Question

Calculate ∆G°rxn at 298 K for the reaction

I₂(g) + Cl₂(g) ⇄ 2 ICl(g)
Kp = 81.9 a) True
b) False
c) Open-ended
d) Not applicable

Answer 1

To find ΔG°rxn for the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) at 298 K with Kp= 81.9, we use the relationship ΔG° = -RTln(K). The calculation yields a ΔG°rxn of approximately -19.5 kJ/mol, indicating spontaneity.

Step-by-step explanation:

To calculate the standard free energy change (ΔG°rxn) for the reaction I2(g) + Cl2(g) ⇌ 2 ICl(g) at 298 K using the given Kp value of 81.9, we apply the equation ΔG° = -RTln(K). Here, R is the universal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and K is the equilibrium constant (81.9 in this case).

ΔG°rxn = -RTln(K)
ΔG°rxn = - (8.314 J/(mol·K)) (298 K) ln(81.9)
ΔG°rxn = - (8.314 J/(mol·K)) (298 K) (ln(81.9))
ΔG°rxn = -19,452 J/mol
ΔG°rxn ≈ -19.5 kJ/mol

This result indicates that the reaction is spontaneous at 298 K, as a negative ΔG°rxn signifies spontaneity.