The value of the equilibrium constant \(K\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) is \(3.0 \times 10^{4}\) at \(700 \mathrm{~K}\). Determine the value of \(K_{c}\) for the reactions (a) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) and \((\mathrm{b}) \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons\) \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\)

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate, leading to a constant concentration of reactants and products. In the given equation, \(2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})\), it reaches equilibrium when the rate at which \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) react to form \(\mathrm{SO}_{3}\) is equal to the rate at which \(\mathrm{SO}_{3}\) decomposes back into \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\).

At this point, despite continuous reaction taking place on a molecular level, there is no observable change in the concentrations of the species involved. Equilibrium does not mean that the concentrations of reactants and products are equal, but that their ratios stay constant over time. This state is dynamic rather than static, meaning that chemical reactions continue to occur, but with no net effect on the concentrations of reactants and products.

The reaction quotient, \(Q\), serves as a predictor to determine the direction a reaction will proceed to achieve equilibrium. For our reaction, the quotient would be expressed as \(Q = \frac{[SO_{3}]^2}{[SO_{2}]^2[O_{2}]}\), similar to the equilibrium constant expression, \(K_c\), but with the concentrations of the reactants and products as they are at any moment rather than at equilibrium.

By comparing the reaction quotient, \(Q\), with the equilibrium constant, \(K\), we can ascertain the reaction's direction. If \(Q < K\), the reaction will move towards the products to reach equilibrium. Conversely, if \(Q > K\), the reaction will shift towards the reactants. When \(Q = K\), the reaction is at equilibrium, and there is no net change.

Le Chatelier's principle is a guiding rule that predicts how a change in conditions can affect the position of equilibrium in a chemical reaction. This principle states that if an external change is applied to a system at equilibrium, the system will adjust itself to partially counteract the change and establish a new equilibrium.

For instance, if the concentration of reactants increases, the reaction will shift towards the products to decrease the concentration of the added reactants. Similarly, changes in pressure and volume in gas-phase reactions or alterations in temperature can cause the equilibrium position to shift in a way that helps offset the imposed change. Understanding this principle allows chemists to manipulate reaction conditions to favor the formation of a desired product.

In gas-phase reactions, the state of equilibrium can be influenced by pressure and volume changes because the reacting gases can exhibit significant changes in their concentration and properties. The reaction we are discussing involves all gaseous reactants and products, and as such, its equilibrium can be described in terms of partial pressures in addition to concentrations.

The effect of pressure on gas-phase reactions can be understood through Le Chatelier's principle. When the pressure increases, the equilibrium shifts towards the side of the reaction with fewer moles of gas, and vice versa. In this case, increasing the pressure would favor the production of \(\mathrm{SO}_{3}\) as 3 moles of reactants form 2 moles of product, thus decreasing the total number of moles and hence the pressure.

The equilibrium constant expression, denoted as \(K_c\) when concerning concentrations or \(K_p\) when concerning partial pressures, relates the concentrations of the products to the reactants at equilibrium, with each raised to the power of their stoichiometric coefficient in the balanced equation.

The equilibrium constant expression for reaction (a) is \(K_c = \frac{[SO_{3}]^2}{[SO_{2}]^2[O_{2}]}\). When calculating \(K_c\) for the reverse reaction (b), we need to invert the equation, thus taking the reciprocal of the equilibrium constant for the forward reaction. This demonstrates the reciprocal nature of equilibrium constants for reverse reactions. In this instance, if \(K_c\) for reaction (a) is \(3.0 \times 10^{4}\), then \(K_c\) for reaction (b) will be the reciprocal of that value.