Question

Determine deltaS_rxn, deltaH_rxn, deltaG_rxn (each in kJ mol^-1 to two decimal places) and determine the spontaneity for the following reaction (at 25C):

CH₃OH (l) + O₂ ---> CO₂ (g) + H₂O

Answer 1

we have:

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

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

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

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

we have the Balanced chemical equation as:

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

deltaHo rxn = 1*Hof(CO2(g)) + 2*Hof(H2O(l)) - 1*Hof( CH3OH(l)) - 2*Hof(O2(g))

deltaHo rxn = 1*(-393.509) + 2*(-285.83) - 1*(-238.66) - 2*(0.0)

deltaHo rxn = -726.509 KJ/mol

we have:

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

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

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

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

we have the Balanced chemical equation as:

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

deltaSo rxn = 1*Sof(CO2(g)) + 2*Sof(H2O(l)) - 1*Sof( CH3OH(l)) - 2*Sof(O2(g))

deltaSo rxn = 1*(213.74) + 2*(69.91) - 1*(126.8) - 2*(205.138)

deltaSo rxn = -183.516 J/mol.K

So we have:

deltaHo rxn = -726.509 KJ/mol

deltaSo rxn = -183.516 J/mol.K

= -0.18352 KJ/mol.K

T = 298 K

Now use:

deltaGo rxn = deltaHo rxn - T*deltaSo rxn

= -726.509 - 298.0*-0.18352

= -671.8212 KJ/mol

Since deltaGo rxn is negative, the reaction is spontaneous

deltaHo rxn = -726.51 KJ/mol

deltaSo rxn = -183.52 J/mol.K

deltaGo rxn = -671.82 KJ/mol

spontaneous