For $$A_2 + B_2 \rightleftharpoons 2AB$$. $$E_a$$ for forward and backward reaction are 180 and 200 kJ mol$$^{-1}$$ respectively. If catalyst lowers $$E_a$$ for both reaction by 100 kJ mol$$^{-1}$$. Which of the following statement is correct?

The energy profile of an elementary reaction relates the activation energies of the forward ($$E_a^{\,f}$$) and backward ($$E_a^{\,b}$$) steps to the enthalpy change $$\Delta H$$ of the reaction through the relation
$$\Delta H = E_a^{\,f} - E_a^{\,b}\; -(1)$$

For the uncatalysed reaction we have
$$E_a^{\,f} = 180\ \text{kJ mol}^{-1},\qquad E_a^{\,b} = 200\ \text{kJ mol}^{-1}$$

Substituting these values in $$(1)$$:
$$\Delta H = 180 - 200 = -20\ \text{kJ mol}^{-1}$$
(The negative sign shows the reaction is exothermic.)

The catalyst lowers the activation energy of both directions by the same amount, 100 kJ mol$$^{-1}$$:
$$E_{a,\ \text{cat}}^{\,f} = 180 - 100 = 80\ \text{kJ mol}^{-1}$$
$$E_{a,\ \text{cat}}^{\,b} = 200 - 100 = 100\ \text{kJ mol}^{-1}$$

Using $$(1)$$ again after catalysis:
$$\Delta H_{\text{cat}} = 80 - 100 = -20\ \text{kJ mol}^{-1}$$
Thus the enthalpy change remains exactly the same.

Now evaluate each statement:

Option A: A catalyst does not change the thermodynamic parameters such as Gibbs free energy change $$\Delta G$$. This is correct.

Option B: A catalyst cannot alter $$\Delta G$$, so it cannot convert a non-spontaneous ($$\Delta G \gt 0$$) reaction into a spontaneous one. This statement is incorrect.

Option C: We found $$\Delta H = -20\ \text{kJ mol}^{-1}$$, not $$+20\ \text{kJ mol}^{-1}$$. This statement is incorrect.

Option D: The enthalpy change is the same before and after catalysis. Therefore this statement is incorrect.

Hence, only Option A is correct.