Nov 10, 2009
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I thought that the value of Keq changes with temperature only...but is this only the case if the equation itself does not change?

So for example, is it true that...

X + Y --> A + B

has a different Keq than

2X + 2Y --> 2A + 2B

?

If so, why is this? Technically you are just increasing the reactant and product concentrations and you are not increasing temperature, right?
 
Nov 10, 2009
10
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0
Status
Pre-Health (Field Undecided)
I know that, with the second reaction, the value of Keq becomes squared. I understand the mathematical details of how Keq would increase, but I instead would like a conceptual explanation. Thanks!
 
Jun 14, 2009
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Pre-Dental
Are you talking about the equilibrium constant Keq, or the rate constant k?

There shouldn't be too much confusion about Keq. Anything that upsets the balance of a reaction in equilibrium (changing concentrations, temperature, pressure, etc) will change Keq. The point is that different reaction conditions establish a different Keq.

"the value of --- changes with temperature only" is more often associated with the rate constant property, k. k is temperature-specific and though not stated, it's implied that it is reaction-specific. When dealing with the same reaction kinetics, all variables that determine the value of k are constant except temperature. If you changed the order of the reactants from 1st order to second order, you indicate a different reaction mechanism, different frequency factor, and different activation energy, therefore a different value for k.

X+Y is a different reaction from 2X+2Y and can have different rate constants if the meaning behind the coefficients is that the first reaction is a simple "X reacts with Y" vs "two X's react with two Y's" You can use your imagination to make up an example of each. If this is the case, they are two different reactions and therefore have two different rate constants.

If you merely want to double the concentrations of the reactants and products, this does not directly affect the value of the rate constant of the forward or reverse reactions (k's don't change), but depending on the order of the reactants and products, may or may not change Keq. It is also not generally appropriate to indicate doubling of concentrations by fiddling with coefficients in the overall reaction expression.
 
Mar 11, 2010
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Pre-Medical
Equilibrium expression is definitely impacted by the stoichiometric coefficients; thus, so is the constant.
 
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BerkReviewTeach

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I thought that the value of Keq changes with temperature only...but is this only the case if the equation itself does not change?

So for example, is it true that...

X + Y --> A + B

has a different Keq than

2X + 2Y --> 2A + 2B

?

If so, why is this? Technically you are just increasing the reactant and product concentrations and you are not increasing temperature, right?
You are absolutely correct that the numerical value of Keq only changes with temperature. However, that value applies for a reaction exactly as written. If you change the phases of any component or as you have done, double all of the stoichiometric terms, then it's a new reaction and thus a new Keq. For instance, if we reverse a reaction, then then we need the recipricol of the original Keq for the new reaction (reverse reaction).

You should note that other changes (pressure, volume, moles of reactant, and moles of product) can shift the equilibrium, but aslong as the temperature remains constant, then the numerical value of Keq does not change.

Are you talking about the equilibrium constant Keq, or the rate constant k?

There shouldn't be too much confusion about Keq. Anything that upsets the balance of a reaction in equilibrium (changing concentrations, temperature, pressure, etc) will change Keq. The point is that different reaction conditions establish a different Keq.
I think you might be mixing up the equilibrium conditions with the numerical value of Keq. A great example involves pH of an aqueous solution at 298K. No matter what we do, add acid or add base, the sum of pH + pOH = 14, as long as the temperature does not change from 298K. This is because the autoionization of water has a Kw = 10^-14 at that temperature.

Adding base will shift the equilibrium so that the [H3O+] will go down and the [OH-] will go up, but the new values after the addition will still multiply out to be 10^-14.
 
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Jun 14, 2009
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Ah, now that I think about it, I was only thinking about Q before equilibrium conditions. Eventually when a reaction reestablishes equilibrium, it will be at Keq again. Thanks for catching that
 

BerkReviewTeach

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Ah, now that I think about it, I was only thinking about Q before equilibrium conditions. Eventually when a reaction reestablishes equilibrium, it will be at Keq again. Thanks for catching that
An honor to do follow up. I've seen some of the answers you've posted and they are excellent. You have a very systematic and logical way of looking at things. I assume you owned this test when you took it. Or if you havent taken it yet, that you will be one of the 40+ people.