Question

Using standard thermochemical data calculate ΔrGθ and K at (a) 298 K and (b) 400 K for the reaction PbO(s) + CO(g) <---> Pb(s) + CO2(g). Assume that the reaction enthalpy is independent of temperature.

ΔrGθ (298 K)=

Κ (298 K)=

ΔrGθ (400 K)=

Κ (400 K)=

Answer 1

a) at 298 K
  Δ_rG ^ϴ =  Σ n_j Δ_f G ^ϴ (j)
n_pb =1 ; n_co2 = 1 ; n_pbo = n_co = -1
Pb + CO₂ - PbO - CO = 0
Δ_rG ^ϴ = Δ_f G ^ϴ (Pb) + Δ_f G ^ϴ (CO₂) - Δ_f G ^ϴ (PbO) - Δ_f G ^ϴ (CO)
Δ_rG ^ϴ = ( - 394.4 kJ mol^-1) - ( - 188.94 kJ mol^-1) - (-137.2 kJ mol^-1)
Δ_rG ^ϴ = -68.26 kJ mol^-1
ln K = - Δ_rG ^ϴ / RT = 68.26*10³ J mol^-1 / (8.314 J mol^-1K^-1) * 298 K = 27.551167
K = e^27.551167 = 9.2325*10¹¹

b) at 498 K
   Δ_rH^ϴ = (-392.81 kJmol^-1) - (-222.3 kJ mol^-1) - (-107.89kJ mol^-1)
   Δ_rH^ϴ  = -62.62 kJ mol^-1
Use Arrhenius equation
ln K (498 K) = ln K(298) -  Δ_rH^ϴ / R [1/498 - 1/ 298
ln K= 27.551167 - (-62.62*10³J mol^-1 /8.314 ) * [ 2*10^-3 - 3.355*10^-3]
ln K = 17.34547
K = 3.4122*10⁷
Δ_rG ^ϴ = -RT ln K = -8.314 J mol^-1 K* 498 K *  17.34547
  Δ_rG ^ϴ = -71.8166 kJ mol^-1