Question

The equilibrium constant \(\left(K_{P}\right)\) for the reaction: $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.40 at \(2000 \mathrm{~K}\). (a) Calculate \(\Delta G^{\circ}\) for the reaction. (b) Calculate \(\Delta G\) for the reaction when the partial $$ \begin{array}{l} \text { pressures are } P_{\mathrm{H}_{2}}=0.25 \mathrm{~atm}, P_{\mathrm{CO}_{2}}=0.78 \mathrm{~atm} \\ P_{\mathrm{H}_{2} \mathrm{O}}=0.66 \mathrm{~atm}, \text { and } P_{\mathrm{CO}}=1.20 \mathrm{~atm} . \end{array} $$

Short answer

(a) \(\Delta G^{\circ} \approx -24.6\, kJ/mol\), (b) \(\Delta G \approx -14.6\, kJ/mol\).

Step-by-step solution

Understanding Equilibrium Constant and Gibbs Free Energy

The relationship between the Gibbs free energy change (\(\Delta G^{\circ}\)) and the equilibrium constant (\(K_P\)) at a given temperature (\(T\)) is given by the formula:\[\Delta G^{\circ}= -RT \ln K_P\]where \(R = 8.314\, J\,mol^{-1}K^{-1}\) is the gas constant and \(T\) is the temperature in Kelvin.

Calculate \(\Delta G^{\circ}\)

Substitute the given values into the formula. Here \(T = 2000\, K\) and \(K_P = 4.40\):\[\Delta G^{\circ} = -(8.314\, J\,mol^{-1}K^{-1})(2000\, K) \ln(4.40)\]First, calculate \(\ln(4.40)\) which is approximately 1.4816. Then substitute back:\[\Delta G^{\circ} = -(8.314)(2000)(1.4816) \approx -24632.87\, J/mol \approx -24.6\, kJ/mol\]

Calculate \(\Delta G\) using Reaction Quotient

Next, we need to calculate \(\Delta G\) using the reaction quotient \(Q\). This is given by:\[Q = \frac{P_{\mathrm{H}_2\mathrm{O}} \times P_{\mathrm{CO}}}{P_{\mathrm{H}_2} \times P_{\mathrm{CO}_2}}\]Given: \(P_{\mathrm{H}_2} = 0.25\, atm\), \(P_{\mathrm{CO}_2} = 0.78\, atm\), \(P_{\mathrm{H}_2\mathrm{O}} = 0.66\, atm\), and \(P_{\mathrm{CO}} = 1.20\, atm\). Substitute these into the expression for \(Q\):\[Q = \frac{0.66 \times 1.20}{0.25 \times 0.78} \approx 4.051\]

Calculate \(\Delta G\)

Finally, use the relationship between \(\Delta G^{\circ}\) and \(\Delta G\) which is given by:\[\Delta G = \Delta G^{\circ} + RT \ln Q\]Substitute the values:\[\Delta G = -24632.87\, J/mol + (8.314\, J/mol\, K)(2000\, K) \ln(4.051)\]Calculate \(\ln(4.051)\) which is approximately 1.399. Then:\[\Delta G = -24632.87 + (8.314)(2000)(1.399) \approx -14558.27\, J/mol \approx -14.6\, kJ/mol\]

Key concepts

Equilibrium Constant

In the world of chemistry, the equilibrium constant, denoted as \(K_P\) for reactions involving gases, plays a crucial role in understanding how reactions reach balance at a given temperature. It helps us determine the ratio of product concentrations to reactant concentrations when a reaction is at equilibrium. For the given reaction, the equilibrium constant \(K_P\) at 2000 K is 4.40, which indicates how much the reaction favors the formation of products (in this case, \(\mathrm{H}_2\mathrm{O}(g)\) and \(\mathrm{CO}(g)\)) over reactants \((\mathrm{H}_2(g)\) and \(\mathrm{CO}_2(g)\)).

When a reaction is at equilibrium, it is dynamic, meaning the concentrations of the reactants and products remain constant, but the molecules continue to react with each other. The beauty of the equilibrium constant is that it remains unchanged for a given reaction at a constant temperature, even when the pressures of the gases involved change.

• This constant helps predict the direction in which a reaction mixture will shift to reach equilibrium.
• It tells us whether the products or reactants are favored when the system is in equilibrium.

This is critical to understanding how changes in reaction conditions affect chemical balances and helps chemists predict the position of equilibrium just by knowing \(K_P\).

Reaction Quotient

The reaction quotient, \(Q\), is a snapshot of a reaction's progress at a particular moment in time, similar to the equilibrium constant \(K_P\). Unlike \(K_P\), \(Q\) can be calculated at any point, not just when the reaction has reached equilibrium. For this specific reaction, we use the partial pressures of the gases to find \(Q\).

It is given by the formula:\[Q = \frac{P_{\mathrm{H}_2\mathrm{O}} \times P_{\mathrm{CO}}}{P_{\mathrm{H}_2} \times P_{\mathrm{CO}_2}}\]Substitute the given pressures to find:\[Q = \frac{0.66 \times 1.20}{0.25 \times 0.78} \approx 4.051\]
This calculation shows us how the current state of the system compares to equilibrium. If \(Q > K_P\), the reaction will proceed in the reverse direction to reach equilibrium. If \(Q < K_P\), it will move forward. In this example, \(Q\) is slightly less than \(K_P\), which indicates a slight shift towards forming more products is needed to reach equilibrium.

• \(Q\) acts as a decision-maker, guiding the reaction towards equilibrium.
• Knowing \(Q\) helps determine whether a reaction will produce more reactants or products as it proceeds.

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. It is essential in the study of gaseous reactions and plays a pivotal role in calculating the equilibrium constant and reaction quotient.

For gases involved in a reaction, the total pressure is the sum of the partial pressures of each gas present. In this problem, the partial pressures are given as:

• \(P_{\mathrm{H}_2} = 0.25\, \mathrm{atm}\)
• \(P_{\mathrm{CO}_2} = 0.78\, \mathrm{atm}\)
• \(P_{\mathrm{H}_2\mathrm{O}} = 0.66\, \mathrm{atm}\)
• \(P_{\mathrm{CO}} = 1.20\, \mathrm{atm}\)

Partial pressures are integral to determining \(Q\), helping us track how far a reaction has progressed and whether it needs adjustment to reach equilibrium.

By adjusting the partial pressures, chemists can influence the position of equilibrium and reaction pathways—demonstrating how practical control of pressures can steer reactions. Understanding and manipulating partial pressures is a skill central to industrial chemistry applications.

Thermodynamics

Thermodynamics is the field of science that deals with energy changes in chemical reactions. Gibbs Free Energy, denoted as \( \Delta G \), is a thermodynamic potential that helps predict the favorability and direction of a chemical reaction.

The formula for the change in Gibbs Free Energy is:\[\Delta G = \Delta G^{\circ} + RT \ln Q\]where \( \Delta G^{\circ} \) is the standard Gibbs Free Energy change, \( R \) is the gas constant, \( T \) is the temperature, and \( Q \) is the reaction quotient.

• Thermodynamics assists in understanding why and how reactions occur.
• It incorporates concepts like temperature, pressure, and energy to predict reaction behavior.

When analyzing \( \Delta G \) within this context, a negative \( \Delta G \) signifies a spontaneous reaction under the given conditions—the more negative, the more favored the reaction. In this exercise, after calculations, the \( \Delta G \) of about \(-14.6\, \mathrm{kJ/mol}\) shows the reaction proceeds forward spontaneously, aligning with the principle that systems tend toward equilibrium.