Question

Could someone help me understand this question, I’m so unsure Wich equation to use and how.

We start with a pure sample of A (g). The following equilibrium is established:
2 A(g) <---> B (g) + C(g)
The total.
pressure is 9.72 atm and the temperature is 25.0°C. The partial pressure
of A (g) is 5.71 atm. Calculate the value of the standard Gibbs free enthalpy change
(in kJ) for this reaction 25.0°C.

Answer 1

Step-by-step explanation:

To solve this problem, we can use the equation:

ΔG° = -RT ln(K)

where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and K is the equilibrium constant.

First, we need to find the value of K. We can use the equation:

K = (P_B × P_C) / (P_A)^2

where P_B and P_C are the partial pressures of B and C, respectively, and P_A is the partial pressure of A.

We are given that the total pressure is 9.72 atm and the partial pressure of A is 5.71 atm. Therefore, we can calculate the partial pressures of B and C:

P_B = (2 × P_A) / (2 + 1) = 3.81 atm

P_C = (1 × P_A) / (2 + 1) = 1.90 atm

Substituting these values into the equation for K, we get:

K = (3.81 × 1.90) / (5.71)^2 = 0.246

Now we can use this value of K to calculate ΔG°:

ΔG° = -RT ln(K)

ΔG° = -(8.314 J/mol·K)(298 K) ln(0.246)

ΔG° = -8,870 J/mol or -8.87 kJ/mol

Therefore, the value of the standard Gibbs free energy change for this reaction at 25°C is -8.87 kJ/mol.