(A) For the given reaction,$\Delta H = -44.12 \ kcal$. Since $\Delta H < 0$,the reaction is exothermic.We know the relationship between enthalpy chang

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(A) For the given reaction,$\Delta H = -44.12 \ kcal$. Since $\Delta H We know the relationship between enthalpy change and activation energies is given by $\Delta H = E_f - E_b$,where $E_f$ is the activation energy of the forward reaction $(E_2)$ and $E_b$ is the activation energy of the backward reaction $(E_1)$.Thus,$\Delta H = E_2 - E_1$.Since $\Delta H$ is negative,$E_2 - E_1 E_2$.

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

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The $\Delta H$ value of the reaction $H_2 + Cl_2 \rightleftharpoons 2HCl$ is $-44.12 \ kcal$. If $E_1$ is the activation energy of the backward reaction and $E_2$ is the activation energy of the forward reaction,then for the above reaction:

A
$E_1 > E_2$
B
$E_1 < E_2$
C
$E_1 = E_2$
D
$\Delta H$ is not related to $E_1$ and $E_2$

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Chemical Kinetics

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Collision theory, Energy of activation and Arrhenius equation

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If we plot a graph between $\log \, K$ and $\frac{1}{T}$ by Arrhenius equation,the slope is

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The $\Delta H$ value of the reaction $H_2 + Cl_2 \rightleftharpoons 2HCl$ is $-44.12 \ kcal$. If $E_1$ is the activation energy of the backward reaction and $E_2$ is the activation energy of the forward reaction,then for the above reaction:

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For two reactions,the values of the pre-exponential factor are the same. However,the difference between their activation energy values is $24.9 \, kJ \, mol^{-1}$. The ratio of their rate constants at $300 \, K$ is ....

Which of the following graphs of $\log \, K \rightarrow 1/T$ can be used to calculate the activation energy?

For a reaction where the activation energies of the forward and reverse directions are equal in value,then:

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

If we plot a graph between $\log \, K$ and $\frac{1}{T}$ by Arrhenius equation,the slope is

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