Question

Use ΔGº =ΔHº -TΔSº to calculate ΔG (in kJ) at 298 K for : 2CO2(g) +4H2O(l) → 2CH3OH(l) + 3O2(g)

If the above reaction could be done at 2081 K, what would be your estimate for ΔGº (in kJ) at this elevated temperature?

Use ΔGº =ΔHº -TΔSº and assume ΔHº and ΔSº are independent of temperature.

(The º is included because it is still for standard conditions, that is, 1 atm for gases and 1 molar for concentrations.)

Answer 1

we have:

Hof(CO2(g)) = -393.509 KJ/mol

Hof(H2O(l)) = -285.83 KJ/mol

Hof(CH3OH(l)) = -238.66 KJ/mol

Hof(O2(g)) = 0.0 KJ/mol

we have the Balanced chemical equation as:

2 CO2(g) + 4 H2O(l) ---> 2 CH3OH(l) + 3 O2(g)

deltaHo rxn = 2*Hof(CH3OH(l)) + 3*Hof(O2(g)) - 2*Hof( CO2(g)) - 4*Hof(H2O(l))

deltaHo rxn = 2*(-238.66) + 3*(0.0) - 2*(-393.509) - 4*(-285.83)

deltaHo rxn = 1453.018 KJ/mol

we have:

Sof(CO2(g)) = 213.74 J/mol.K

Sof(H2O(l)) = 69.91 J/mol.K

Sof(CH3OH(l)) = 126.8 J/mol.K

Sof(O2(g)) = 205.138 J/mol.K

we have the Balanced chemical equation as:

2 CO2(g) + 4 H2O(l) ---> 2 CH3OH(l) + 3 O2(g)

deltaSo rxn = 2*Sof(CH3OH(l)) + 3*Sof(O2(g)) - 2*Sof( CO2(g)) - 4*Sof(H2O(l))

deltaSo rxn = 2*(126.8) + 3*(205.138) - 2*(213.74) - 4*(69.91)

deltaSo rxn = 161.894 J/mol.K

So we have:

deltaHo rxn = 1453.018 KJ/mol

deltaSo rxn = 161.894 J/mol.K

= 0.16189 KJ/mol.K

T = 2081 K

Now use:

deltaGo rxn = deltaHo rxn - T*deltaSo rxn

= 1453.018 - 2081.0*0.16189

= 1116.2 KJ/mol

Answer: 1116.2 KJ/mol