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smetin92

? help ?

I posted this a week ago...

This is a classic question that comes up often. The issue is that to increase the pressure for a fixed mass of gas at constant temperature you MUST decrease the volume. Hence the concentration is being afffected and you know that concentration does not affect the equilibrim constant.

However to see the logic mathematically suffice to represent the partial pressures at equilibrium of N2O4 and 2NO2 in the equation below as x and y respectively.

N2O4 <==> 2NO2

by the equlibrium law Kp = y2/x

Now if we double the pressure y becomes 2y and x becomes 2x

The equilibrium expression = (2y)2/2x = 4y2/2x = 2y2/x

In other words the effect of increasing the pressure has disturbed the equilibrium and the numerator (y) is now double what it should be. The equilibrium respoonds by decreasing y and increasing x, i.e. it moves towards the side of fewer moles (Le Chatelier) in order to restore the value of Kp.

Hence a change in pressure causes a response from equilibria with an unequal number of moles of gas on either side BUT ONLY to retore the value of the equilibrium constant.

Irrelevant note: If I have spelled equilibrium incorrectly occasionally it's because I find it THE most irritating word to type - it usually comes out equilibrum...

Original post by charco

Hence a change in pressure causes a response from equilibria with an unequal number of moles of gas on either side BUT ONLY to retore the value of the equilibrium constant.

Irrelevant note: If I have spelled equilibrium incorrectly occasionally it's because I find it THE most irritating word to type - it usually comes out equilibrum...

Irrelevant note: If I have spelled equilibrium incorrectly occasionally it's because I find it THE most irritating word to type - it usually comes out equilibrum...

I can understand your dislike of equilibrium, but what is wrong with "restore"?

Original post by Pigster

I can understand your dislike of equilibrium, but what is wrong with "restore"?

I posted that eight years ago ...

... the 's' has probably fallen off or eroded away.

Original post by charco

I posted this a week ago...

This is a classic question that comes up often. The issue is that to increase the pressure for a fixed mass of gas at constant temperature you MUST decrease the volume. Hence the concentration is being afffected and you know that concentration does not affect the equilibrim constant.

However to see the logic mathematically suffice to represent the partial pressures at equilibrium of N2O4 and 2NO2 in the equation below as x and y respectively.

N2O4 <==> 2NO2

by the equlibrium law Kp = y2/x

Now if we double the pressure y becomes 2y and x becomes 2x

The equilibrium expression = (2y)2/2x = 4y2/2x = 2y2/x

In other words the effect of increasing the pressure has disturbed the equilibrium and the numerator (y) is now double what it should be. The equilibrium respoonds by decreasing y and increasing x, i.e. it moves towards the side of fewer moles (Le Chatelier) in order to restore the value of Kp.

Hence a change in pressure causes a response from equilibria with an unequal number of moles of gas on either side BUT ONLY to retore the value of the equilibrium constant.

Irrelevant note: If I have spelled equilibrium incorrectly occasionally it's because I find it THE most irritating word to type - it usually comes out equilibrum...

This is a classic question that comes up often. The issue is that to increase the pressure for a fixed mass of gas at constant temperature you MUST decrease the volume. Hence the concentration is being afffected and you know that concentration does not affect the equilibrim constant.

However to see the logic mathematically suffice to represent the partial pressures at equilibrium of N2O4 and 2NO2 in the equation below as x and y respectively.

N2O4 <==> 2NO2

by the equlibrium law Kp = y2/x

Now if we double the pressure y becomes 2y and x becomes 2x

The equilibrium expression = (2y)2/2x = 4y2/2x = 2y2/x

In other words the effect of increasing the pressure has disturbed the equilibrium and the numerator (y) is now double what it should be. The equilibrium respoonds by decreasing y and increasing x, i.e. it moves towards the side of fewer moles (Le Chatelier) in order to restore the value of Kp.

Hence a change in pressure causes a response from equilibria with an unequal number of moles of gas on either side BUT ONLY to retore the value of the equilibrium constant.

Irrelevant note: If I have spelled equilibrium incorrectly occasionally it's because I find it THE most irritating word to type - it usually comes out equilibrum...

Thanks for a great explanation! Do you mind if you also explain specifically why Kc is affected by temperature but pressure isn't?

(edited 7 years ago)

N2O4 <==> 2NO2

by the equlibrium law Kp = y2/x

Now if we double the pressure y becomes 2y and x becomes 2x

The equilibrium expression = (2y)2/2x = 4y2/2x = 2y2/x

WRONG. (2y)2/2x = 2y/x NOT 2y2/x. So you've proven that kc does not change with pressure instead.

by the equlibrium law Kp = y2/x

Now if we double the pressure y becomes 2y and x becomes 2x

The equilibrium expression = (2y)2/2x = 4y2/2x = 2y2/x

WRONG. (2y)2/2x = 2y/x NOT 2y2/x. So you've proven that kc does not change with pressure instead.

he just didn't have to the power of. (2y)^2/2x = 4y^2/2x = 2y^2/x

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