Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Big-Daddy on April 10, 2013, 08:32:32 AM
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Does changing the pressure significantly change the Kc or Kp of a reaction, or does it only change the immediate reaction quotient Q to help reach equilibrium faster? i.e. will the final yield be the same with increased pressure, or will it change?
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I think you're confusing partial pressure of a reactant with the overall pressure (in a reaction vessel). The overall pressure can shift an equilibrium just like temperature can. (Consider the effect of pressure on an equilibrium like H2O(l) ::equil:: H2O (g) Atmospheric pressure changes the boiling point, no?
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I see. So like temperature, changing pressure changes both K and Q. Is there a quantitative way to understand this? (similar to the ΔG=ΔG°+R·T·loge(Q) relation we discussed recently, perhaps)
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Sorry - I misunderstood the question. Pressure can shift equilibrium, but for most pressure changes, equilibrium constant itself is reasonably independent of pressure, particularly for ideal gasses, no matter if the pressure change is due to change in volume or addition of inert gas. If gasses deviate significantly from ideal behavior though (as at really high pressures), this may not be the case.
Also as a quick aside, be careful using delta G in situations with pressure changes at constant volume. It's not encountered much in chemistry labs outside of industry, but in these situations Helmholtz energy would be used.
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Sorry - I misunderstood the question. Pressure can shift equilibrium, but for most pressure changes, equilibrium constant itself is reasonably independent of pressure, particularly for ideal gasses, no matter if the pressure change is due to change in volume or addition of inert gas. If gasses deviate significantly from ideal behavior though (as at really high pressures), this may not be the case.
Also as a quick aside, be careful using delta G in situations with pressure changes at constant volume. It's not encountered much in chemistry labs outside of industry, but in these situations Helmholtz energy would be used.
What about Kp? Is it the partial pressures which are always fixed in the gaseous sytem, or the mole fractions? (Because if it is partial pressure, then Kp must not have any dependence on total pressure, because the partial pressures will always be reset, but if it is, then Kp would be dependent on changes to total pressure.)
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Also as a quick aside, be careful using delta G in situations with pressure changes at constant volume. It's not encountered much in chemistry labs outside of industry, but in these situations Helmholtz energy would be used.
That's an interesting comment. Can you describe a situation and how you'd use a Helmholtz approach?
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@BigDaddy
Perhaps this will help: http://www.everyscience.com/Chemistry/Physical/Equilibria/c.1062.php
@Curiouscat
Unfortunately not in any great detail. A chemical engineer would probably be a better person to ask, but I believe a large-scale reaction in a pressurized vessel would be subjected to Helmholz free energy instead of Gibbs energy. (In a pressurized vessel where volume cannot change, there is no expansion work, so the enthalpic term essentially reduces to a change in internal energy.)
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@Curiouscat
Unfortunately not in any great detail. A chemical engineer would probably be a better person to ask, but I believe a large-scale reaction in a pressurized vessel would be subjected to Helmholz free energy instead of Gibbs energy. (In a pressurized vessel where volume cannot change, there is no expansion work, so the enthalpic term essentially reduces to a change in internal energy.)
Ok, thanks. So far as I remember, in all pressure vessel calculations I've used tabulations of Keq as a function of T. The rest being gas laws and a material / energy balance.
The Keq function itself should not change as a function of T no matter if derived via the Helmholtz or Gibbs approach.
Which is why the Helmholtz approach you mentioned seemed somewhat novel to me (as a Chemical Engineer :) )
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@BigDaddy
Perhaps this will help: http://www.everyscience.com/Chemistry/Physical/Equilibria/c.1062.php
Ah thanks, I see. So Kp is indeed based on partial pressures rather than mole fractions, meaning it is not really pressure dependent - the ratio will always be held constant.
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@curiouscat
To make sure I was answering the original question correctly, I did a fair amount of searching to see how Helmholz energy relates to the equilibrium constant Didn't find much. I suspect the relationship would be similar to that of the Gibbs energy. Just don't have a lot of experience using Helmholz energy. If anyone knows the answer, I'd love to hear it as well.
@BD
Assuming gasses behave ideally, yes.
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@BD
Assuming gasses behave ideally, yes.
Thanks. But the equilibrium yield is what we predict by Le Chatelier's Principle, and this is affected by pressure? (i.e. let's say we've got N2 + 3H2 :rarrow: 2NH3 and I increase the pressure; do I finally end up with more NH3 at equilibrium, or not?)
In fact, let's take this from the perspective of Kc so I can understand properly. If we define the final amount of NH3 (partial pressure or concentration) as the yield, then this will move in the opposite direction to the reaction quotient Q - i.e. if we make Q larger (e.g. adding to the products), then the final yield of products will decrease and yield of reactants will decrease, whereas if we make Q smaller (e.g. adding to the reactants), the final yield of products will increase and yield of reactants will decrease. But then how is the equilibrium constant being maintained?
e.g. N2 + 3H2 :rarrow: 2NH3
Let's say I add more moles of N2 to the mixture. We can assume the gas is ideal so Kp and Kc are independent of pressure and stay constant. Now the value of Q has become smaller (as we have added reactant), and to bring this back up to Kc and Kp the reaction is driven forward. But what does this mean for the equilibrium amounts of products and reactant in the solution? The equilibrium ratio will always stay the same (that is, equal to the equilibrium constants) but what about the amount (i.e. number of moles)? And analogously, what if I added moles of NH3 rather than N2?
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Thanks. But the equilibrium yield is what we predict by Le Chatelier's Principle, and this is affected by pressure? (i.e. let's say we've got N2 + 3H2 :rarrow: 2NH3 and I increase the pressure; do I finally end up with more NH3 at equilibrium, or not?)
Yes. More NH3.
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Yes. More NH3.
Thanks. To clarify, is this correct:
My initial feeling is that if we add more reactant and/or product (this can be raising concentration or raising partial pressure, doesn't matter; raising pressure raises partial pressure for obvious reasons) then we will finally have more reactant and product at the end than we would have had if we hadn't added anything; but if we add more reactant, the percentage of reactant that gets converted to product increases, whereas if we add product, the percentage of reactant that gets converted to product decreases, and this is what is meant by "yield". For pressure, if the LHS has more moles of gas, then when we cancel out pressure terms in the Kp equation we will find that increasing pressure means that Q will decrease, so in this case the equilibrium is driven towards the products to increase Q back to Kc and in doing so the percentage yield of products increases (as does the partial pressures of both reactants and products), whereas if the RHS has more moles then increasing pressure means Q will increase, so the equilibrium shifts towards the reactants and the percentage yield of products decreases.
Is what I've referred to as "percentage yield" essentially the same as when a textbook or exam paper says "yield" in this context?
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I tried reading what you wrote a couple of times without really understanding all of it. Sorry.
But I think you're right in general. You just restated Le Chatelier in a grossly convoluted way.
Perhaps you might want to use equations and numbers. That might be more precise.
In general, I take yield to mean the fraction of feed that got converted to the desired product. In the absence of side-products and recycle streams fractional yield should be the same as fractional conversion.
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Thanks. To clarify, is this correct:
My initial feeling is that if we add more reactant and/or product (this can be raising concentration or raising partial pressure, doesn't matter; raising pressure raises partial pressure for obvious reasons) then we will finally have more reactant and product at the end than we would have had if we hadn't added anything;
That itself seems inconsistent. When increasing pressure you are not adding anything new at all. In this case raising pressure will not finally lead to "more reactant" at all. (Assuming a decreasing moles reaction)
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Thanks. To clarify, is this correct:
My initial feeling is that if we add more reactant and/or product (this can be raising concentration or raising partial pressure, doesn't matter; raising pressure raises partial pressure for obvious reasons) then we will finally have more reactant and product at the end than we would have had if we hadn't added anything;
That itself seems inconsistent. When increasing pressure you are not adding anything new at all. In this case raising pressure will not finally lead to "more reactant" at all. (Assuming a decreasing moles reaction)
I was writing the general argument with relation to concentration. When dealing with gas, we would say partial pressures instead? (Which would be increased if you increased the total pressure or if you added more moles).
I would love to do this using equations instead of words. Let me try and clarify what I wrote: yeah your definition of yield, the fraction of reactants converted to products, seems good.
If we add either reactants or products, then our equilibrium concentrations of both reactants and products will be higher than before but the equilibrium ratio of concentrations of products over reactants will be maintained. Correct or not?
If we increase the pressure (while calculating Kp), or add gaseous moles of either reactant or product, our equilibrium partial pressures of both reactants and products will be higher than before but the equilibrium ratio of partial pressures of products over reactants will be maintained (as per Kp being constant). Correct or not?
Henceforth I will assume arguments applying to concentration and Kc also apply to partial pressure and Kp.
If we add reactant, the reaction will be driven towards the products (as Q<Kc) so equilibrium yield will increase. If we add product, the reaction will be driven towards the reactants (as Q>Kc) so equilibrium yield will decrease. Correct or not?
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If we add either reactants or products, then our equilibrium concentrations of both reactants and products will be higher than before but the equilibrium ratio of concentrations of products over reactants will be maintained. Correct or not?
Correct, to a good approximation. For the next question as well, if I'm reading it correctly.
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If we add either reactants or products, then our equilibrium concentrations of both reactants and products will be higher than before but the equilibrium ratio of concentrations of products over reactants will be maintained. Correct or not?
Correct, to a good approximation. For the next question as well, if I'm reading it correctly.
So all of those ones are right?
The problem with yield is that because the calculations can be decently long it's difficult to use the quantitative aspect to visualize the system qualitatively.