offline) - No. 17)
Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to :
(R = 8.314 J mol–1 K–1)
(R = 8.314 J mol–1 K–1)
12
6
4
8
Giải thích
We know, from arrhenius equation,
k = A.$${e^{{{ - {E_a}} \over {RT}}}}$$
$$ \therefore $$ k1 = A.$${e^{{{ - {E_{{a_1}}}} \over {RT}}}}$$ ......(1)
k2 = A.$${e^{{{ - {E_{{a_2}}}} \over {RT}}}}$$ ......(2)
On dividing equation (2) by (1), we get
$${{{k_2}} \over {{k_1}}} = {e^{{{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}}}$$
$$ \Rightarrow $$ $$\ln \left( {{{{k_2}} \over {{k_1}}}} \right) = {{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}$$ = $${{10,000} \over {8.314 \times 300}}$$ = 4
k = A.$${e^{{{ - {E_a}} \over {RT}}}}$$
$$ \therefore $$ k1 = A.$${e^{{{ - {E_{{a_1}}}} \over {RT}}}}$$ ......(1)
k2 = A.$${e^{{{ - {E_{{a_2}}}} \over {RT}}}}$$ ......(2)
On dividing equation (2) by (1), we get
$${{{k_2}} \over {{k_1}}} = {e^{{{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}}}$$
$$ \Rightarrow $$ $$\ln \left( {{{{k_2}} \over {{k_1}}}} \right) = {{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}$$ = $${{10,000} \over {8.314 \times 300}}$$ = 4