Page 83 - E-MODUL PEMBELAJARAN TERMOKIMIA BERBASIS PBL

Ditanya: Besarnya perubahan entalpi pelarutan urea

Jawab:
15
• Massa air = V . ρair • Mol = = = 0,25
60
Massa air = 200 mL . 1 g/mL = 200 gram

∆H = - = - ∆T = - 200 4,2 J.gr −1 ˚ −1 −7˚C
0,25
∆H = +23.520 J/mol = +23,52 kJ/mol

18. Diketahui:
Massa air dingin = 20 gram Massa air panas = 45 gram

∆T air dingin = 10˚C ∆T air panas = -25˚C
c = 4,18 J.gr ˚C Suhu kalorimeter = 7˚C
-1
-1
Ditanya: C (kapasitas kalorimeter)

Jawab: q lepas = q terima
q air panas = q air dingin + q kalorimeter

m air panas . c . ∆T air panas = m air dingin . c . ∆T air dingin + C. Suhu kalorimeter
-1
-1
-1
-1
45gram . 4,18 J.gr ˚C .-25˚C = 20gram . 4,18 J.gr ˚C .10˚C + C. 7˚C
-4.702,5 = 836 + 7C
7C = -4.702,5 – 836
7C = -5.538,5

C = -791,2

20. Diketahui:

C═C = +611 Kj C─C = +347 kJ
H─H = +436 kJ C─H = +414 kJ

Ditanya: ∆H reaksi CH2═CH2 +H2 → C2H6
Jawab:

H H
H ─ C ═ C ─ H + H ─ H → H ─ C ─ C ─ H

H H H H

∆H reaksi = ∑ D pemutusan ikatan - ∑ D pembentukan ikatan
∆H reaksi = ((4 x C─H) + (1 x C═C) + (1 x H─H)) – ((1 x C─C) + (6 x C─H))

∆H reaksi = ((4 x 414 kJ) + (1 x 611 kJ) + (1 x 436 kJ )) – ((1 x 347 kJ) + (6 x 414 kJ))
∆H reaksi = (1.656 kJ + 611 kJ + 436 kJ) – (347 kJ + 2.484 kJ))

∆H reaksi = 2.703 kJ – 2.831 kJ

∆H reaksi = -128 kJ

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