JEE MAIN - Chemistry (2020 - 9th January Morning Slot - No. 7)
For the following reactions
$$A\buildrel {700K} \over \longrightarrow {\mathop{\rm Product}\nolimits} $$
$$A\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{catalyst}^{500K}} {\mathop{\rm Product}\nolimits} $$
it was found that Ea is decreased by 30 kJ/mol in the presence of catalyst.
If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):
$$A\buildrel {700K} \over \longrightarrow {\mathop{\rm Product}\nolimits} $$
$$A\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{catalyst}^{500K}} {\mathop{\rm Product}\nolimits} $$
it was found that Ea is decreased by 30 kJ/mol in the presence of catalyst.
If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):
198 kJ/mol
135 kJ/mol
105 kJ/mol
75 kJ/mol
Explanation
K1 = A$${e^{ - {{{E_a}} \over {R \times 700}}}}$$
K2 = A$${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$$
$$ \because $$Rate is same
$$ \therefore $$ Rate constant will also be same
K1 = K2
A$${e^{ - {{{E_a}} \over {R \times 700}}}}$$ = A$${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$$
$$ \Rightarrow $$ $${{{\left( {{E_a} - 30} \right)} \over {R \times 500}}}$$ = $${{{{E_a}} \over {R \times 700}}}$$
$$ \Rightarrow $$ 5Ea = 7Ea – 210
$$ \Rightarrow $$ 210 = 2Ea
$$ \Rightarrow $$ Ea = 105 kJ/mole
$$ \therefore $$ Activation energy in the presence of catalyst = 105 – 30 = 75 kJ/mol
K2 = A$${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$$
$$ \because $$Rate is same
$$ \therefore $$ Rate constant will also be same
K1 = K2
A$${e^{ - {{{E_a}} \over {R \times 700}}}}$$ = A$${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$$
$$ \Rightarrow $$ $${{{\left( {{E_a} - 30} \right)} \over {R \times 500}}}$$ = $${{{{E_a}} \over {R \times 700}}}$$
$$ \Rightarrow $$ 5Ea = 7Ea – 210
$$ \Rightarrow $$ 210 = 2Ea
$$ \Rightarrow $$ Ea = 105 kJ/mole
$$ \therefore $$ Activation energy in the presence of catalyst = 105 – 30 = 75 kJ/mol
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