For the equilibrium, \(\mathrm{A(g) \rightleftharpoons B(g)}\), \(\Delta H\) 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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- \(E_f = 60\) kJ/mol ; \(E_b = 100\) kJ/mol
- \(E_f = 30\) kJ/mol ; \(E_b = 70\) kJ/mol
- \(E_f = 80\) kJ/mol ; \(E_b = 120\) kJ/mol
- \(E_f = 70\) kJ/mol ; \(E_b = 30\) kJ/mol
The Correct Option is C
Solution and Explanation
Given \(\frac{E_f}{E_b} = \frac23\), so \(E_b = 1.5E_f\).
\(E_f - 1.5E_f = -40\) gives \(-0.5E_f = -40\), so \(E_f = 80\) kJ/mol and \(E_b = 120\) kJ/mol.
Answer: (3) \(E_f = 80\) kJ/mol ; \(E_b = 120\) kJ/mol
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