Question

Calculate ΔG for the reaction when the partial pressures are PH2 = 0.28 atm, PCO2 = 0.71 atm, PH2O = 0.60 atm, and PCO = 1.12 atm.

Answer 1

Since the reaction would be,

H₂ (g) + CO₂ (g) ------> H₂O(g) + CO(g)

Now using the formula for gibbs free energy,

ΔG = ΔG° + RTln(Q)

For a gas phase reaction of the type

aA(g) + bB(g) ⇌ cC(g) + dD(g),

Q = P_C^cP_D^d/P_A^aP_B^b

Therefore ,

ΔG=ΔG^o+RTln(P_C^cP_D^d/P_A^aP_B^b)

here, PA = P_H2 = 0.28 atm

  P_B = P_CO2 = 0.71 atm

  P_C = P _H2O = 0.60 atm

  P_D = P_CO = 1.12 atm

ΔG° = 0 (assuming equilibrium condition)

R = 0.0827  L atm mol^-1K^-1

T = 298K (Room temperature)

Now substituting the values,

ΔG = 0 + (0.0827 * 298 * ln (0.60 * 1.12 / 0.28 * 0.71))

ΔG = 0.0827 * 298 * ln ( 14.11 )

ΔG = 0.0827 * 298 * 2.646

ΔG = 65.231 Jmol^-1