(A) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{T_2 - T_1}{T_1 T_2})$.Given that the rate doubles,$\frac{k_2}{k_1} = 2$

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(A) Using the Arrhenius equation: $\ln(\frac{k_2}{k_1}) = \frac{E_a}{R} (\frac{T_2 - T_1}{T_1 T_2})$.Given that the rate doubles,$\frac{k_2}{k_1} = 2$,$T_1 = 298 \text{ K}$,and $T_2 = 308 \text{ K}$.Substituting the values: $\ln(2) = \frac{E_a}{8.314} (\frac{308 - 298}{298 \times 308})$.$0.693 = \frac{E_a}{8.314} (\frac{10}{91784})$.$E_a = \frac{0.693 \times 8.314 \times 91784}{10}$.$E_a \approx 52897 \text{ J mol}^{-1} = 52.897 \text{ kJ mol}^{-1}$.

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

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The rate of the chemical reaction doubles for an increase of $10 \text{ K}$ in absolute temperature from $298 \text{ K}$. What will be the activation energy?

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A
$52.897 \text{ kJ mol}^{-1}$
B
$51.897 \text{ kJ mol}^{-1}$
C
$42.897 \text{ kJ mol}^{-1}$
D
$41.897 \text{ kJ mol}^{-1}$

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

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$A$ plot of $ \frac{1}{T} $ vs. $ \ln k $ for a reaction gives the slope $ -1 \times 10^{4} \ K $. The energy of activation for the reaction is (Given $ R = 8.314 \ J \ K^{-1} \ mol^{-1} $)

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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 $(A)$?

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When the temperature changes from $20\,^oC$ to $50\,^oC$,the rate of reaction becomes three times. The activation energy $(E_a)$ for the reaction is $.... \, kJ \, mol^{-1}$ $(R = 8.314 \, J \, K^{-1} \, mol^{-1})$.

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In the Arrhenius plot of $\log_{10} k$ versus $1 / T$,find the value of the intercept on the $y$-axis.

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The rate of the chemical reaction doubles for an increase of $10 \text{ K}$ in absolute temperature from $298 \text{ K}$. What will be the activation energy?

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

$A$ plot of $ \frac{1}{T} $ vs. $ \ln k $ for a reaction gives the slope $ -1 \times 10^{4} \ K $. The energy of activation for the reaction is (Given $ R = 8.314 \ J \ K^{-1} \ 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 $(A)$?

When the temperature changes from $20\,^oC$ to $50\,^oC$,the rate of reaction becomes three times. The activation energy $(E_a)$ for the reaction is $.... \, kJ \, mol^{-1}$ $(R = 8.314 \, J \, K^{-1} \, mol^{-1})$.

In the Arrhenius plot of $\log_{10} k$ versus $1 / T$,find the value of the intercept on the $y$-axis.

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