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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Mangoezzz on October 23, 2017, 09:40:22 PM

Title: Pressure in relation to Kc value in equilibrium
Post by: Mangoezzz on October 23, 2017, 09:40:22 PM

So the book states pressure doesn't affect the value of Kc

However, I'm struggling because if equilibrium shifts to favours a direction of one of the 2 reactions in order to decrease the pressure. Then to maintain the state of equilibrium it needs to consistently have more of that molecule and therefore a higher concentration which means the Kc value should change.

2SO₂ + O₂  ::equil:: 2SO₃

it will favour the forward reaction as there are fewer moles on that side and therefore a higher concentration as the reactants concentration will decrease to maintain the new adaption in reducing the pressure. So why doesn't Kc change?

Title: Re: Pressure in relation to Kc value in equilibrium
Post by: mjc123 on October 24, 2017, 04:44:30 AM

K_c does not change (at constant temperature). The concentrations change. If you increase the pressure, you increase the concentrations, and the equilibrium shifts so that the reaction quotient equals K_c, but the value of K_c doesn't change. That's the idea of an equilibrium constant.

Title: Re: Pressure in relation to Kc value in equilibrium
Post by: XeLa. on October 24, 2017, 08:11:22 AM

I think that mjc summed it up nicely, but maybe I could clarify a little further...

Quote from: Mangoezzz on October 23, 2017, 09:40:22 PM

it will favour the forward reaction as there are fewer moles on that side and therefore a higher concentration as the reactants concentration will decrease to maintain the new adaption in reducing the pressure. So why doesn't Kc change?

I feel like you almost have the concept down-packed here (albeit only a half of it), but remember that atoms can't be created or destroyed in a reaction. So if the concentration of sulfur trioxide is to increase to relieve stress in system (added pressure), then wouldn't it make sense that the concentration of reactants must decrease hencewith? This increase and decrease is affected by the K_c value, not the other way around!

N.b., Remember that K_c is a measure of the rate of reaction. Even if you think about it logically, changing pressure and concentration of species shouldn't affect it, should it?

I hope this helps,
XeLa

Title: Re: Pressure in relation to Kc value in equilibrium
Post by: Vidya on October 28, 2017, 09:37:44 AM

Quote from: Mangoezzz on October 23, 2017, 09:40:22 PM

So the book states pressure doesn't affect the value of Kc

However, I'm struggling because if equilibrium shifts to favours a direction of one of the 2 reactions in order to decrease the pressure. Then to maintain the state of equilibrium it needs to consistently have more of that molecule and therefore a higher concentration which means the Kc value should change.

2SO₂ + O₂  ::equil:: 2SO₃

it will favour the forward reaction as there are fewer moles on that side and therefore a higher concentration as the reactants concentration will decrease to maintain the new adaption in reducing the pressure. So why doesn't Kc change?

You know that Kp and Kc are related to each other.If system shifts in the direction of low pressure to maintain Kp ...it means Kc is also maintained.Now let us try to understand how you have increased the pressure
1) By decreasing the volume ...then it means you have increased the concentrations of reactants /products .Now increase is more in the denominator as there are two reactants so reaction shifts in the direction of the products.
2) If you have added more moles of the reactants and increased the pressure then again it will shift in the direction products.