(A) Given,$\frac{E_f}{E_b} = \frac{2}{3}$.We know that for a reaction,$\Delta H = E_f - E_b$.Given $\Delta H = -40 \ kJ/mol$,so $-40 = E_f - E_b$,whic

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$E_f = 80 \ kJ/mol; E_b = 120 \ kJ/mol$

(A) Given,$\frac{E_f}{E_b} = \frac{2}{3}$.We know that for a reaction,$\Delta H = E_f - E_b$.Given $\Delta H = -40 \ kJ/mol$,so $-40 = E_f - E_b$,which implies $E_b = E_f + 40$.Substituting this into the ratio: $\frac{E_f}{E_f + 40} = \frac{2}{3}$.Cross-multiplying gives $3E_f = 2(E_f + 40)$.$3E_f = 2E_f + 80$.$E_f = 80 \ kJ/mol$.Then,$E_b = 80 + 40 = 120 \ kJ/mol$.

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

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For the equilibrium $A_{(g)} \rightleftharpoons B_{(g)}$,$\Delta H$ is $-40 \ kJ/mol$. If the ratio of the activation energies of the forward $(E_f)$ and reverse $(E_b)$ reactions is $\frac{2}{3}$,then:

JEE MAIN 2015 All JEE MAIN

A
$E_f = 80 \ kJ/mol; E_b = 120 \ kJ/mol$
B
$E_f = 60 \ kJ/mol; E_b = 100 \ kJ/mol$
C
$E_f = 30 \ kJ/mol; E_b = 70 \ kJ/mol$
D
$E_f = 70 \ kJ/mol; E_b = 30 \ kJ/mol$

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Collision theory, Energy of activation and Arrhenius equation

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For the equilibrium $A_{(g)} \rightleftharpoons B_{(g)}$,$\Delta H$ is $-40 \ kJ/mol$. If the ratio of the activation energies of the forward $(E_f)$ and reverse $(E_b)$ reactions is $\frac{2}{3}$,then:

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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:

Which one of the following is true for an exothermic reaction $A \rightleftharpoons B$, if $E_f$ and $E_b$ are the activation energies of forward and backward reactions respectively?

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