Question

For the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\), the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) are 4 and \(8 \mathrm{~atm}\) respectively. The value of \(K_{p}\) for this reaction is: (a) \(14 \mathrm{~atm}\) (b) \(16 \mathrm{~atm}\) (c) \(18 \mathrm{~atm}\) (d) \(12 \mathrm{~atm}\)

Short answer

The value of \(K_p\) is \(16 \mathrm{~atm}\), so the answer is (b).

Step-by-step solution

Identify the Expression for the Equilibrium Constant

The equilibrium constant for the reaction involving gases is given by the expression for the partial pressures. For the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\), the expression for the equilibrium constant \(K_p\) is:\[ K_p = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO}_2}} \]This is based on the law of mass action where only gases and aqueous solutions are considered.

Substitute the Partial Pressures into the Expression

The partial pressures given are \(P_{\mathrm{CO}_2} = 4 \mathrm{~atm}\) and \(P_{\mathrm{CO}} = 8 \mathrm{~atm}\). Substitute these values into the equilibrium expression:\[ K_p = \frac{(8)^2}{4} \]

Calculate the Value of the Equilibrium Constant

Evaluate the expression obtained in Step 2:\[ K_p = \frac{64}{4} = 16 \]Thus, the equilibrium constant \(K_p\) for the reaction is \(16 \mathrm{~atm}\).

Key concepts

Equilibrium Calculations

Equilibrium calculations are pivotal when analyzing chemical reactions at equilibrium. These calculations help us determine the equilibrium constant, which reflects the extent of a reaction. To begin with, knowing the balanced chemical equation is crucial. It guides us in forming the expression that represents equilibrium conditions. The equilibrium constant, represented as \( K_p \) for gas-phase reactions, is derived from partial pressures of the chemicals involved in the reaction. For instance, in the reaction \( \mathrm{C(s)} + \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \), each participant's partial pressure affects the calculation. A step-by-step approach is recommended: first, write the equilibrium expression, then plug in known values, and finally compute the result. This calculation helps in predicting the position of equilibrium and determining whether the forward or reverse reaction is favored.

Partial Pressures

The concept of partial pressures is pivotal in the context of gaseous equilibria. They represent the pressure exerted by a single gas in a mixture, assuming that it alone occupies the entire volume. Understanding partial pressures is essential because, in equilibrium calculations involving gases, we need these values to calculate the equilibrium constant. For instance, partial pressures are denoted as \( P_{\text{gas}} \). For the reaction \( \mathrm{C(s)} + \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \), we consider \( P_{\mathrm{CO}_2} = 4 \mathrm{~atm} \) and \( P_{\mathrm{CO}} = 8 \mathrm{~atm} \). These values are then plugged into the equilibrium expression. To understand partial pressures better, remember: add up all partial pressures to get the total pressure in a system; each gas's contribution to the total pressure is its partial pressure. It's a foundational concept in predicting how gases will behave at equilibrium.

Law of Mass Action

The law of mass action is a fundamental principle in chemistry regarding the behavior of equilibrium reactions. It postulates that the rate of a reaction is directly proportional to the product of the concentrations of the reactants, raised to the power of their stoichiometric coefficients. This forms the basis for the expression of the equilibrium constant. For gas-phase reactions involving partial pressures, the law of mass action transforms these concentrations to pressures. For the reaction \( \mathrm{C(s)} + \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) \), the equilibrium expression is given by \[ K_p = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO}_2}} \]. The law dictates how only the gaseous and aqueous components' concentrations or pressures contribute to the expression, leaving out solids and liquids. Understanding and applying the law of mass action allows us to predict and quantitatively describe chemical equilibria, crucial for chemical engineering, biological processes, and environmental science.