Page 82 - E-MODUL PEMBELAJARAN TERMOKIMIA BERBASIS PBL

Jawab:

(1) N2(g) + 3O2(g) + H2(g) → 2HNO3 (aq) ∆H = -207,1 kJ

(2) N2O5(g) + H2O(g) → 2HNO3(aq) ∆H = +218,5 kJ dibalik

(3) 2H2(g) + O2(g) → 2 H2O (g) ∆H = -286,4 kJ

Eliminasi persamaan reaksi 1 dan 2

(1) N2(g) + 3O2(g) + H2(g) → 2HNO3 (aq) ∆H = -207,1 kJ
(2) 2HNO3(aq) → N2O5(g) + H2O(g) ∆H = -218,5 kJ

+
(4) N2(g) + 3O2(g) + H2(g) → N2O5(g) + H2O(g) ∆H = -425,6 kJ

Eliminasi persamaan reaksi 4 dan 3

(4) N2(g) + 3O2(g) + H2(g) → N2O5(g) + H2O(g) ∆H = -425,6 kJ dikali 2

(3) 2H2(g) + O2(g) → 2 H2O (g) ∆H = -286,4 kJ dibalik

(4) 2N2(g) + 6O2(g) + 2H2(g) → 2N2O5(g) + 2H2O(g) ∆H = -851,2 kJ

(3) 2H2O (g) → 2H2(g) + O2(g) ∆H = +286,4 kJ

+
(5) 2N2(g) + 5O2(g) → 2N2O5(g) ∆H = -564,8 kJ
Jadi, nilai ∆H reaksi 2N2(g) + 5/2O2(g) → 2N2O5(g) adalah sebesar -564,8 kJ

16. Diketahui:
-1
∆Hf CO(g) = -110,5 kJ.mol
-1
∆Hf CaCO3(s) = -1.206 kJ.mol
-1
∆Hf CH3OH(g) = -200,66 kJ.mol .
Ditanya: perubahan entalpi pada reaksi CO(g) + 2H2(g) → CH3OH(g)

Jawab:
o
∆H Reaksi = ∑∆H f produk - ∑∆H f reaktan
o
o
o
o
∆H Reaksi = (1 . ∆H f CH3OH ) – (1 . ∆H f CO + 2. ∆H f H2)
∆H Reaksi = (1 (-200,66 kJ.mol )) – (1.(-110,5 kJ.mol ) + 5/2.0))
–1
–1
∆H Reaksi = -200,66 kJ.mol - (-110,5 kJ.mol ) = -90,16 kJ.mol –1
–1
–1
–1
Jadi, nilai ∆H pada reaksi CO(g) + 2H2(g) → CH3OH(g) yakni -90,16 kJ.mol

17. Diketahui:
Massa urea = 15 gram V air = 200mL
Mr urea = 60 ∆T = -7˚C

c = 4,2 J.gr ˚C ρair = 1 gram/mL
-1
-1

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