(B) The Arrhenius equation is given by $k = A \cdot e^{-E_a / RT}$.Taking the natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Given:

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(B) The Arrhenius equation is given by $k = A \cdot e^{-E_a / RT}$.Taking the natural logarithm on both sides: $\ln k = \ln A - \frac{E_a}{RT}$.Given: $E_a = 94.14 \, kJ \, mol^{-1} = 94140 \, J \, mol^{-1}$,$T = 310 \, K$,$k = 10^{-2} \, s^{-1}$,and $R = 8.314 \, J \, K^{-1} \, mol^{-1}$.Substituting the values: $\ln(10^{-2}) = \ln A - \frac{94140}{8.314 \times 310}$.$-4.605 = \ln A - 36.51$.$\ln A = 36.51 - 4.605 = 31.905$.$A = e^{31.905} \approx 7.2 \times 10^{13} \, s^{-1}$.

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

Medium

The activation energy of a reaction is $94.14 \, kJ \, mol^{-1}$ and the rate constant at $310 \, K$ is $10^{-2} \, s^{-1}$. What is the value of the frequency factor $(A)$?

A
$3.6 \times 10^{14} \, s^{-1}$
B
$7.2 \times 10^{13} \, s^{-1}$
C
$5.6 \times 10^{10} \, s^{-1}$
D
$2.7 \times 10^{11} \, s^{-1}$

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Class 12

Chemistry Questions

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

Consider the following graph of the kinetic energy distribution among molecules at temperature $T$. If the temperature is increased,how would the resulting graph differ from the one above?

Medium

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The rate constant of a chemical reaction at a very high temperature will approach:

MediumKVPY 2015

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

Medium

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$A$ reaction takes place in three steps with individual rate constant and activation energy,

Step Rate constant and Activation energy

$Step \ 1$ $k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$

$Step \ 2$ $k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$

$Step \ 3$ $k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$

Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.

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Consider the following plots of rate constant versus $\frac{1}{T}$ for four different reactions. Which of the following orders is correct for the activation energies of these reactions?

DifficultJEE MAIN 2020

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Step	Rate constant and Activation energy
$Step \ 1$	$k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$
$Step \ 2$	$k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$
$Step \ 3$	$k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$

The activation energy of a reaction is $94.14 \, kJ \, mol^{-1}$ and the rate constant at $310 \, K$ is $10^{-2} \, s^{-1}$. What is the value of the frequency factor $(A)$?

Similar Questions

Consider the following graph of the kinetic energy distribution among molecules at temperature $T$. If the temperature is increased,how would the resulting graph differ from the one above?

The rate constant of a chemical reaction at a very high temperature will approach:

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

Consider the following plots of rate constant versus $\frac{1}{T}$ for four different reactions. Which of the following orders is correct for the activation energies of these reactions?

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