(D) The Arrhenius equation is given by: $\log K = \log A - \frac{E_a}{2.303 RT}$ Given: $A = 1.45 \times 10^{11} s^{-1}$,$E_a = 35 \times 10^3 \, cal\

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(D) The Arrhenius equation is given by: $\log K = \log A - \frac{E_a}{2.303 RT}$ Given: $A = 1.45 \times 10^{11} s^{-1}$,$E_a = 35 \times 10^3 \, cal\, mol^{-1}$,$T = 300 + 273 = 573 \, K$,$R = 1.987 \approx 2 \, cal\, K^{-1} mol^{-1}$ Substituting the values: $\log K = \log (1.45 \times 10^{11}) - \frac{35000}{2.303 \times 2 \times 573}$ $\log K = 11.161 - \frac{35000}{2640.038} \approx 11.161 - 13.257 = -2.096$ $K = \text{antilog}(-2.096) = 7.94 \times 10^{-3} s^{-1}$

Class 12 Chemistry Chemical Kinetics Collision theory, Energy of activation and Arrhenius equation

Medium

The rate constant for a first order reaction at $300\, ^\circ C$ for which $E_a$ is $35\, kcal\, mol^{-1}$ and frequency constant $(A)$ is $1.45 \times 10^{11} s^{-1}$ will be:

A
$10 \times 10^{-2} s^{-1}$
B
$5.37 \times 10^{10} s^{-1}$
C
$5 \times 10^{-4} s^{-1}$
D
$7.94 \times 10^{-3} s^{-1}$

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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The rate constant for a first order reaction at $300\, ^\circ C$ for which $E_a$ is $35\, kcal\, mol^{-1}$ and frequency constant $(A)$ is $1.45 \times 10^{11} s^{-1}$ will be:

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$A$ reaction rate constant is given by $k = 1.2 \times 10^{14} e^{-(25000/RT)} \, s^{-1}$. It means

In a reaction,the rate constant $K_1$ is twice the rate constant $K_2$ of another reaction. Find the relationship between the activation energies $E_1$ and $E_2$ of the two reactions at the same temperature.

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