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### Topic: Equilibrium partial pressures  (Read 5114 times)

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#### Pelle

• Guest ##### Equilibrium partial pressures
« on: September 28, 2005, 01:39:57 PM »
I am sitting with some exercises and having trouble with the calculation of equilibrium partial pressures.

The excercise is from Zumdahl's "Chemical Principles" 4th ed, ch. 6 ex. 31.

"At 1100 K, Kp= 0.25 atm-1 for the following reaction:

2SO2(g) + O2(g) <-> 2SO3(g)

Calculate the equilibrium partial pressures of SO2, O2 and SO3 produced from an initial mixture in which
P(SO2)=P(O2)=0.5 atm and P(SO3)=0"

Now, I get that the reaction will move right and that the equilibrium equation is:

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

I have tried to solve the resulting equation but I go wrong somewhere and I don't know where.

I am assuming that, at equilibrium, the new partial pressures are:

SO2: 0.5 - 2x
O2: 0.5-x
SO3: 2x

I believe I am doing something fundamentally wrong, but don't know where. I am pulling my hair so I would really appreciate some help.