Question

The following equilibrium constants were determined at \(1123 \mathrm{~K}:\) $$ \begin{array}{l} \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g) \quad K_{P}^{\prime}=1.3 \times 10^{14} \\ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows \mathrm{COCl}_{2}(g) \quad K_{P}^{\prime \prime}=6.0 \times 10^{-3} \end{array} $$ Write the equilibrium constant expression \(K_{P}\), and calculate the equilibrium constant at \(1123 \mathrm{~K}\) for $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{COCl}_{2}(g) $$

Short answer

The equilibrium constant \( K_P \) is \( 4.68 \times 10^9 \).

Step-by-step solution

Understand the given reactions

We are provided with two reactions and their equilibrium constants: \( \mathrm{C}(s) + \mathrm{CO}_{2}(g) \leftrightarrows 2 \mathrm{CO}(g) \) with \( K_{P}^{\prime} = 1.3 \times 10^{14} \) and \( \mathrm{CO}(g) + \mathrm{Cl}_{2}(g) \leftrightarrows \mathrm{COCl}_{2}(g) \) with \( K_{P}^{\prime\prime} = 6.0 \times 10^{-3} \). We are tasked with finding the equilibrium constant for the overall reaction.

Identify the overall reaction

The target reaction is \( \mathrm{C}(s) + \mathrm{CO}_{2}(g) + 2 \mathrm{Cl}_{2}(g) \leftrightarrows 2 \mathrm{COCl}_{2}(g) \). This reaction can be obtained by summing the two given reactions.

Add the reactions

Add the two reactions given: \( \mathrm{C}(s) + \mathrm{CO}_{2}(g) \leftrightarrows 2 \mathrm{CO}(g) \) and \( \mathrm{CO}(g) + \mathrm{Cl}_{2}(g) \leftrightarrows \mathrm{COCl}_{2}(g) \). Doubling the second reaction and adding to the first gives \( \mathrm{C}(s) + \mathrm{CO}_{2}(g) + 2 \mathrm{Cl}_{2}(g) \leftrightarrows 2 \mathrm{COCl}_{2}(g) \).

Calculate the overall equilibrium constant \(K_P\)

When reactions are added, their equilibrium constants multiply. Therefore, the overall equilibrium constant \(K_P\) is \( K_{P}= K_{P}^{\prime} \times (K_{P}^{\prime\prime})^2 \). Substitute the known values: \( K_{P} = 1.3 \times 10^{14} \times (6.0 \times 10^{-3})^2 \).

Compute the final result

Calculate \( K_{P}^{\prime\prime}^2 = (6.0 \times 10^{-3})^2 = 36 \times 10^{-6} \). Thus, \( K_P = 1.3 \times 10^{14} \times 36 \times 10^{-6} \). Final calculation gives \( K_P = 4.68 \times 10^{9} \).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products. When a reaction reaches this state, we say the reaction has achieved equilibrium. At equilibrium, the concentrations of reactants and products remain constant over time because their rates are balanced.
If you consider a generic reaction involving substances A and B producing substances C and D, it can be expressed as:

• A + B ⇌ C + D

At equilibrium, the expression for the equilibrium constant in terms of partial pressures (if gases are involved) is represented as:

• \( K_P = \frac{{P_C^c \cdot P_D^d}}{{P_A^a \cdot P_B^b}} \)

Where \( P \) refers to the partial pressures of the substances, and \( a, b, c, \) and \( d \) are the stoichiometric coefficients from the balanced reaction equation.
It is important to note that solids and pure liquids are not included in the equilibrium constant expression. The chemical equilibrium is dynamic because, even though the concentrations of products and reactants remain constant, the reactions continue to occur at the molecular level.

Thermodynamics

Thermodynamics in chemical reactions involves the study of energy changes that occur during a chemical reaction. One key aspect of thermodynamics related to chemical equilibrium is the concept of Gibbs free energy, \( \Delta G \). Gibbs free energy determines the spontaneity of a reaction. When \( \Delta G \) is negative, the process is spontaneous, indicating that the forward reaction is favored.

• If \( \Delta G > 0 \), the reaction is non-spontaneous in the forward direction.
• If \( \Delta G = 0 \), the system is at equilibrium.

The relationship between Gibbs free energy and the equilibrium constant is given by the equation:

• \( \Delta G = -RT \ln K \)

Where \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin. Understanding this relationship helps infer how changes in temperature can shift the position of equilibrium.
Furthermore, Le Chatelier's Principle provides insight into how a system at equilibrium responds to external changes. According to this principle, if a system at equilibrium is subjected to a change in concentration, pressure, or temperature, the system will adjust itself to counteract that change and establish a new equilibrium position. This principle is an essential part of thermodynamics and applies to various chemical processes.

Reaction Equations

Reaction equations are symbolic representations of chemical reactions. They illustrate how reactants are transformed into products. A balanced reaction equation has equal numbers of each atom on both sides of the equation, maintaining the principle of the conservation of mass.
Consider the equation from the problem:

• \( \mathrm{C}(s) + \mathrm{CO}_{2}(g) + 2 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{COCl}_{2}(g) \)

This equation shows carbon reacting with carbon dioxide and chlorine gas to form phosgene.
Balancing is crucial because it ensures matter is neither created nor destroyed, only transformed. Coefficients are used in front of compounds to balance the equation, not by altering subscripts in the compounds themselves.
Reaction equations also aid in deriving equilibrium constant expressions. They provide the stoichiometric coefficients, which serve as exponents in the expression for \( K_P \). Thus, understanding how to write and balance reaction equations is fundamental in calculating equilibrium constants and analyzing chemical reactions.