Problem 119 - Chemistry An Atoms First Approach

Question

For the equilibrium $$\mathrm{A}(g)+2 \mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$$ the initial concentrations are \([\mathrm{A}]=[\mathrm{B}]=[\mathrm{C}]=0.100 \mathrm{atm}\) Once equilibrium has been established, it is found that \([\mathrm{C}]=\) 0.040 atm. What is \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The standard Gibbs free energy change (ΔG°) for the given reaction at 25°C is approximately -6.73 kJ/mol.

Step 1: Determine the equilibrium concentrations

Given the initial concentrations of A, B, and C are all 0.100 atm, we can write the change in concentrations as they approach equilibrium as follows: A: 0.100 - x B: 0.100 - 2x C: 0.040 + x At equilibrium, we're given that the concentration of C is 0.040 atm. So, we can set up the following equation to solve for x: 0.040 + x = 0.100

Step 2: Solve for x

Solving for x, we find that x = 0.060. Now, we can determine the equilibrium concentrations of A and B: A: 0.100 - x = 0.100 - 0.060 = 0.040 atm B: 0.100 - 2x = 0.100 - 2(0.060) = 0.020 atm

Step 3: Calculate the reaction quotient Q and equilibrium constant K

Using the equilibrium concentrations, we can calculate the reaction quotient Q: Q = [C] / ([A][B]^2) = (0.040) / ((0.040)(0.020)^2) = 50 Since the system is at equilibrium, Q is equal to the equilibrium constant K: K = 50

Step 4: Determine ΔG° using K

We can find the standard Gibbs free energy change (ΔG°) using the relationship between K and ΔG°: ΔG° = -RT ln(K) Where R is the gas constant, 8.314 J/(mol K), and T is the temperature in Kelvin, which is 25°C + 273.15 = 298.15 K. So, ΔG° = - (8.314 J/(mol K)) (298.15 K) ln(50)

Step 5: Calculate ΔG°

Now, we calculate the value of ΔG°: ΔG° ≈ - (8.314 J/(mol K)) (298.15 K) ln(50) ≈ -6733 J/mol We can convert this to kJ/mol: ΔG° ≈ -6.73 kJ/mol Thus, the standard Gibbs free energy change (ΔG°) for this reaction at 25°C is approximately -6.73 kJ/mol.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G \)) is a fundamental thermodynamic quantity that helps predict the direction of chemical reactions. It represents the maximum reversible work that a thermodynamic system can perform at constant temperature and pressure without any exchange of heat or matter.

At equilibrium, the change in Gibbs Free Energy (\( \Delta G \)) is zero. The standard change in Gibbs Free Energy (\( \Delta G^{\circ} \)) can be related to the equilibrium constant (\( K \)) using the equation:

• \( \Delta G^{\circ} = -RT \ln(K) \)

where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.

A negative \( \Delta G^{\circ} \) indicates a spontaneous process under standard conditions, while a positive value means the reaction is non-spontaneous.

Equilibrium Constant

The Equilibrium Constant (\( K \)) expresses the ratio of concentrations of products to reactants at the state of equilibrium for a reversible chemical reaction. It is specific to a given temperature and does not change unless the temperature changes.

For the reaction: \( \text{A(g) + 2B(g) } \rightleftharpoons \text{ C(g)} \), the equilibrium constant expression is:

• \( K = \frac{[C]}{[A][B]^2} \)

If \( K \) is large, the reaction heavily favors products. Conversely, if \( K \) is small, reactants are favored. In this exercise, the calculated \( K \) was 50, indicating a strong formation of the product C at equilibrium.

Reaction Quotient

The Reaction Quotient (\( Q \)) is calculated in the same way as the equilibrium constant (\( K \)), but it can be done at any point in a reaction. It helps indicate which direction a reaction will proceed to reach equilibrium.

When calculating Q:

• If \( Q < K \), the reaction will move forward, producing more products to reach equilibrium.
• If \( Q > K \), the reaction will shift in the reverse direction, forming more reactants.
• If \( Q = K \), the reaction is at equilibrium.

In the given exercise, since the system is at equilibrium, the calculated Q is equal to the equilibrium constant K, both being 50.

Thermodynamics

Thermodynamics is the study of energy transformations and its principles apply to chemical reactions as well. It helps us understand how energy is transferred or changed in chemical processes.

In terms of chemical reactions:

• The First Law states energy cannot be created or destroyed, only transformed.
• The Second Law explains that spontaneous processes increase the entropy of the universe.

The analysis of a reaction through thermodynamics involves these principles and concepts like Gibbs Free Energy, which ties in entropy and enthalpy changes to predict reaction spontaneity.