Exercise 8 | Chemical Thermodynamics

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 K2K1 = H0rxnR x 1T1 - 1T2 = H0rxnR x T2 - T1T1T2

 

Situation 1 : T1 = 400°C = 673K and K1 = 58.0

Situation 2 : T2 = 500°C = 773K and K2 = ?


ln K258.0 = - 9.10 × 1038.314 x  100673 × 773

ln K258.0 = -0.217

K2 = 46.7