Questions: What is the free energy for the reaction 3 A(g) + B(g) → 2 C(g) at 1275 K with ΔG° = 82.8 kJ/mol, given the partial pressures of A, B, and C are 11.5 atm, 8.60 atm, and 0.520 atm respectively?

Calculate the free energy for the reaction 3A(g) + B(g) → 2C(g) at 1275 K with given partial pressures.

Recall the relationship between free energy and standard free energy

The relationship between the free energy change (ΔG) and the standard free energy change (ΔG°) for a reaction is given by:

ΔG = ΔG° + RT ln Q

Where:

• R is the gas constant (8.314 J/(mol·K))
• T is the temperature in Kelvin
• Q is the reaction quotient

Calculate the reaction quotient Q

For the reaction 3A(g) + B(g) → 2C(g), the reaction quotient is:

Q = $\frac{P_C^2}{P_A^3 \times P_B}$

Substituting the given partial pressures:

• P_A = 11.5 atm
• P_B = 8.60 atm
• P_C = 0.520 atm

Q = $\frac{(0.520)^2}{(11.5)^3 \times 8.60}$ = $\frac{0.2704}{13,944.8}$ = 1.939 × 10^(-5)

Calculate the free energy change (ΔG)

Given:

• ΔG° = 82.8 kJ/mol = 82,800 J/mol
• T = 1275 K
• R = 8.314 J/(mol·K)
• Q = 1.939 × 10^(-5)

ΔG = ΔG° + RT ln Q ΔG = 82,800 + (8.314 × 1275 × ln(1.939 × 10^(-5)))

RT = 8.314 × 1275 = 10,600.35 J/mol ln(1.939 × 10^(-5)) = -10.85

ΔG = 82,800 + (10,600.35 × (-10.85)) ΔG = 82,800 - 115,014 ΔG = -32,214 J/mol = -32.2 kJ/mol

\(\boxed{\Delta G = -32.2 \text{ kJ/mol}}\)

\(\boxed{\Delta G = -32.2 \text{ kJ/mol}}\)