Question

(a) What is the standard Gibbs free energy of the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}\) (g) when \(\mathrm{K}=1.00\) ? (b) From data available in Appendix \(2 \mathrm{~A}\), estimate the temperature at which \(K=1.00\). (c) At this temperature, a cylinder is filled with \(\mathrm{CO}(\mathrm{g})\) at \(10.00\) bar, \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(10.00\) bar, \(\mathrm{H}_{2}(\mathrm{~g})\) at \(5.00\) bar, and \(\mathrm{CO}_{2}(\mathrm{~g})\) at \(5.00\) bar. What will be the partial pressure of each of these gases when the system reaches equilibrium? (d) If the cylinder were filled instead with \(\mathrm{CO}(\mathrm{g})\) at \(6.00\) bar, \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(4.00\) bar, \(\mathrm{H}_{2}(\mathrm{~g})\) at \(5.00 \mathrm{bar}\), and \(\mathrm{CO}_{2}(\mathrm{~g})\) at \(10.00\) bar, what would the partial pressures be at equilibrium?

Short answer

a) \(\Delta G^{\circ} = 0\, J/mol\). b) Temperature cannot be determined from the given information. c) Partial pressures remain unchanged when the system reaches equilibrium. d) Equilibrium partial pressures cannot be determined without additional data, but the qualitative shift in equilibrium can be predicted based on the initial reaction quotient.

Step-by-step solution

Determine Standard Gibbs Free Energy Change

Use the equation for the standard Gibbs free energy change given by \[ \Delta G^{\circ} = -RT \ln(K) \]Given that the equilibrium constant \(K=1.00\), we have\[ \Delta G^{\circ} = -RT \ln(1) \]Since \(\ln(1) = 0\), it follows that\[ \Delta G^{\circ} = 0 \cdot J/mol. \]

Estimate the Temperature for K=1

If \(K=1\) at equilibrium, the standard Gibbs free energy change \(\Delta G^{\circ}\) is 0, and this usually occurs at a specific temperature. From thermodynamic tables, find the temperature at which the free energy change for the reaction is zero or use the van't Hoff equation to solve for temperature. However, without specific thermodynamic data provided, we cannot calculate the exact temperature.

Calculate Partial Pressures at Equilibrium for Given Pressures

Since we do not have the specific change in Gibbs energy as a function of temperature, we can only say that at equilibrium when \(K=1\), the reaction quotient \(Q\) equals the equilibrium constant \(K\). Thus no net reaction occurs, and the partial pressures will remain the same (10 bar for CO and H2O, 5 bar for H2 and CO2) when the system reaches equilibrium.

Determine the Reaction Quotient for the Changed Conditions

The reaction quotient \(Q\) is calculated using the initial pressures of the gases in the following reaction:\[ CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \]The reaction quotient \(Q_p\) is given by\[ Q_p = \frac{P_{CO_2} \cdot P_{H_2}}{P_{CO} \cdot P_{H_2O}} \]Using the given pressures for the second scenario yields\[ Q_p = \frac{(10.00 \text{ bar}) \cdot (5.00 \text{ bar})}{(6.00 \text{ bar}) \cdot (4.00 \text{ bar})} \]Calculate \(Q_p\) to determine the direction in which the reaction will proceed.

Calculating the Shift in Equilibrium

Since we determined that at equilibrium \(K=1\), if \(Q_p > 1\), the reaction will shift to the left to reach equilibrium. If \(Q_p < 1\), the reaction will shift to the right. Based on our calculated \(Q_p\) from the initial conditions and knowing that \(K=1\), predict which direction the reaction will favor.

Establish Equilibrium Partial Pressures

As the system shifts towards equilibrium, the partial pressures of the reacting gases will change. Without specific data on reaction enthalpies and entropy, we cannot provide exact final partial pressures. However, the qualitative shift in the system can be determined from the direction that was established in the previous step. The pressures will adjust accordingly until the reaction quotient equals the equilibrium constant (\(Q = K\)).

Key concepts

Equilibrium Constant

The equilibrium constant, represented as \( K \), plays a pivotal role in understanding chemical reactions at equilibrium. It is a quantitative measure of the proportion of reactants and products in a balanced chemical equation at a given temperature. When the value of \( K \) is known, it allows for the prediction of the extent to which a reaction will proceed before reaching equilibrium.

For the reaction given by \( CO(g) + H2O(g) \rightleftharpoons CO2(g) + H2(g) \), when \( K = 1.00 \), it indicates that the concentrations (or partial pressures in the case of gases) of the reactants and products are equal at equilibrium under standard conditions. This means the reaction is perfectly balanced, and no net change is observed in the concentrations of reactants and products over time. Consequently, the standard Gibbs free energy change \( \Delta G^{\circ} \) is zero for this scenario, as nicely demonstrated in the solution's step 1.

Partial Pressure

Partial pressure is a term often used when we deal with gases in a mixture, such as the components in the given chemical reaction. The partial pressure of a gas is simply the pressure that the gas would exert if it alone occupied the entire volume of the mixture at the same temperature.

Thus, for a gas in a mixture, the partial pressure is the pressure that the gas contributes to the total pressure of the mixture. The total pressure is the sum of the partial pressures of all gases present. In the context of this exercise, when the system reaches equilibrium with \( K = 1 \), the partial pressures of the reactant gases (CO and H2O) and the product gases (CO2 and H2) do not change, maintaining the values provided at the onset of the reaction. This is because \( K \) indicates the system is already at equilibrium, as explained in steps 3 and 6 of the solution.

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a snapshot of a reaction at a specific moment in time and is calculated the same way as the equilibrium constant \( K \), except the concentrations (or partial pressures for gases) are not necessarily those at equilibrium. It indicates the direction in which a reaction must shift to reach equilibrium.

If \( Q < K \), the reaction proceeds forward to create more products. If \( Q > K \), the reaction goes in reverse to form more reactants. When \( Q = K \), the reaction is at equilibrium, and no net change occurs in the concentrations of reactants and products. This concept is vital in predicting how changes in conditions, such as changes in partial pressures, affect the direction in which the reaction proceeds, as shown in steps 4 and 5 of the provided solution. Understanding \( Q \) helps in calculating the shifts in partial pressures until equilibrium is re-established.