Jawaban:

terurai = 50% x 2 mol = 1 mol

perbandingan mol terurai = perbandingan koefisien

2SO3(g) =>2SO2(g) + O2(g)

Mula2 : 2 mol

Terurai : 1 mol 1 mol 0,5 mol

setbang : 1 mol 1 mol 0,5 mol

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Volume = 1 L

konsentrasi = mol / volume

Kc => tetapan kesetimbangan konsentrasi

Kc = produk/reaktan

= [SO2]^2[O2] / [SO3]^2

masukan zat dalam keadaan setimbang

= [1]^2[0,5]/[1]^2

= 0,5

===================================

mol total dalam keadaan setimbang

= mol SO3 + mol SO2 + mol O2

= 1 × 1 + 0,5

= 2,5 mol

P° = tekanan total

P = tekanan parsial

P SO3 = P° x mol SO3/mol total

= 5 atm x 1/2,5

= 2 atm

P SO2 = P° x mol SO2/mol total

= 5 atm x 1/2,5

= 2 atm

P O2 = P° x mol O2/mol total

= 5 atm x 0,5/2,5

= 1 atm

Kp = tetapan kesetimbangan tekanan

Kp = produk/reaktan

= (P SO2)^2 (P O2) / (P SO3)^2

= (2)^2)(1) / (2)^2

= 4 x 1 / 4

= 1

^ dibaca pangkat