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ΔH°f CO2 = - 395 kJ
ΔH°f H2O = - 285 kJ
ditanya: ΔH°f C2H2 ...?
jawab: ΔH
ΔH = {Σ ΔH°f produk - Σ ΔH°f reaktan}
ΔH = {(4 . ΔH°f CO2 + 2 . ΔH°f H2O) - (2 . ΔH°f C2H2)}
- 2600 = {(4 . (-395) + 2 . (-285)) - (2 . ΔH°f C2H2)}
- 2600 = {(-1580 - 570) - (2 . ΔH°f C2H2)}
-2600 = -2150 - (2 . ΔH°f C2H2)
-2600 + 2150 = -2 . ΔH°f C2H2
- 450 = -2 . ΔH°f C2H2
ΔH°f C2H2 = - 450 / - 2
ΔH°f C2H2 = + 225 kJ