Final answer:
The reaction where X has less energy than Y is favored to move to the right, as it is an exothermic process with the products having lower energy than the reactants. This direction is enthalpically driven and is energetically favorable, especially at low temperatures where entropy's role is minimized.
Step-by-step explanation:
If X has less energy than Y, the favoured direction of the reaction is to the right, where X is converted into Y. This is because, during a reaction that proceeds to the right, there is a release of heat energy (q), making it an exothermic process. The change in enthalpy (ΔH) for this reaction is negative, indicating that energy is released and the products (Y) have lower energy than the reactants (X). This type of reaction is energetically favorable, as it is driven by the enthalpy change; it goes 'downhill' in terms of energy. Conversely, to drive the reaction to the left, energy must be input, as it would be going 'uphill', making it endothermic and less favorable.
Furthermore, an entropically unfavorable change is witnessed, with the entropy decreasing as the system goes from a higher-energy state to a lower-energy state. At low temperatures, the entropic contribution to the free energy change is minimized, which can further favor the reaction proceeding to the right. As a reaction goes towards the products, the system loses energy, making the exothermic direction more spontaneous and thermodynamically favorable.