When the pressure is decreased, what happens to the equilibrium position of a reaction (to which direction ...
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When the pressure is decreased, what happens to the equilibrium position of a reaction (to which direction ...

[From: ] [author: ] [Date: 13-05-04] [Hit: ]
According to Le Chateliers Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. That means that if you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible. It can do this by favouring the reaction which produces the fewer molecules. If there are the same number of molecules on each side of the equation,......
... to which direction does it shift?


Thank you in advance. I really need to know how to do this.

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Changing pressure

This only applies to systems involving at least one gas.

The facts-
Equilibrium constants aren't changed if you change the pressure of the system. The only thing that changes an equilibrium constant is a change of temperature.
The position of equilibrium may be changed if you change the pressure. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made.
That means that if you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible. It can do this by favouring the reaction which produces the fewer molecules. If there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium.

Explanation-
Where there are different numbers of molecules on each side of the equation

Let's look at the same equilibrium we've used before. This one would be affected by pressure because there are 3 molecules on the left but only 2 on the right. An increase in pressure would move the position of equilibrium to the right.
Because this is an all-gas equilibriium, it is much easier to use Kp:
Once again, it is easy to suppose that, because the position of equilibrium will move to the right if you increase the pressure, Kp will increase as well. Not so!
To understand why, you need to modify the Kp expression.
Remember the relationship between partial pressure, mole fraction and total pressure?
Replacing all the partial pressure terms by mole fractions and total pressure gives you this:
If you sort this out, most of the "P"s cancel out - but one is left at the bottom of the expression.
Now, remember that Kp has got to stay constant because the temperature is unchanged. How can that happen if you increase P?
To compensate, you would have to increase the terms on the top, xC and xD, and decrease the terms on the bottom, xA and xB.

Increasing the terms on the top means that you have increased the mole fractions of the molecules on the right-hand side. Decreasing the terms on the bottom means that you have decreased the mole fractions of the molecules on the left.

That is another way of saying that the position of equilibrium has moved to the right - exactly what Le Chatelier's Principle predicts. The position of equilibrium moves so that the value of Kp is kept constant.
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