Consider the following system at equilibrium at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \quad \Delta G^{\circ}=-92.50 \mathrm{~kJ} $$ What will happen to the ratio of partial pressure of \(\mathrm{PCl}_{5}\) to partial pressure of \(\mathrm{PCl}_{3}\) if the temperature is raised? Explain completely.

Based on the given reaction and the Gibbs free energy equation, there are two possible scenarios when the temperature is increased: 1) If the reaction is exothermic (ΔH < 0), the equilibrium constant decreases with increasing temperature, causing the reaction to shift towards the reactants (PCl3 and Cl2). In this case, the ratio of the partial pressures of PCl5 to PCl3 would decrease. 2) If the reaction is endothermic (ΔH > 0), the equilibrium constant increases with increasing temperature, causing the reaction to shift towards the products (PCl5). In this case, the ratio of the partial pressures of PCl5 to PCl3 would increase. Without specific values for ΔH and ΔS, we cannot provide a definitive answer to whether the ratio of partial pressures of PCl5 to PCl3 would increase or decrease with an increase in temperature. However, we've provided an explanation for both scenarios.