Exercise 8 | Entropy, Free Energy, and Equilibrium
General Chemistry 3 - Exercise 8
Calculate ΔH0rxn for the following reaction: H2 (g) + I2 (g) 2 HI (g)
Calculate the value of the equilibrium constant K at 500°C.
Data: ΔH0f [H2 (g)] = 0 J.mol-1
ΔH0f [I2 (g)] = 62.4 kJ.mol-1
ΔH0f [HI (g)] = 26.5 kJ.mol-1
K [400°] = 58.0
ΔH0rxn = 2 x ΔH0f [HI (g)] – ΔH0f [H2 (g)] – ΔH0f [I2 (g)]
ΔH0rxn = 2 x 26.5 – 62.4
ΔH0rxn = - 9.40 kJ.mol-1
van’t Hoff Equation:
ln = x = x
Situation 1 : T1 = 400°C = 673K and K1 = 58.0
Situation 2 : T2 = 500°C = 773K and K2 = ?
ln = - x
ln = -0.217
K2 = 46.7