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JEE Main levelOne or more correct optionsDifficulty 2 of 3ck142

Two reactions have the same pre-exponential factor; their activation energies are $E_{1} = 40$ and $E_{2} = 80 kJ mo l^{- 1}$ . Take $R = 8.314 J mo l^{- 1} K^{- 1}$ . Which statements are correct?

A $E_{2}$ decreases as the temperature is raised, which makes reaction 2 catch up
BAt 300 K, $k_{1} / k_{2}$ is greater than $1 0^{6}$
COn raising the temperature from 300 K to 310 K, $k_{2}$ increases by a larger factor than $k_{1}$
DAt very high temperature $k_{1} / k_{2}$ tends to 1

Answer: B, C, D

k_{1} / k_{2} = e^{(E_{2} - E_{1}) / RT} = e^{40000/ (8.314 \times 300)} = 9.22 e + 06 > 1 0^{6}

. The 300 to 310 K factors are 1.68 for reaction 1 and 2.81 for reaction 2, so the second statement is true. With equal

A

,

k_{1} / k_{2} \to 1

as

T \to \infty

.

E_{a}

is taken constant, so the last statement is false.

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