Final answer:
To determine the free energy of the reaction under non-standard conditions, we calculate the reaction quotient Q using the given partial pressures and substitute it along with the provided ΔG°, temperature, and the gas constant R into the extended Gibbs free energy equation.
Step-by-step explanation:
To calculate the free energy for the reaction 3 A (g) + B (g) → 2 C (g) at 1120 K with the given partial pressures of A, B, and C, we need to utilize the Gibbs free energy equation:
ΔG = ΔG° + RT ln Q
Where:
- ΔG° is the standard free energy change, which is given as 60.9 kJ/mol.
- R is the universal gas constant, 8.314 J/(mol⋅K).
- T is the temperature, 1120 K.
- Q is the reaction quotient, which is defined as the product of the partial pressures of the products raised to the power of their stoichiometric coefficients divided by the product of the partial pressures of the reactants raised to the power of their stoichiometric coefficients.
In this case:
Q = (PC)2 / ((PA)3 ⋅ PB)
Substitute the given partial pressures into Q:
Q = (0.510 atm)2 / ((11.5 atm)3 ⋅ 8.60 atm)
Now, substitute the values of Q, R, T, and ΔG° into the equation and solve for ΔG:
ΔG = 60900 J/mol + (8.314 J/(mol⋅K) ⋅ 1120 K ⋅ ln(Q))
The result will be the Gibbs free energy change for the reaction at the given conditions. This value will indicate whether the reaction is spontaneous under these non-standard state conditions.