Question

The reduction of metal oxides is often accomplished using carbon monoxide as a reducing agent. Carbon (coke) and carbon dioxide are usually present, leading to the following reaction: $$ \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Using data from Appendix \(C\), calculate the equilibrium constant for this reaction at \(298 \mathrm{~K}\) and at \(2000 \mathrm{~K}\), assuming that the enthalpies and entropies of formation do not depend upon temperature.

Short answer

The equilibrium constants for the given reaction at the two temperatures are: at 298 K, K = 1.36, and at 2000 K, K = 9.34 × 10^5.

Step-by-step solution

Write down the balanced equation for the reaction

Given reaction: \[C(s) + CO_2(g) \rightleftharpoons 2 CO(g) \]

Find the enthalpy change (ΔH) for the reaction

From Appendix C, we can find the standard enthalpy change of formation (ΔH_f°) for the involved species: - for CO_2(g), ΔH_f° = -393.5 kJ/mol - for CO(g), ΔH_f° = -110.5 kJ/mol Using these values, we can calculate the enthalpy change (ΔH) for the reaction: ΔH = Σ (ΔH_f° of products) - Σ (ΔH_f° of reactants) ΔH = (2 × (-110.5)) - ((1 × 0) + (1 × (-393.5))) = 172.5 kJ/mol

Find the entropy change (ΔS) for the reaction

From Appendix C, we can find the standard molar entropy (S°) for the involved species: - for C(s), S° = 5.74 J/mol·K - for CO_2(g), S° = 213.7 J/mol·K - for CO(g), S° = 197.7 J/mol·K Using these values, we can calculate the entropy change (ΔS) for the reaction: ΔS = Σ (S° of products) - Σ (S° of reactants) ΔS = (2 × (197.7)) - ((1 × 5.74) + (1 × 213.7)) = 168.96 J/mol·K

Find the Gibbs free energy change (ΔG) for the reaction at each temperature

We can use the following equation to find the Gibbs free energy change (ΔG) for the reaction at a given temperature (T): ΔG = ΔH - TΔS For 298 K: ΔG_298 = 172.5 kJ/mol - (298 K × 168.96 J/mol·K × (1 kJ / 1000 J)) ΔG_298 = -32.53 kJ/mol For 2000 K: ΔG_2000 = 172.5 kJ/mol - (2000 K × 168.96 J/mol·K × (1 kJ / 1000 J)) ΔG_2000 = -163.4 kJ/mol

Calculate the equilibrium constant (K) for the reaction at each temperature

We can use the following equation to relate ΔG to the equilibrium constant (K) for the reaction: ΔG = -RT ln(K), where R is the universal gas constant (8.314 J/mol·K) For 298 K: -32.53 kJ/mol = -(8.314 J/mol·K × 298 K) ln(K_298) K_298 = 1.36 For 2000 K: -163.4 kJ/mol = -(8.314 J/mol·K × 2000 K) ln(K_2000) K_2000 = 9.34 × 10^5 The equilibrium constants for the reaction are K_298 = 1.36 and K_2000 = 9.34 × 10^5.

Key concepts

Chemical Thermodynamics

Chemical thermodynamics involves the study of energy changes in chemical reactions and the transfer of heat and work between a system and its surroundings. It's a field that intertwines physics and chemistry to predict the direction of chemical processes and consider efficiency and spontaneity.

The fundamental laws of thermodynamics define how energy is transferred and conserved in these reactions. In the context of our exercise, we are looking at a redox reaction where carbon monoxide reduces metal oxides. Thermodynamics allows us to calculate the feasibility of such reactions under different conditions, like at 298 K and at 2000 K. By understanding these concepts, students are better equipped to predict reaction behavior and manipulate conditions for desired outcomes.

Gibbs Free Energy

Gibbs free energy (\( G \)) is a concept from thermodynamics that represents the amount of work that can be extracted from a system at constant temperature and pressure. It's a measure of a system's tendency to do work and thus a predictor of the spontaneity of a reaction. A negative value of Gibbs free energy indicates a spontaneous process.

In the exercise's step-by-step solution, we use the equation \[ \Delta G = \Delta H - T\Delta S \] to calculate the Gibbs free energy change at specific temperatures. This relationship shows the balance between enthalpy, entropy, and temperature to determine if a reaction will proceed without input of additional energy. A grasp of this equation is vital for students to understand how reactions can be driven or restrained by changing conditions, particularly in industrial processes.

Standard Enthalpy Change

The standard enthalpy change (\( \.Delta H^{\circ} \) is a measure of heat evolved or absorbed in a reaction at standard conditions (1 bar and typically 298 K) for each mole of substance involved. It's a key component of the equilibrium constant calculation as shown in our exercise.

During the solution, the values for \(.Delta H_f^{\circ}\) of CO and CO₂ are used to calculate the overall enthalpy change of the reaction. This is crucial because it tells us how much energy is released or consumed when a given chemical change happens under standard conditions. The understanding of enthalpy change allows students to analyze and design reactions, important for creating energy-efficient processes.

Standard Molar Entropy

Standard molar entropy (\( S^{\circ} \) is a thermodynamic property referring to the entropy content of one mole of substance under standard conditions of 1 bar pressure. The entropy represents a measure of disorder or randomness in a system, with larger values indicating higher degrees of freedom for the particles involved.

In our exercise, the standard molar entropy of reactants and products is used to compute the entropy change (\( \.Delta S \) for the reaction. This is an essential part of determining Gibbs free energy and thus the likelihood of the reaction occurring spontaneously. For students, understanding how entropy influences chemical reactions is pivotal for any foray into reaction thermodynamics, as it underlies the fundamental principles of evolution and equilibrium in chemical systems.