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):
Solution
K<sub>1</sub> = A${e^{ - {{{E_a}} \over {R \times 700}}}}$
<br><br>K<sub>2</sub> = A${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$
<br><br>$\because$Rate is same
<br><br>$\therefore$ Rate constant will also be same
<br><br>K<sub>1</sub> = K<sub>2</sub>
<br><br>A${e^{ - {{{E_a}} \over {R \times 700}}}}$ = A${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$
<br><br>$\Rightarrow$ ${{{\left( {{E_a} - 30} \right)} \over {R \times 500}}}$ = ${{{{E_a}} \over {R \times 700}}}$
<br><br>$\Rightarrow$ 5E<sub>a</sub> = 7E<sub>a</sub> – 210
<br><br>$\Rightarrow$ 210 = 2E<sub>a</sub>
<br><br>$\Rightarrow$ E<sub>a</sub> = 105 kJ/mole
<br><br>$\therefore$ Activation energy in the presence of
catalyst = 105 – 30 = 75 kJ/mol
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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