Question

Part A

What is ΔGrxno (in kJ) at 542 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)

Part B

At what temperature (in K) does the above reaction become spontaneous?

Answer 1

A)

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

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/K

deltaH = -86.6 KJ/mol

deltaS = -149.1 J/mol.K

= -0.1491 KJ/mol.K

T = 542 K

we have below equation to be used:

deltaG = deltaH - T*deltaS

deltaG = -86.6 - 542.0 * -0.1491

deltaG = -5.79 KJ/mol

Answer: -5.79 KJ/mol

B)

we have below equation to be used:

deltaG = deltaH - T*deltaS

for reaction to be spontaneous, deltaG should be negative

that is deltaG<0

since deltaG = deltaH - T*deltaS

so, deltaH - T*deltaS < 0

-86.6- T *-0.1491 < 0

T *0.1491 < 86.6

T < 581 K

Answer: the reaction is spontaneous for T< 581 K