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:

We have the gaseous equilibrium $$A(g) \rightleftharpoons B(g)$$ in which the enthalpy change is given as $$\Delta H = -40\; \text{kJ mol}^{-1}$$. A negative value tells us that the forward reaction is exothermic, that is, the products lie lower on the energy scale than the reactants.

For any elementary reaction we picture a potential-energy diagram with a single transition state. Let the reactants be our zero level of energy. Then:

• The energy of the transition state above the reactants is the forward activation energy, so its height is $$E_f$$.

• Because the products are $$\Delta H$$ lower than the reactants, their energy is $$\Delta H$$ (here $$-40\; \text{kJ mol}^{-1}$$).

• The reverse activation energy $$E_b$$ is the difference between the transition-state energy and the energy of the products.

Hence, starting from these definitions, we write

$$E_b = (\text{transition-state energy}) - (\text{product energy}) = E_f - \Delta H.$$

Re-arranging this very useful relation we also get

$$\Delta H = E_f - E_b.$$

Next, the question states that the ratio of the forward to the reverse activation energies is

$$\frac{E_f}{E_b} = \frac{2}{3}.$$

Using the relation $$E_b = E_f - \Delta H$$ and substituting the numerical value $$\Delta H = -40\; \text{kJ mol}^{-1}$$, we obtain

$$E_b = E_f - (-40) = E_f + 40.$$

Now we substitute this expression for $$E_b$$ into the given ratio:

$$\frac{E_f}{E_b} = \frac{E_f}{E_f + 40} = \frac{2}{3}.$$

Cross-multiplying gives every algebraic step clearly:

$$3\,E_f = 2\,(E_f + 40).$$

$$3E_f = 2E_f + 80.$$

$$3E_f - 2E_f = 80.$$

$$E_f = 80\; \text{kJ mol}^{-1}.$$

Once $$E_f$$ is known we return to $$E_b = E_f + 40$$:

$$E_b = 80 + 40 = 120\; \text{kJ mol}^{-1}.$$

Thus the forward and reverse activation energies are

$$E_f = 80\; \text{kJ mol}^{-1}, \qquad E_b = 120\; \text{kJ mol}^{-1}.$$

Comparing with the options provided, these values correspond exactly to Option C.

Hence, the correct answer is Option C.