(B) The rate constant $k$ is calculated using the Arrhenius equation: $k = A e^{-E_a / RT}$.Given: $A = 4 \times 10^{10}$,$E_a = 54.7 \, kJ/mol = 5470

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(B) The rate constant $k$ is calculated using the Arrhenius equation: $k = A e^{-E_a / RT}$.Given: $A = 4 \times 10^{10}$,$E_a = 54.7 \, kJ/mol = 54700 \, J/mol$,$T = 527 + 273 = 800 \, K$,$R = 8.314 \, J \cdot K^{-1} \cdot mol^{-1}$.Substituting the values:$k = 4 \times 10^{10} \times e^{-54700 / (8.314 \times 800)}$$k = 4 \times 10^{10} \times e^{-8.224}$$k = 4 \times 10^{10} \times 2.68 \times 10^{-4}$$k \approx 10.7 \times 10^{6} \, s^{-1}$.

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

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At $527 \, ^{\circ}C$ temperature,the activation energy is $54.7 \, kJ/mol$. The value of the Arrhenius factor is $4 \times 10^{10}$. The rate constant will be:

AIIMS 2019 All AIIMS

A
$12.28 \times 10^{11}$
B
$10.7 \times 10^{6}$
C
$12.28 \times 10^{17}$
D
$14.58 \times 10^{-13}$

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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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At $527 \, ^{\circ}C$ temperature,the activation energy is $54.7 \, kJ/mol$. The value of the Arrhenius factor is $4 \times 10^{10}$. The rate constant will be:

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