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ΔH°f CO2 = - 393,5 kJ/mol
ΔH°f H2O = - 285,8 kJ/mol
ditanya: ΔH°f CH4 ...?
jawab: ΔH
ΔH°c CH4
CH4 + O2 → CO2 + H2O
CH4 + 2O2 → CO2 + 2H2O ΔH = - 890 kJ/mol
ΔH = {Σ ΔH°f produk - Σ ΔH°f reaktan}
ΔH = {(1 . ΔH°f CO2 + 2 . ΔH°f H2O) - (1 . ΔH°f CH4)}
- 890 = {(1 . (-393,5) + 2 . (-285,8)) - (1 . ΔH°f CH4)}
- 890 = {(-393,5 - 571,6) - (1 . ΔH°f CH4)}
- 890 = - 956,1 - (1 . ΔH°f CH4)
- 890 + 956,1 = -1 . ΔH°f CH4
+ 75,1 / -1 = ΔH°f CH4
ΔH°f CH4 = - 75,1 kJ/mol