Question

1. Methanol is a high-octane fuel used in high performance racing engines. 2 CH3OH(l) + 3O2(g) → 2CO2(g) + 4 H20(g) a) Calculate ∆H० and ∆S० using thermodynamic data, and then ∆G

Answer 1

The reaction given in the question is:

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

The values of ΔH°formation and ΔS° of the reactants and products given in the reaction based on the thermodynamics data is:

ΔH°formation values of CH3OH (l) is -238.4 kJ/mol, CO2(g) is -393.52 kJ/mol, H2O (g) is -241.83 kJ/mol and O2 (g) is 0.

The S° values of CH3OH (l) is 127.19 J/molK, CO2(g) is 213.79 J/molK, H2O (g) is 188.84 J/moleK, and O2 (g) is 205.15 J/molK.

Now the values of ΔH° and ΔS° are,

ΔH°rxn = 2 * ΔH°formation CO2 (g) + 4 * ΔH°formation H2O (g) - 2*ΔH°formation CH3OH (l)

ΔH°rxn = 2 * (-393.52) + 4 (-241.83) -2 * (-238.4)

ΔH°rxn = -1277.56 kJ/mole

ΔS°rxn = 2 * S° CO2 (g) + 4 * S° H2O (g) - 2*S° CH3OH (l) - 3 * S° O2 (g)

ΔS°rxn = 2 * 213.79 + 4 * 188.84 - 2 * 127.19 - 3*205.15

ΔS°rxn = 313.11 J/mole/K

Now the formula for calculating ΔG°rxn is,

ΔG°rxn = ΔH°rxn - TΔS°rxn

ΔG°rxn = -1277.56 * 1000 J/mole - 298 * 313.11 J/mole

ΔG°rxn = -1370.86 kJ/mol