Dalam wadah bervolume 2liter diisi 0,7 mol gas SO2 dan 0,4 mol gas O2 sehingga terjadi reaksi kesetimbangan: 2SO2(g) + O2 (g) = 2SO3(g) delta H = +189 kj/mol Ji

2SO2 + O2 => 2SO3
M 0.7. 0.4
T. 0.6. 0.3. 0.6
S. 0.1. 0.1. 0.6

[SO2] = 0.1/2 = 0.05 Molar
[O2] =0.1/2 = 0.05 Molar
[SO3] = 0.6/2 = 0.3 Molar

Kc= [SO3]^2/([SO2]^2[O2])
= (0.3)^2/((0.05)^2(0.05))
= 720

~Keseimbangan Kimia~

Data
V = 2 L
Gas SO₂ = 0,7 mol
Gas O₂ = 0,4 mol
Gas SO₂ saat seimbabg = 0,1

Ditanya
Kc =__?

Jawab
1) Tentukan proses reaksinya
.............2SO₂ + ....O₂ ⇆ 2SO₃
Awal = 0,7..........0,4...........-
R = ......0,6..........0,3...........0,6
S = .......0,1...........0,1............0,6

2) Tentukan konsenterasi mol
[SO₂] = Mol setimbang / V = 0,1 / 2 = 0,05 mol
[O₂] = Mol setimbang / V = 0,1 / 2 = 0,05 mol
[SO₃] = Mol Setimbang / V = 0,6 / 2 = 0,3 mol

3) Tentukanlah Kc nya
Kc = [SO₃]² / [O₂] [SO₂]²
Kc = [0,3]² / [0,05] [0,05]²
Kc = [0,09] / [0,00125]
Kc = 720