Question

Consider the following reaction:2SO3(g) 2SO2(g) + O2(g)If 0.108 moles of SO3, 0.390 moles of SO2, and 0.282 moles of O2 are at equilibrium in a 13.2 L container at 1.30×103 K, the value of the equilibrium constant, Kp, is .

Answer 1

The equilibrium constant Kp = 29.68

Step-by-step explanation:

2SO3(g) ⇄ 2SO2(g) + O2(g)

Kp = (pSO2)²(pO2)/(pSO3)²

We can use ideal gas formula to find pressure of each: PV = nRT thus P=nRT/V,

with n=moles at equilibrium; R=0.082057L.atm.mol⁻¹.K⁻¹ (gas constant); T=1.30*10^3K; V=13.2L

After substituting each values:

P(SO3) = 0.873 atm

P(SO2) = 3.15 atm

P(O2) = 2.28 atm

Kp = (3.15²×2.28)/(0.873)² = 29.68