An equilibrium mixture contains 0.60 g solid carbon and the gases carbon dioxide and carbon monoxide at partial pressures of 2.60 atm and 2.89 atm, respectively. Calculate the value of \(K_{\mathrm{p}}\) for the reaction \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\).

Chemical equilibrium is a dynamic state in which the rates of the forward and reverse reactions in a closed system are equal, resulting in no net change in the concentration of reactants and products over time. Although reactions continue to occur, the amounts of each substance remain constant. This balance is not static but rather a continuous adjustment of molecular processes that are invisible to the naked eye.

In the context of our equation \( \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \), equilibrium is reached when the rate at which carbon and carbon dioxide react to form carbon monoxide is equal to the rate at which carbon monoxide decomposes back into carbon and carbon dioxide. When these rates equalize, the partial pressures no longer change, and the system is at equilibrium.

Partial pressure refers to the pressure exerted by a single gas component in a mixture of gases. It's proportional to the mole fraction of the gas in the mixture and the total pressure of the gas mixture. For a gas at equilibrium, partial pressure is crucial for understanding how gases behave in reaction dynamics.

In our case, the equilibrium involves gases carbon dioxide (\( \mathrm{CO}_{2} \)) and carbon monoxide (\( \mathrm{CO} \)). The partial pressures given in the problem statement, 2.60 atm for \( \mathrm{CO}_{2} \) and 2.89 atm for \( \mathrm{CO} \), directly influence the equilibrium constant calculation. It's also important to note that solids and liquids do not exert partial pressures, which is why the carbon solid is not included in the partial pressure calculations.

The reaction quotient (Q) is a measure that tells us how far a reaction has proceeded towards equilibrium at any given moment. It has the same form as the equilibrium constant expression but unlike \( K \), which only applies at equilibrium, Q can be calculated using initial concentrations or pressures of reactants and products.

Q is compared to the equilibrium constant \( K \) to predict the direction in which the reaction will proceed to reach equilibrium. If \( Q < K \), the forward reaction is favored to form more products. If \( Q > K \), the reverse reaction is favored, reducing the amount of products. When \( Q = K \), the reaction is at equilibrium. In this exercise, we are directly calculating the equilibrium constant \( K_\mathrm{p} \), but understanding Q is essential for grasping the nature of the chemical equilibria.

Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. It is based on the conservation of mass and the law of definite proportions. Stoichiometric calculations involve using balanced chemical equations to determine the amounts of substances consumed and produced.

For the equilibrium reaction in our exercise, stoichiometry dictates that one mole of carbon reacts with one mole of carbon dioxide to produce two moles of carbon monoxide, as shown by the balanced chemical equation. This stoichiometric ratio determines how the partial pressures of the reactants and products are related in the equilibrium expression. Notably, the coefficients from the balanced equation become the powers for each component's partial pressure within the equilibrium constant expression.