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Topic: Kp and partial pressure  (Read 11143 times)

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Offline sparkly

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Kp and partial pressure
« on: March 23, 2009, 11:38:58 AM »
This is the last one. The book doesnt even mention how to do partial pressure problems in this chapter. I can apply some logic to the problem. Since the Kp is less than one, the reactants are favored, so I guess 24% of the CCl4 converts. But I don't know what to do with the partial pressures. Is there a basic tutorial you guys can direct me to?


10 At 700 K the equilibrium constant for the reaction

CCl4(g) <===> C(s) + 2 Cl2(g)

is Kp = 0.76. A flask is charged with 2.00 atm CCl4, which then reaches equilibrium at 700 K.

(a) What fraction of the CCl4 is converted into C and Cl2;

(b) What are the partial pressures of CCl4 and Cl2 at equilibrium?

Offline Vidya

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Re: Kp and partial pressure
« Reply #1 on: March 24, 2009, 10:59:25 AM »
This is simple.
You know the formula of Kp .
Let us say partial pressure of Cl2 is X.
now make an ICE table and get the value of X using Kp value .
This will also help you in solving next part of the question.

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