Page 81 - E-MODUL PEMBELAJARAN TERMOKIMIA BERBASIS PBL

Ditanya: ∆H Reaksi

Jawab: 2C2H2 (g) + 5 O2 (g) → 4 CO2 (g) + 2 H2O (g)
o
o
∆H Reaksi = ∑∆H f produk - ∑∆H f reaktan
o
o
o
o
∆H Reaksi = (4 . ∆H f CO2 + 2. ∆H f H2O) – (2 . ∆H f C2H2 + 5 . ∆H f O2)
∆H Reaksi = (4 . (-393,5 kJ) + 2. (-242 kJ)) – (2 . 227 kJ) + 5 . 0 kJ)
∆H Reaksi = (-1.574 kJ - 484 kJ) - (454 kJ + 0 kJ)
∆H Reaksi = -2.058 kJ – 454 kJ
∆H Reaksi = -2.512 kJ

52 1
• n C2H2 = = mol = 0,5 mol
26 2
Maka kalor yang dibebaskan pada pembakaran 52 gram C2H2 adalah:
= 0,5 x -2.512 /2

= -628 kJ

14. Diketahui:

–1
Reaksi pembakaran C2H2(g) ΔH = –1.256 kJ.mol
–1
o
∆H f CO2 (g) = –393,5 kJ.mol
o
∆H f H2O (g) = –242 kJ.mol
–1
o
Ditanya: ∆H f C2H2
Jawab: C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(g)

o
o
∆H Reaksi = ∑∆H f produk - ∑∆H f reaktan
o
∆H Asetilena = (2 . ∆H f CO2 + 1. ∆H f H2O) – (1 . ∆H f C2H2 + 5/2. ∆H f O2)
o
o
o
–1
–1
–1
o
–1.256 kJ.mol = (2 (–393,5 kJ.mol ) + 1(–242 kJ.mol )) – (1.∆H f C2H2 + 5/2.0)
o
–1.256 kJ.mol = (-787 kJ.mol + (-242 kJ.mol )) – ∆H f C2H2
–1
–1
–1
∆H f C2H2 = -1.029 kJ.mol + 1.256 kJ.mol –1
–1
o
o –1
∆H f C2H2 = +227 kJ.mol
o
–1
Jadi, nilai ∆H f C2H2 yakni +227 kJ.mol

15. Diketahui:
N2(g) + 3O2(g) + H2(g) → 2HNO3 (aq) ∆H = -207,1 kJ
N2O5(g) + H2O(g) → 2HNO3(aq) ∆H = +218,5 kJ
2H2(g) + O2(g) → 2 H2O (g) ∆H = -286,4 kJ
Ditanya: ∆H reaksi 2N2(g) + 5/2O2(g) → 2N2O5(g)

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