Question

Consider the reaction shown below. PbCO3(s) PbO(s) + CO2(g) Calculate the equilibrium pressure of CO2 in the system at the following temperatures. (a) 100°C atm (b) 420°C atm Note: To find the value of the equilibrium constant at each temperature you must first find the value of G0 at each temperature by using the equation G0 = H0 - TS0 For this reaction the values are H0 = +88.3 kJ/mol and S0= 151.3 J/mol*K

Answer 1

PbCO 3 Delta H_f= -699.1 Delta G_f= -625.5 Delta S=161
CO2 Delta H_f = -393.5 Delta G_f= -394.4 Delta S=213.6
PbO= Delta H_f = -217.3 Delta G_f = -187.9 Delta S=68.7

H₀ = +88.3 kJ/mol and S₀= 151.3 J/mol*K

G₀= H₀ - TΔS₀   ......AT 100°C

  ₌ 88300 - 373.15*151.3 = 31.842 Kj /mol

ΔG = Δ G₀ + RT ln K = Δ G₀ + RT ln P(REST ARE SOLIDS SO CONC CAN BE TAKEN 1)

AT EQ ΔG = 0

-31842 = 8.314 * 373.15 ln P

P = 3.487 * 10^-5 atm

G₀= H₀ - TΔS₀   ......AT 420°C

  ₌ 88300 - 693.15*151.3 = -16.573 Kj /mol

ΔG = Δ G₀ + RT ln K = Δ G₀ + RT ln P(REST ARE SOLIDS SO CONC CAN BE TAKEN 1)

AT EQ ΔG = 0

16573 = 8.314 * 693.15 ln P

ln P = 2.8759

P = 17.74 atm