Problem 77 - Chemistry: Principles and Reactions

Question

At \(1200 \mathrm{~K}\), an equilibrium mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) gases contains 98.31 mol percent CO and some solid carbon. The total pressure of the mixture is \(1.00 \mathrm{~atm}\). For the system $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) $$ calculate (a) \(P_{\mathrm{Co}}\) and \(P_{\mathrm{CO}_{2}}\) (b) \(K\) (c) \(\Delta G^{\circ}\) at \(1200 \mathrm{~K}\)

Step-by-Step Solution

Answer

The partial pressures of CO and CO2 are 0.9831 atm and 0.0169 atm, respectively. The equilibrium constant, K, is 56.93, and the change in Gibbs free energy at 1200 K is approximately -70.0 kJ/mol.

Step 1: Calculate the partial pressures of CO and CO2

Given that the CO mixture contains 98.31 mol % CO, that means CO2 is the remaining 1.69 mol % of the gas mixture. The total pressure is 1.00 atm, so to determine the partial pressures we can multiply the mol percentages by the total pressure: \(P_{\mathrm{CO}} = 0.9831 \times 1.00 \mathrm{~atm} = 0.9831 \mathrm{~atm}\) \(P_{\mathrm{CO_2}} = 0.0169 \times 1.00 \mathrm{~atm} = 0.0169 \mathrm{~atm}\)

Step 2: Write the equilibrium expression for the reaction

The equilibrium expression for the given reaction is: \(K = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO_2}} \times P_{\mathrm{C}}} = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO_2}}}\) Since carbon is in a solid state, its pressure can be considered as 1, so we have: \(K = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO_2}}}\)

Step 3: Calculate the equilibrium constant (\(K\))

Now we will use the partial pressures from step 1 to calculate the equilibrium constant: \(K = \frac{(0.9831)^2}{0.0169} = 56.93\)

Step 4: Calculate the change in Gibbs free energy (\(\Delta G^{\circ}\)) at 1200 K

The relationship between \(\Delta G^{\circ}\) and \(K\) is given by the equation: \(\Delta G^{\circ} = -R \times T \times \ln{K}\) Where \(R\) is the gas constant (8.314 J/mol K) and \(T\) is the temperature (1200 K). Now we will substitute the values into the equation: \(\Delta G^{\circ} = -8.314 \times 1200 \times \ln{56.93} = -70046.72 \mathrm{~J/mol}\) So, the change in Gibbs free energy at 1200 K is approximately -70.0 kJ/mol.

Key Concepts

Partial Pressure

When looking at gas mixtures, the term "partial pressure" refers to the pressure exerted by a single type of gas on its own, even though it's in a mixture with other gases. It thinks about what the pressure would be if that gas were all by itself in the container.
To find the partial pressures of \(CO\) and \(CO_2\) in our equilibrium mixture, you have to know their proportion in the mix. In this case, carbon monoxide (CO) makes up 98.31% while carbon dioxide (CO_2) makes up 1.69% of the mixture. Given the total pressure is 1 atm, you can calculate the partial pressures:

• \(P_{CO} = 0.9831 \, \text{atm}\)
• \(P_{CO_2} = 0.0169 \, \text{atm}\)

This essentially gives you a breakdown of how much pressure each gas contributes. It's crucial because these individual pressures play a key role in chemical equilibrium, influencing the reaction's balance.
Partial pressures help you comprehend and calculate how gases behave, especially when they interact with solids, like carbon here.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a valuable indicator in chemistry. It tells us about the ratio of product concentrations to reactant concentrations when the system reaches equilibrium.
In the chemical reaction \(\mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g)\), the equilibrium constant focuses on the relationship between \(CO_2\) and \(CO\). Since carbon (C) is a solid, it doesn’t affect the equilibrium expression and is thus set to 1.

• Formula for \(K\): \(K = \frac{(P_{CO})^2}{P_{CO_2}}\)
• Plugging in values, \(K = \frac{(0.9831)^2}{0.0169}\)
• Result: \(K = 56.93\)

This calculated value of \(K\) reveals that, at equilibrium, there is more \(CO\) being produced than \(CO_2\) consumed, confirming the forward reaction's dominance under the given conditions. Understanding \(K\) lets chemists predict the direction the reaction tends toward, and how the balance may shift if conditions change.

Gibbs Free Energy

Gibbs Free Energy, denoted as \(\Delta G^{\circ}\), is a concept used to predict whether a chemical process is spontaneous. Essentially, it tells us if a reaction takes place on its own. For the given reaction at equilibrium, \(\Delta G^{\circ}\) ties to the equilibrium constant \(K\) through the formula:

• \(\Delta G^{\circ} = -R \times T \times \ln{K}\)
• With \(R = 8.314 \, \text{J/mol K}\) and \(T = 1200 \, \text{K}\), and substituting \(K\) as 56.93, calculate:
• \(\Delta G^{\circ} = -8.314 \times 1200 \times \ln{56.93}\)
• Result: Approx. \(-70046.72 \, \text{J/mol}\), or \(-70.0 \, \text{kJ/mol}\)

The negative \(\Delta G^{\circ}\) indicates that the reaction is spontaneous under these conditions; \(CO_2\) and carbon naturally convert into \(CO\). Gibbs Free Energy gives deeper insight into energy changes, making it invaluable for predicting reaction courses regardless of external interventions.