Question

For the equilibrium, $A ( g ) \rightleftharpoons B ( g ), \Delta H$ is $-40\, kJ\, / mol$. If the ratio of the activation energies of the forward $\left(E_{ f }\right)$ and reverse $\left(E_{ b }\right)$ reactions is $\frac{2}{3}$ then:

• A. $E_{f} = 60\, kJ/mol$ ; $E_{b} = 100 \,kJ /mol$
• B. $E_{f} = 30 \,kJ/mol$ ; $E_{b} = 70 \,kJ /mol$
• C. $E_{f} = 80 \,kJ/mol$ ; $E_{b} = 120 \,kJ /mol$
• D. $E_{f} = 70\,kJ/mol$ ; $E_{b} = 30 \,kJ/mol$

Answer 1

Answer: (C)

$A ( g ) \rightleftharpoons B ( g ) \quad \Delta H =-40 kJ$

Since, $\frac{E_{ f }}{E_{ b }}=\frac{2}{3}$, therefore, $E_{ f }=\frac{2 x}{5}$ and $E_{ b }=\frac{3 x}{5}$

$E_{ b }-E_{ f }=+40$

$\frac{3 x}{5}-\frac{2 x}{5}=+40 \Rightarrow \frac{x}{5}=40 \Rightarrow x=200$

Therefore,

$E_{ b }=\frac{3 x}{5}=\frac{3 \times 200}{5}=120\, kJ\, mol ^{-1}$

$E_{ f }=\frac{2 x}{5}=\frac{2 \times 200}{5}=80 \,kJ\, mol ^{-1}$