Acetic acid can be manufactured by combining methanol with carbon monoxide, an example of a carbonylation reaction: CH3OH(l) + CO(g) ⇌ CH3COOH(l). (a) Calculate the equilibrium constant for the reaction at 25 °C. (b) At what temperature will this reaction have an equilibrium constant equal to 1? (You may assume that ΔH° and ΔS° are temperature independent, and you may ignore any phase changes that might occur.)

Verified step by step guidance

Step 1: Identify the reaction and write the balanced chemical equation: CH3OH(l) + CO(g) ⇌ CH3COOH(l).

Step 2: For part (a), use the standard Gibbs free energy change equation: ΔG° = -RT ln(K), where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin. Calculate ΔG° using ΔG° = ΔH° - TΔS°.

Step 3: Obtain ΔH° and ΔS° values for the reaction from standard thermodynamic tables. Use these values to calculate ΔG° at 25 °C (298 K).

Step 4: Rearrange the equation ΔG° = -RT ln(K) to solve for the equilibrium constant K: K = e^(-ΔG°/RT). Substitute the calculated ΔG° and the given temperature to find K.

Step 5: For part (b), set K = 1 and use the equation ΔG° = -RT ln(K) to find the temperature. Since ln(1) = 0, ΔG° = 0 at equilibrium. Use ΔG° = ΔH° - TΔS° = 0 to solve for the temperature T.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (K)

The equilibrium constant (K) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. It is calculated using the formula K = [products]/[reactants], where the concentrations are raised to the power of their coefficients in the balanced equation. A larger K value indicates a reaction that favors products, while a smaller K suggests a preference for reactants.

Gibbs Free Energy (ΔG)

Gibbs free energy (ΔG) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. The relationship between ΔG and the equilibrium constant is given by the equation ΔG° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin. A negative ΔG indicates that a reaction is spontaneous, while a positive ΔG suggests non-spontaneity.

Van 't Hoff Equation

The Van 't Hoff equation relates the change in the equilibrium constant (K) of a reaction to the change in temperature. It is expressed as ln(K2/K1) = -ΔH°/R(1/T2 - 1/T1), where ΔH° is the standard enthalpy change, R is the gas constant, and T1 and T2 are the initial and final temperatures, respectively. This equation is essential for determining how temperature affects the position of equilibrium and the value of K.

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Related Practice

Textbook Question

(a) For each of the following reactions, predict the sign of ΔH° and ΔS° without doing any calculations. (i) 2 Mg(s) + O₂ (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I₂(g) (iii) Na₂(g) ⇌ 2 Na(g) (iv) 2 V₂O₅(s) ⇌ 4 V(s) + 5 O₂(g)

Textbook Question

(b) Based on your general chemical knowledge, predict which of these reactions will have K>1. (i) 2 Mg(s) + O₂ (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I₂(g) (iii) Na₂(g) ⇌ 2 Na(g) (iv) 2 V₂O₅(s) ⇌ 4 V(s) + 5 O₂(g)

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Textbook Question

(c) In each case, indicate whether K should increase or decrease with increasing temperature. (i) 2 Mg(s) + O₂ (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I₂(g) (iii) Na₂(g) ⇌ 2 Na(g) (iv) 2 V₂O₅(s) ⇌ 4 V(s) + 5 O₂(g)

Textbook Question

The oxidation of glucose (C₆H₁₂O₆) in body tissue produces CO₂ and H₂O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C₂H₅OH) and CO2.

(a) Using data given in Appendix C, compare the equilibrium constants for the following reactions:

C₆H₁₂O₆(s) + 6 O₂(g) ⇌ 6 CO₂(g) + 6 H₂O(l)

C₆H₁₂O₆(s) ⇌ 2 C₂H₅OH(l) + 2 CO₂(g)

Textbook Question

The oxidation of glucose (C₆H₁₂O₆) in body tissue produces CO₂ and H₂O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C₂H₅OH) and CO2.

(b) Compare the maximum work that can be obtained from these processes under standard conditions.

C₆H₁₂O₆(s) + 6 O₂(g) ⇌ 6 CO₂(g) + 6 H₂O(l)

C₆H₁₂O₆(s) ⇌ 2 C₂H₅OH(l) + 2 CO₂(g)

Textbook Question

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane (C₂H₆), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: 2 CH₄(g) → C₂H₆(g) + H₂(g) In practice, this reaction is carried out in the presence of oxygen: 2 CH₄(g) + 12 O₂(g) → C₂H₆(g) + H₂O(g) (b) Is the difference in ΔG° for the two reactions due primarily to the enthalpy term (ΔH) or the entropy term (-TΔS)?