Problem 22 - Chemistry and Chemical Reactivity

Question

The equilibrium constant \(K\) for the reaction $$ \mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) $$ is \(6.66 \times 10^{-12}\) at \(1000 \mathrm{K}\). Calculate \(K\) for the reaction $$ 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}_{2}(\mathrm{g}) $$

Step-by-Step Solution

Answer

The equilibrium constant \( K \) is \( 2.25 \times 10^{23} \).

Step 1: Understand the given reactions

The problem provides two related reactions. The first reaction is \( \mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}(g) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \) with an equilibrium constant \( K_1 = 6.66 \times 10^{-12} \). You need to find the equilibrium constant for the reaction \( 2 \mathrm{CO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}_{2}(\mathrm{g}) \).

Step 2: Write the reverse reaction of the given reaction

The reverse reaction of the first one is \( \mathrm{CO}(g) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}_{2}(\mathrm{g}) \). The equilibrium constant \( K_2 \) for the reverse reaction is the reciprocal of \( K_1 \). Calculate \( K_2 = \frac{1}{K_1} = \frac{1}{6.66 \times 10^{-12}} \).

Step 3: Calculate K for the target reaction

The reaction of interest is \( 2 \mathrm{CO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}_{2}(\mathrm{g}) \). Notice this reaction is double the reverse of the original reaction. Therefore, the equilibrium constant \( K_3 \) for this reaction is \( (K_2)^2 \). Calculate \( K_3 = \left(\frac{1}{6.66 \times 10^{-12}}\right)^2 \).

Step 4: Perform the final calculation

Calculate \( K_3 = \left(6.66 \times 10^{-12}\right)^{-2} = 2.25 \times 10^{23} \). This is the equilibrium constant for the reaction \( 2 \mathrm{CO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}_{2}(\mathrm{g}) \).

Key Concepts

Chemical Equilibrium

In a chemical reaction, equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentration of reactants and products remains constant over time. Equilibrium doesn't necessarily mean that the reactants and products are in equal amounts, but rather that their ratios remain constant. This stable state can be quantified using the equilibrium constant, denoted as \( K \).
The equilibrium constant is a numerical value that characterizes the relationship between the concentrations of reactants and products at equilibrium for a particular chemical reaction. It is derived from the balanced chemical equation and varies with temperature. For instance, in the exercise, the equilibrium constant \( K_1 \) for the reaction \( \mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}(g) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \) helps determine the stability and composition of the system at equilibrium at a specified temperature, here 1000 K.
Key things to remember:

• Equilibrium does not mean equal concentrations but instead a stable ratio of concentrations.
• The equilibrium constant is dependent on temperature.
• The larger the value of \( K \), the more products are favored at equilibrium.

Reaction Manipulation

When manipulating reactions, often the equilibrium constant of one reaction is used to find the equilibrium constant for a related reaction. This involves direct mathematical manipulation of \( K \).
For example, reversing a chemical reaction involves taking the reciprocal of the known equilibrium constant. If you need to double or halve reaction coefficients, the new \( K \) value is raised to the power corresponding to the factor by which you've multiplied the coefficients. In our exercise, the equilibrium constant \( K_2 \) for the reverse reaction \( \mathrm{CO}(g) + \frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}_{2}(\mathrm{g}) \) is \( \frac{1}{K_1} \).
Manipulating the equation to form the required reaction \( 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\rightleftarrows 2 \mathrm{CO}_{2}(\mathrm{g}) \), we multiply the reverse reaction by a factor of 2. This means squaring \( K_2 \) results in \( K_3 = (K_2)^2 \).
Key aspects include:

• Reversing a reaction involves taking the reciprocal of \( K \).
• If reaction coefficients are multiplied by a factor \( n \), raise \( K \) to the power of \( n \).
• These transformations allow calculation of \( K \) for different but related reactions.

Thermodynamics

Thermodynamics governs the principles of heat, energy, and work in chemical systems. Understanding these principles is crucial in analyzing chemical equilibrium.
The equilibrium constant ties directly into the thermodynamic concept of Gibbs Free Energy, which determines the spontaneity of a reaction. If the \( \Delta G \) of a reaction is negative, the process can proceed spontaneously. The relationship between \( \Delta G \) and the equilibrium constant \( K \) is given by:
\[ \Delta G^\circ = -RT \ln K \]
Where \( \Delta G^\circ \) is the standard Gibbs free energy change, \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( K \) is the equilibrium constant.
High values of \( K \) indicate a process that strongly favors the formation of products at equilibrium, suggesting a strong tendency towards spontaneity over the course of the reaction. In the exercise provided, the computation of \( K_3 = 2.25 \times 10^{23} \) reflects the significant favoring of products, indicating an exergonic process under the given conditions.
Important thermodynamic perspectives include:

• Equilibrium and Gibbs Free Energy are intimately connected.
• A high equilibrium constant suggests a favorable formation of products.
• Gibbs Free Energy and \( K \) allow prediction of reaction direction and spontaneity.