Question

At 1125 K, ∆G° = 16.0 kJ/mol for the reaction 3 A (g) + B (g) → 2 C (g). If the partial pressures of A, B, and C are 0.620 atm, 2.50 atm, and 8.20 atm respectively, what is the free energy for this reaction?

Answer 1

To calculate the free energy (ΔG) for the reaction at the given conditions, you can use the following equation:

ΔG = ΔG° + RT * ln(Q)

Where:
ΔG° = Standard free energy change (given as 16.0 kJ/mol)
R = Gas constant (8.314 J/(mol K))
T = Temperature in Kelvin (1125 K)
Q = Reaction quotient

First, let's calculate the reaction quotient (Q):

Q = (P_C)^2 / (P_A)^3 * P_B

Where:
P_A, P_B, and P_C are the partial pressures of A, B, and C, respectively.

Now, let's calculate Q:

Q = (8.20 atm)^2 / (0.620 atm)^3 * 2.50 atm

Next, convert the gas constant R from J/(mol K) to kJ/(mol K) by dividing it by 1000:

R = 8.314 J/(mol K) / 1000 = 0.008314 kJ/(mol K)

Now, we can plug the values into the equation for ΔG:

ΔG = 16.0 kJ/mol + 0.008314 kJ/(mol K) * 1125 K * ln(Q)

Calculate ln(Q):

ln(Q) = ln[(8.20 atm)^2 / (0.620 atm)^3 * 2.50 atm]

After calculating ln(Q), you can find the value of ΔG in kJ/mol.