At 1100 K, Kp = 0.25 atm-1 for the following reaction. 2 SO2(g) + O2(g) 2 SO3(g)?

Question

Calculate the equilibrium partial pressures of SO2, O2, and SO3 produced from an initial mixture in which PSO2 = PO2 = 0.50 atm and PSO3 = 0.
PSO2=
PO2=
PSO3 =

Answer 1

Kp = 0.25 /atm = [SO3(g)] / {[SO2(g)]*[O2(g)]}
Let the equilibrium partial pressures of SO3 be X atm
.............2 SO2(g) + O2(g) <==> 2 SO3(g)
Initial:...0.50............0.50..............0..
equal:..0.50-X........0.50-0.5X.....X..
Hence: 0.25 = X^2/ {(0.50-X)^2*(0.50-0.5X)}
(1-2X)^2(1-X) = 32X^2
1 - 4X + 4X^2 -X + 4X^2 -4X^3 = 32X^2
4X^3 + 24X^2 + 5X - 1 = 0
X = 0.1243
Hence the equilibrium partial pressures of SO3: 0.12 atm;
the equilibrium partial pressures of SO2: 0.38 atm;
and the equilibrium partial pressures of O2: 0.44 atm.

Answer 2

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