In: Chemistry
What is ΔGrxno (in kJ) at 647 K for the following reaction? PbO(g) + CO2(g) → PbCO3(s)
PbO: ΔHfo = -219.0 kJ/mol and So = 66.5 J/K mol)
PbCO3(s): ΔHfo = -699.1 kJ/mol and So = 131.0 J/K mol)
CO2: ΔHfo = -393.5 kJ/mol and So = 213.6 J/K mol)
we have:
Hof(PbO(g)) = -219.0 KJ/mol
Hof(CO2(g)) = -393.5 KJ/mol
Hof(PbCO3(s)) = -699.1 KJ/mol
we have the Balanced chemical equation as:
PbO(g) + CO2(g) ---> PbCO3(s)
deltaHo rxn = 1*Hof(PbCO3(s)) - 1*Hof( PbO(g)) - 1*Hof(CO2(g))
deltaHo rxn = 1*(-699.1) - 1*(-219.0) - 1*(-393.5)
deltaHo rxn = -86.6 KJ/mol
we have:
Sof(PbO(g)) = 66.5 J/mol.K
Sof(CO2(g)) = 213.6 J/mol.K
Sof(PbCO3(s)) = 131.0 J/mol.K
we have the Balanced chemical equation as:
PbO(g) + CO2(g) ---> PbCO3(s)
deltaSo rxn = 1*Sof(PbCO3(s)) - 1*Sof( PbO(g)) - 1*Sof(CO2(g))
deltaSo rxn = 1*(131.0) - 1*(66.5) - 1*(213.6)
deltaSo rxn = -149.1 J/mol.K
deltaH = -86.6 KJ/mol
deltaS = -149.1 J/mol.K
= -0.1491 KJ/mol.K
T = 647 K
we have below equation to be used:
deltaG = deltaH - T*deltaS
deltaG = -86.6 - 647.0 * (-0.1491)
deltaG = 9.87 KJ/mol
Answer: 9.87 KJ/mol