For the reaction $$ \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ \(K=2.2 \times 10^{11}\) at \(473 \mathrm{~K}\) and \(4.6 \times 10^{8}\) at \(533 \mathrm{~K}\), Calculate \(\Delta G^{\circ}\) at both temperatures.

The concept of chemical equilibrium is pivotal in understanding how chemical reactions occur and how they can be quantified. Equilibrium is attained when the rates of the forward and reverse reactions in a closed system become equal, resulting in no net change in the concentration of reactants and products over time. This state can be visualized on a molecular level where, despite ongoing exchange, the amounts of each substance remain constant.

It is essential to note that equilibrium does not mean the reactants and products are present in equal amounts. Instead, the specific ratio depends on several factors including the nature of the reactants, the temperature, and the presence of catalysts. In the context of the given exercise, the reaction between carbon monoxide and hydrogen gas to form methane and water vapor reached chemical equilibrium at the specified conditions.

The equilibrium constant, represented as K, quantifies the ratio of product concentrations to reactant concentrations at equilibrium for a given reaction. This value is dimensionless and considers the activities or effective concentrations of the species involved. In the provided exercise, we have two different values of K for the same reaction at different temperatures, which illustrates the temperature-dependence of the equilibrium constant.

Mathematically, for a general reaction \( aA + bB \leftrightarrows cC + dD \), the equilibrium constant is given by \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \), where the square brackets denote the concentration of each substance. The values of K given in the exercise tell us how the position of equilibrium changes with temperature for the reaction in question. A higher K value means a greater concentration of products at equilibrium.

The reaction temperature is a crucial factor influencing the position of the chemical equilibrium and the value of the equilibrium constant. As observed in the exercise, the value of K changes with the temperature of the reaction. According to Le Chatelier's principle, an increase in temperature for an exothermic reaction will shift the equilibrium to favor the reactants, resulting in a lower equilibrium constant. Conversely, for an endothermic reaction, an increase in temperature will shift the equilibrium towards the products, hence a higher K.

In applying this to the Gibbs free energy equation \( \Delta G^{\circ} = -RT \ln K \), which relates the standard Gibbs free energy change with the temperature and equilibrium constant, it's clear how temperature affects the spontaneity of a reaction. The different values of \(\Delta G^{\circ}\) calculated at 473 K and 533 K reflect the influence of reaction temperature on the free energy change and thereby the extent to which a reaction is favorable under different conditions.