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Topic: Why does a change in pressure not change K (equilibrium constant)?  (Read 3187 times)

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Offline miaou5

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K can only be changed with a change in temperature, according to my prof, but I don't understand how K wouldn't also change with a change in pressure. Here's what I mean:

X (g) + Y (g) --> Z (g)

If I increase the pressure, this will shift the reaction toward the side with fewer gaseous moles (the right). The equilibrium constant is expressed as:

K = [Z]/([X][Y])

Thus the numerator will increase, and the denominator will decrease. So wouldn't K increase overall? I understand how K stays the same with changes in concentration, since any concentration changes are offset by the necessary changes in the numerator or denominator. But in the case of pressure, there's nothing to numerically offset the numerator increase/denominator decrease. So how would K stay the same?

Offline curiouscat

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Re: Why does a change in pressure not change K (equilibrium constant)?
« Reply #1 on: January 29, 2013, 05:35:23 AM »
There remains a Total Pressure term in the denominator that compensates.

Offline miaou5

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Re: Why does a change in pressure not change K (equilibrium constant)?
« Reply #2 on: January 29, 2013, 06:24:59 AM »
Thanks so much! Do first year college students not learn about this pressure value? I've never heard anything about including pressure terms in K, but maybe that's just something our professor never told us.  :-\

Offline curiouscat

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Re: Why does a change in pressure not change K (equilibrium constant)?
« Reply #3 on: January 29, 2013, 06:51:08 AM »
Thanks so much! Do first year college students not learn about this pressure value? I've never heard anything about including pressure terms in K, but maybe that's just something our professor never told us.  :-\

What's your equation for K?


Offline miaou5

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Re: Why does a change in pressure not change K (equilibrium constant)?
« Reply #5 on: January 30, 2013, 02:54:02 AM »
Given the equation X (g) + Y (g) --> Z (g), we've been taught that K = [Z]/([X][Y]) (the only equation we've been given). This is what I don't quite understand: if I increase the pressure and shift the reaction to the right, [Z] will increase and both [X] and [Y] will decrease, hence causing the overall value of K to increase. (There's nothing numerically to balance these changes out and keep K constant.) However, I also know that K isn't supposed to change with anything except for T, so I'm a bit confused.

(Also, I checked the Chegg site but unfortunately I think I need a prescription to view the materials. Thank you anyway though!)

Offline curiouscat

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Re: Why does a change in pressure not change K (equilibrium constant)?
« Reply #6 on: January 30, 2013, 03:10:00 AM »
Given the equation X (g) + Y (g) --> Z (g), we've been taught that K = [Z]/([X][Y]) (the only equation we've been given). This is what I don't quite understand: if I increase the pressure and shift the reaction to the right, [Z] will increase and both [X] and [Y] will decrease, hence causing the overall value of K to increase. (There's nothing numerically to balance these changes out and keep K constant.) However, I also know that K isn't supposed to change with anything except for T, so I'm a bit confused.

(Also, I checked the Chegg site but unfortunately I think I need a prescription to view the materials. Thank you anyway though!)

What is [Z]?

Offline miaou5

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Re: Why does a change in pressure not change K (equilibrium constant)?
« Reply #7 on: January 30, 2013, 08:44:29 PM »
Oops, my mistake--the brackets should be parentheses. So Kp = (Z)/((X)(Y)), where (Z) equals the partial pressure of Z, (Y) equals the partial pressure of Y, and (Z) equals the partial pressure of Z. So If pressure is increased, the reaction shifts to the left and (X) and (Y) decrease & (Z) increases, which at least appears to cause Kp to increase overall. I'm not sure where the total pressure term is?

Again, big thanks!

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