Question

Calculate \(\Delta \mathrm{G}^{\circ}\) and \(\mathrm{K}_{\mathrm{p}}\) at \(25^{\circ} \mathrm{C}\) for $$ \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) $$ \(\Delta \mathrm{G}^{\circ} \mathrm{f}\) for \(\mathrm{H}_{2}(\mathrm{~g}), \mathrm{CO}_{2}(\mathrm{~g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), and \(\mathrm{CO}(\mathrm{g})\) are 0, \(-94,2598,-54.6357\), and \(-32.8079 \mathrm{Kcal} / \mathrm{mole}\) respectively.

Short answer

The standard Gibbs free energy change (ΔG°) for the given reaction is -6.8162 Kcal/mole. At 25°C, the equilibrium constant (K_p) is approximately 1133.469.

Step-by-step solution

Write down the given data

Given the standard Gibbs free energy of formation (ΔG°f) for the reactants and products: - H₂(g) = 0 Kcal/mole - CO₂(g) = -94.2598 Kcal/mole - H₂O(g) = -54.6357 Kcal/mole - CO(g) = -32.8079 Kcal/mole Temperature: 25°C = 298.15 K. Reactants: CO(g) + H₂O(g) Products: CO₂(g) + H₂(g)

Calculate ΔG° for the reaction

We can calculate ΔG° for the reaction using the following formula: ΔG° = Σ(ΔG°f of products) - Σ(ΔG°f of reactants) Plugging in the given data: ΔG° = [ΔG°f(CO₂(g)) + ΔG°f(H₂(g))] - [ΔG°f(CO(g)) + ΔG°f(H₂O(g))] ΔG° = [(-94.2598) + (0)] - [(-32.8079) + (-54.6357)] ΔG° = -94.2598 + 32.8079 + 54.6357 ΔG° = -6.8162 Kcal/mole

Convert ΔG° to joules/mole

We need to convert ΔG° from Kcal/mole to J/mole. The conversion factor is: 1 Kcal = 4184 J So, ΔG° = -6.8162 Kcal/mole × 4184 J/Kcal = -28523.49 J/mole.

Calculate Kp using ΔG°

Now we will use the relationship between ΔG°, the gas constant R, and Kp: ΔG° = -RTln(Kp) Where: ΔG° = -28523.49 J/mole R = 8.314 J/mol K T = 298.15 K Rearranging the equation to solve for Kp: Kp = exp(-ΔG°/RT) Plugging in the values: K_p = exp(28523.49 J/mol / (8.314 J/mol K × 298.15 K)) K_p ≈ 1133.469

Present the results

The standard Gibbs free energy change (ΔG°) for the reaction is -6.8162 Kcal/mole, and the equilibrium constant (K_p) at 25°C is approximately 1133.469.

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as K, is a fundamental concept in the study of chemical reactions and thermodynamics. It reflects the ratio of concentrations of products to reactants at equilibrium. In simpler terms, it tells you how far a reaction proceeds. The equilibrium constant can be expressed in terms of partial pressures for gaseous reactions using the symbol Kp.
For our reaction, the equilibrium constant Kp is calculated using the relationship between the standard Gibbs free energy change (ΔG°) and Kp. The formula \[\Delta G^\circ = -RT \ln(K_p)\]helps us determine Kp when we know ΔG°.
In this exercise, after calculating ΔG° as -28523.49 J/mole, we used this formula to find Kp, resulting in Kp ≈ 1133.469. A large Kp value indicates that the reaction favors the formation of products at equilibrium, which in this case are CO₂(g) and H₂(g).

Standard Conditions

Standard conditions are crucial for ensuring consistency in thermodynamic calculations. They typically involve a temperature of 25°C (298.15 K), a pressure of 1 atm, and concentrations of solutes at 1 M. These conditions are standardized to allow comparison of results between different experiments and reactions.
In the context of our exercise, standard Gibbs free energy changes, \(\Delta G^\circ\), are reported under these conditions. This means that the values provided for the Gibbs free energy of formation (\(\Delta G^\circ_f\)) for each compound (e.g., H₂(g), CO₂(g), etc.) are considered at these standard conditions. By adhering to these conditions, scientists can accurately predict how reactions will behave in real-world scenarios.

Thermodynamics

Thermodynamics is a branch of physical science that deals with heat and temperature and their relation to energy and work. In the context of chemical reactions, thermodynamics is crucial for understanding how and why chemicals react as they do.
One key aspect of thermodynamics is the concept of Gibbs free energy (\(\Delta G\)), which drives reactions. If \(\Delta G\) is negative, as it is in our example (-6.8162 Kcal/mole), it indicates that the reaction is spontaneous under standard conditions. The sign and magnitude of \(\Delta G\) inform us whether a reaction will proceed naturally or require energy input.

• Positive \(\Delta G\): Non-spontaneous reaction
• Negative \(\Delta G\): Spontaneous reaction

Being able to predict spontaneity is essential for applications in chemistry and engineering, where controlling reaction pathways can be critical.

Chemical Reaction

A chemical reaction involves transforming reactants into products through the breaking and forming of chemical bonds. In our exercise, the reaction involves gases: CO(g) and H₂O(g) reacting to form CO₂(g) and H₂(g).
The reaction\[\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})\]is an example of a redox reaction, where electron transfer takes place. Here, carbon in CO is oxidized to CO₂, while hydrogen in H₂O is reduced to H₂. This is just one of many types of chemical reactions, each with distinct characteristics and rules.
Understanding the details of chemical reactions is foundational in predicting how changes in conditions, such as pressure or temperature, can affect the reaction's behavior and outcomes.