(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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Class 12

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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What is the relationship between the temperature $T$ and the activation energy $E_a$ in the Arrhenius equation?

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The reaction rate at a given temperature becomes slower,then

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For the reaction,following data is given,
$A \rightarrow B$; $K_1 = 10^{15} \exp \left( \frac{-2000}{T} \right)$
$C \rightarrow D$; $K_2 = 10^{14} \exp \left( \frac{-1000}{T} \right)$
The temperature at which $K_1 = K_2$ is ........... $K$ $(exp. = e)$

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Explain the rate of reaction with energy of activation and temperature with the help of the Arrhenius equation and state its importance.

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The time required for $10 \%$ completion of a first order reaction at $298 \,K$ is equal to that required for its $25 \%$ completion at $308 \,K$. If the value of $A$ is $4 \times 10^{10} \,s^{-1}$,calculate $k$ at $318 \,K$ and $E_a$.

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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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What is the relationship between the temperature $T$ and the activation energy $E_a$ in the Arrhenius equation?

The reaction rate at a given temperature becomes slower,then

For the reaction,following data is given,$A \rightarrow B$; $K_1 = 10^{15} \exp \left( \frac{-2000}{T} \right)$$C \rightarrow D$; $K_2 = 10^{14} \exp \left( \frac{-1000}{T} \right)$The temperature at which $K_1 = K_2$ is ........... $K$ $(exp. = e)$

Explain the rate of reaction with energy of activation and temperature with the help of the Arrhenius equation and state its importance.

The time required for $10 \%$ completion of a first order reaction at $298 \,K$ is equal to that required for its $25 \%$ completion at $308 \,K$. If the value of $A$ is $4 \times 10^{10} \,s^{-1}$,calculate $k$ at $318 \,K$ and $E_a$.

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For the reaction,following data is given,
$A \rightarrow B$; $K_1 = 10^{15} \exp \left( \frac{-2000}{T} \right)$
$C \rightarrow D$; $K_2 = 10^{14} \exp \left( \frac{-1000}{T} \right)$
The temperature at which $K_1 = K_2$ is ........... $K$ $(exp. = e)$