(A) Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$ Given: $k_2 = 2k_1$,$T_1 = 300

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(A) Using the Arrhenius equation: $\log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$ Given: $k_2 = 2k_1$,$T_1 = 300 \ K$,$T_2 = 310 \ K$,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$. $\log 2 = \frac{E_a}{2.303 \times 8.314} \left( \frac{310 - 300}{300 \times 310} \right)$ $0.301 = \frac{E_a}{19.147} \left( \frac{10}{93000} \right)$ $E_a = \frac{0.301 \times 19.147 \times 93000}{10} \approx 53598 \ J \ mol^{-1} \approx 53.6 \ kJ \ mol^{-1}$.

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

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The rate of a reaction doubles,when the temperature is changed from $300 \ K$ to $310 \ K$. Activation energy of the reaction is....... $(R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 2=0.301)$

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A
$53.6 \ kJ \ mol^{-1}$
B
$48.6 \ kJ \ mol^{-1}$
C
$58.5 \ kJ \ mol^{-1}$
D
$60.5 \ kJ \ mol^{-1}$

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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For the gaseous reactions,calculate the approximate value of temperature at which $k_1 = k_2$. $[\ln\, 10 = 2.3]$.
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.314\, T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.314\, T}$

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In respect of the equation $k = A e^{-E_a/RT}$ in chemical kinetics,which one of the following statements is correct?

DifficultAIEEE 2003

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Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11} \text{ s}^{-1}) e^{\frac{-28000 \text{ K}}{T}}$. The activation energy of the reaction is . . . . . . $\text{kJ mol}^{-1}$. (Nearest Integer) Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

DifficultJEE MAIN 2026

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The relation between the fraction of molecules $\left( \frac{N_E}{N_T} \right)$ and temperature is explained by the Boltzmann and Maxwell distribution graph.

Difficult

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The rate constant for a first order reaction is given by the following equation:
$\ln k = 33.24 - \frac{2.0 \times 10^{4} \, K}{T}$
The activation energy for the reaction is given by $... \, kJ \, mol^{-1}$. (In nearest integer)
(Given: $R = 8.3 \, J \, K^{-1} \, mol^{-1}$)

MediumJEE MAIN 2022

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The rate of a reaction doubles,when the temperature is changed from $300 \ K$ to $310 \ K$. Activation energy of the reaction is....... $(R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 2=0.301)$

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For the gaseous reactions,calculate the approximate value of temperature at which $k_1 = k_2$. $[\ln\, 10 = 2.3]$.$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.314\, T}$$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.314\, T}$

In respect of the equation $k = A e^{-E_a/RT}$ in chemical kinetics,which one of the following statements is correct?

Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11} \text{ s}^{-1}) e^{\frac{-28000 \text{ K}}{T}}$. The activation energy of the reaction is . . . . . . $\text{kJ mol}^{-1}$. (Nearest Integer) Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

The relation between the fraction of molecules $\left( \frac{N_E}{N_T} \right)$ and temperature is explained by the Boltzmann and Maxwell distribution graph.

The rate constant for a first order reaction is given by the following equation:$\ln k = 33.24 - \frac{2.0 \times 10^{4} \, K}{T}$The activation energy for the reaction is given by $... \, kJ \, mol^{-1}$. (In nearest integer)(Given: $R = 8.3 \, J \, K^{-1} \, mol^{-1}$)

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For the gaseous reactions,calculate the approximate value of temperature at which $k_1 = k_2$. $[\ln\, 10 = 2.3]$.
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.314\, T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.314\, T}$

The rate constant for a first order reaction is given by the following equation:
$\ln k = 33.24 - \frac{2.0 \times 10^{4} \, K}{T}$
The activation energy for the reaction is given by $... \, kJ \, mol^{-1}$. (In nearest integer)
(Given: $R = 8.3 \, J \, K^{-1} \, mol^{-1}$)