Question

Gaseous hydrogen iodide is placed in a closed 1.0 L container at 425°C, where it partially decomposes to hydrogen and iodine in the reaction below. 2 HI (g) ⇌ H2 (g) + I2 (g). At equilibrium it is found that: PHI = 3.53 x 10−3 atm PH2 = 4.79 x 10−4 atm PI2 = 4.79 x 10−4 atm What is the standard free energy change (ΔG°) for this reaction?

Answer 1

Answer:

ΔG° = 23192 J/mol

Explanation:

For the reaction:

2 HI (g) ⇌ H₂(g) + I₂(g)

It is possible to find Kp thus:

Kp = P(H₂) P(I₂) / P(HI)²

Replacing:

Kp = 4,79x10⁻⁴ × 4,79x10⁻⁴ / (3,53x10⁻³)²

Kp = 0,0184

As moles of products (2) are the same moles of reactants, Kp = K

It is possible to find ΔG° from K, thus:

ΔG° = -RT ln K

where R is 8,314472J/molK, temperature in kelvin is 425 + 273,15 = 698,15K and K is 0,0184

Replacing:

ΔG° = -8,314472J/molK × 698,15K  ln 0,0184

ΔG° = 23192 J/mol

I hope it helps!