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What is ΔGrxno (in kJ) at 647 K for the following reaction? PbO(g) + CO2(g) →...

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)

Solutions

Expert Solution

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

queen_honey_blossom answered 2 years ago
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