Question

Calculate ΔS°for the combustion of methanol. 2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g) Substance CH3OH(g) O2(g)...

Calculate Δfor the combustion of methanol.

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

Substance CH3OH(g) O2(g) CO2(g) H2O(g)
(J/K·mol) 240 205.138 213.74 188.825

-42.6 J/K

-87.3 J/K

42.6 J/K

87.3 J/K


Homework Answers

Answer #1

The given reaction is

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

The entropy change is given as

ΔS0 = Σn.S0(products) – Σn.S0(reactants) where n = number of moles

= [(2 mole)*S0(CO2,g) + (4 mole)*S0(H2O,g)] – [(2 mole)*S0(CH3OH,g) + (3 mole)*S0(O2,g)]

= [(2 mole)*(213.74 J/mol.K) + (4 mole)*(188.825 J/mol.K)] – [(2 mole)*(240 J/mol.K) + (3 mole)*(205.138 J/mol.K)]

= (427.48 + 755.3) J/K – (480 + 615.414) J/K

= 1182.78 J/K – 1095.414 J/K

= 87.366 J/K ≈ 87.3 J/K (ans).

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