Question

The balanced equation for the combustion of methanol is 2CH3OH(l)+3O2(g)→2CO2(g)+4H2O(g). Calculate ΔH∘rxn, ΔS∘rxn, and  ΔG∘rxn at 25 ∘C. Is it spontaneous or nonspontaneous?

Answer 1

Balanced equation:

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

ΔH for the reaction at 25°C

ΔH = Enthalpy change of products - Enthalpy change of reactants

= [(2 mol) * (-393.509 kJ/mol) + (4 mol)*(-241.83 kJ/mol)] - [(2 mol) * (238.4 kJ/mol) + (3 mol) *(0 kJ/mol)] = - 1277.5 kJ

ΔS for the reaction at 25°C

ΔS = entropy change in products - entropy change in reactants

= [(2 x 213.7) + (4 x 188.8)] - [(2 x 127.2) + (3 x 205.1)] = 312.9 J K⁻¹

ΔG for the reaction at 25°C

ΔG = ΔH - (T * ΔS)

= - 1277.5 x 10³ J - (298 x 312.9 J K⁻¹) = -1370.7 kJ

AS DELTA G is less than zero it is SPONTANEOUS REACTION

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