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.

Question

Chemistry

Messiah Welch

5 May, 19:58

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.

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Theodore York · Answer 1

The equilibrium constant Kp = 29.68

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