Effect of increasing temperature on Ka

[OneManShow]

So lets say we have
A + B --> X + Y + heat
(exothermic rxn)

The question is asking what would happen to Ka if temperature increases.
My answer was that Ka should increase since there are more products now, but the answer is Ka decreases.

Is it because heat is not considered in the Ka expression, and that an increase in heat just shifts the rxn to the left side? Just wanted to check.

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[Catburr]

Is it because heat is not considered in the Ka expression, and that an increase in heat just shifts the rxn to the left side? Just wanted to check.

Yup. By directly adding heat, you are directly adding a 'product', which will cause the equilibrium to shift toward reactants.

Important thing to note here is that Ka is the equilibrium constant, not the "present state of the reaction constant," so it represents the state of things when equilibrium is reached. You throw in a bunch of heat, its harder for a exothermic reaction to make products, so equilibrium will be reached with more reactants and less products present than before you messed with it.

If it's a reversible reaction (as acid-base reactions generally are), you get the same effect if you throw in some Y. Reaction goes backward, and at equilibrium you have more reactants, and thus a lower Ka.

 

[GreenRabbit]

If it's a reversible reaction (as acid-base reactions generally are), you get the same effect if you throw in some Y. Reaction goes backward, and at equilibrium you have more reactants, and thus a lower Ka.

You were right until here. Ka doesn't change if u throw in products, but it does vary with temperature.

 

[gunj122]

nvm.

 

[Catburr]

Rofl, guess I'm already losing this stuff a month after not studying anymore. Thanks for the fix. I better stick to questions about books/practice tests I still have lying around. 🙂

 

[GreenRabbit]

..and just to add a little more on that, Ka varies ONLY with temperature. it will not change if concentration, pressure, or volume are changed. PV=nRT equation explains it all.

I don't think the ideal gas law affects acid dissociation.

 

[banner18]

You were right until here. Ka doesn't change if u throw in products, but it does vary with temperature.

Why not? LeChat's principle would predict that the reaction would be favored to the right. So adding products will change the Q but not the Keq?

I'm getting really messed up when relearning themo vs. kinetic control in orgo, what are the hard and fast rules about factors that affect Keq. Is it only temperature, pressure and volume?

 

[gunj122]

I don't think the ideal gas law affects acid dissociation.

yea you're right. sorry guys, i meant to say equilibrium equation where it equals [products]/[reactants].

this website really cleared up any problems for me. it talks about what happens when you add reactant, change pressure, or change temperature.

http://www.chemguide.co.uk/physical/equilibria/change.html

 

[movax]

First, divorce kinetics and thermodynamics in your brain (this was messing me up for awhile). One tells you how fast a reaction will go, the other tells you whether it will happen or not.

Also when you say Ka, do you mean the Keq of the rxn? Thinking of Ka as an acid dissociation constant or something...

Now, for Kconsts. (Keq, Ksp, etc). Suffering through the EK practice exams on these yesterday, these constants only change with temperature. There are big books somewhere that list specific Keq/Ksp for specific reactions at specific temperatures. The only variable in those books are temperature. They just say 'for temps X, Y, and Z the Keq of A + B -> blah are J, K and L'. I'm relatively certain this statement is correct for our baby MCAT-level equilibria.

So in A + B -> X + Y + Heat, and assuming everything is gaseous/aqueous, Keq = [X][Y] / [A]. Increasing temp as previous posters have stated shove this reaction to the left based on Le Chatlier's principle (the only reason we can apply this is because the equation is already in dynamic equilibrium. I just got burned on this on a practice test because I didn't read closely enough to realize that there was no equilibrium. You can't use Le Chatlier's if you don't have equilibrium.).

So, shoving reaction to the left would increase concentrations of [A] and . In my head, if I had say [1][2] / [3][4] and I now have [1][2] / [6][8], Keq decreases. (2/12 > 2/48).

This is also (I think) a good point to relearn why if we mess with concentrations, the Keq remains the same. That website that was linked is real helpful. Keq wants to remain constant when concentrations are changed at a constant temperature. The concentrations of the other products / reactants will adjust to maintain the same Keq.

Finally a catalyst doesn't do **** except lower Ea. Since it is not a change in temperature, Keq stays the same. Since it speeds up both the forward AND the reverse reaction the same, the position of equilibrium also stays the same.

After typing this and going over this in my head I feel confident saying that for our MCAT-level science, nothing except temperature will mess with our Kx values...just watch out and make sure the system in question is in some kind of equilibrium before trying to use Le Chatlier. A common trap for this is solubility questions (so Ksp), where the question stem disguises the fact that the solution may no longer be saturated (i.e. "the solid was filtered, then..."). Equilibrium is lost, so be careful.

 

[GreenRabbit]

His mention of thermodynamic vs kinetic control concerns additions to conjugated double bonds when there is more than one possible product. Thermodynamic control generally occurs in high temperature when the reversible reaction is possible (and equilibrium occurs quickly). Kinetic control occurs at low temperature when the fastest product will predominate. Which product is the thermodynamic product and which is the kinetic product depends on the structure of the product and the mechanism of the reaction.

 

[movax]

His mention of thermodynamic vs kinetic control concerns additions to conjugated double bonds when there is more than one possible product. Thermodynamic control generally occurs in high temperature when the reversible reaction is possible (and equilibrium occurs quickly). Kinetic control occurs at low temperature when the fastest product will predominate. Which product is the thermodynamic product and which is the kinetic product depends on the structure of the product and the mechanism of the reaction.

Right, I understand that as it pertains to Organic reactions (thanks for that refresher though, clarifies that even further. Kinetic predominates at lower temp, thermo at higher but identifying each one is of course reaction specific as you said).

I think for inorganic though, its a good idea to separate the concepts in your brain initially, since those questions tend to skew more towards 'will this change affect x, what would change when y varies, etc.' and sometimes there are red herrings describing a kinetics-related affect for a question asking about thermodynamics.