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COMEDK 2025 Afternoon Solved Paper

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Question : 99 of 180

Marks: +1, -0

(i) and (ii) are 2 chemical reactions carried out at TK.

\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{W}

\mathrm{k}_1

as rate constant.

\mathrm{X} \rightarrow \mathrm{Z}+\mathrm{W}

k_2

as rate constant.

The Activation energy for reaction (ii) is 3 times that of reaction (i). What is the expression for

k_1

?

$k_1=-2 k_2 e^{2 E a_1 / R T}$
$k_1=k_2 e^{2 E a_1 / R T}$
$k_1=2 k_2 e^{E a_1 / R T}$
$k_1=k_2 / 2 e^{-E a_1 / R T}$

Solution:

Concept:

The Arrhenius equation relates rate constant to activation energy and temperature.

Explanation:

For reaction (i), the rate constant is

k_1 = A e^{-E_{a1} / (RT)}

.

For reaction (ii), the activation energy is

E_{a2} = 3E_{a1}

, so

k_2 = A e^{-3E_{a1} / (RT)}

.

The pre-exponential factor

A

is the same for both reactions (same reactant

\mathrm{X}

).

Take the ratio:

\frac{k_1}{k_2} = \frac{A e^{-E_{a1}/(RT)}}{A e^{-3E_{a1}/(RT)}} = e^{(3E_{a1} - E_{a1})/(RT)} = e^{2E_{a1}/(RT)}

.

Thus,

k_1 = k_2 \, e^{2E_{a1}/(RT)}

.

Answer:

k_1 = k_2 \, e^{2E_{a1}/(RT)}

which matches option B.

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