Question

At 975 k, ∆g° = 32.3 kj/mol for the reaction 3 a (g) b (g) →2 c (g). if the partial pressures of a, b, and c are 11.5 atm, 8.60 atm, and 0.520 atm respectively,what is the free energy for this reaction?

Answer 1

To find the free energy (ΔG) for the reaction given the partial pressures and standard free energy change (ΔG°), we use the Gibbs free energy equation, incorporating the universal gas constant and the temperatures, and then calculate the reaction quotient Q based on the partial pressures.

Step-by-step explanation:

To calculate the free energy (ΔG) for the reaction of 3A (g) + B (g) → 2C (g) given the partial pressures of the gases A, B, and C at a certain temperature, we use the Gibbs free energy equation ΔG = ΔG° + RTlnQ, where R is the universal gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient which depends on the partial pressures of the reactants and products. The standard free energy change (ΔG°) is given as 32.3 kJ/mol at 975 K. Q is calculated using the partial pressures of the gases as Q = (PC)2 / (PA)3PB, so we substitute PA = 11.5 atm, PB = 8.60 atm, and PC = 0.520 atm into the equation to find Q. Then we plug the values of Q, ΔG°, R, and T into the Gibbs free energy equation to find ΔG. Note that the value of R must be converted from J/mol·K to kJ/mol·K by dividing it by 1000 to match the units of ΔG°.