Calculate ΔS°for the combustion of methanol.
2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)
Substance | CH3OH(g) | O2(g) | CO2(g) | H2O(g) |
S°(J/K·mol) | 240 | 205.138 | 213.74 | 188.825 |
-42.6 J/K |
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-87.3 J/K |
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42.6 J/K |
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87.3 J/K |
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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