equilibrium value

[stevvo111]

So I found a website that said when you "double an equation" your Keq is squared.

Now I always thought the Keq only varied with temperature....not concentration?? And when you "double" the equation you are essentially doubling the concentration right?

Am I missing something here? I mean I understand if you double everything you do get a Keq value that is the square of the original, but just not totally getting why this is the case. Since you have double of everything and since this change is not related to temperature, I would have assumed the Keq would not change... but surprisingly it does... why? Does Keq change to things other than temperature?

A+B--->C+D

new equation

2A+2B---->2C+2D


Any help?!

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[cheesier]

You're not changing concentration in this case. You are changing stoichiometric coefficients which means the whole reaction changes, therefore Keq changes. You're right that if you had just double the concentration of everything in the first reaction, Keq would not have changed, but concentration and coefficients aren't related.

 

[theCLE]

So I found a website that said when you "double an equation" your Keq is squared.

Now I always thought the Keq only varied with temperature....not concentration?? And when you "double" the equation you are essentially doubling the concentration right?

Am I missing something here? I mean I understand if you double everything you do get a Keq value that is the square of the original, but just not totally getting why this is the case. Since you have double of everything and since this change is not related to temperature, I would have assumed the Keq would not change... but surprisingly it does... why? Does Keq change to things other than temperature?

A+B--->C+D

new equation

2A+2B---->2C+2D


Any help?!

I can't think of a time when you would want to have 2's in front of each reactant and product. To my knowledge, the lowest common denominator is best. why is this important? can you post the webpage? edit: I mean, I understand you're trying to figure out when Keq changes. If it means anything, I think Keq changes when pressure changes too. This is because of a change in temperature though. I suppose you could relate any of this to PV=nRT if they say temp changes?

 

[stevvo111]

I found notes on this website. lol , it's kinda sketch, but I figured I'd ask.

https://docs.google.com/viewer?a=v&...pVsQLU&sig=AHIEtbTvCamwIcdYxdphseUqaa_E0keimA

 

[cheesier]

You would multiply everything by 2 if you wanted to add the equation with another equation and find the Keq. For instance, let's say you have:

A + B --> C + D

2C + 2D --> E

And want to find the Keq for 2A + 2B --> E

You'd have to change the first equation so that you can sum it with the second, and its Keq would change.

 

[DNA]

Gosh, it's been so long since I even looked at this stuff, but like cheesier explained, it has to do which changing the stoichiometric coefficients.

Consider the basic chemical equation:

2A + B --> 3C

You would right the Keq for this reaction as:

Keq1= [C}^3 / [A]^2

But let's say, you double the coefficients to: 4A + 2B --> 6C

Then Keq expression would then be as follows:

Keq2 = [C]^6 / [A]^4 ^2

If you actually divided the second expression by the first, you'll see that essentially, all you're doing is squaring the original Keq1expression.

Also, keep in mind that the coefficients tell us absolutely nothing about the concentration of the reactants or products. It simply describes the chemical reaction itself taking place. You're probably wondering what exactly is the purpose in needing an equilibrium expression for the second reaction when you can easily just simplify it by divided by the less common denominator (in this case 2). We totally can do that, but if the question is asking you for the equilibrium expression for specific reaction: 4A + 2B --> 6C, then you better provide Keq for that reaction as well, regardless if it can be simplified.

 

[DNA]

So I found a website that said when you "double an equation" your Keq is squared.

Now I always thought the Keq only varied with temperature....not concentration?? And when you "double" the equation you are essentially doubling the concentration right?

Am I missing something here? I mean I understand if you double everything you do get a Keq value that is the square of the original, but just not totally getting why this is the case. Since you have double of everything and since this change is not related to temperature, I would have assumed the Keq would not change... but surprisingly it does... why? Does Keq change to things other than temperature?

A+B--->C+D

new equation

2A+2B---->2C+2D


Any help?!

By the way, you are right in realizing that temperature is the only thing that effects the equilibrium constant. But you need to realize here that doubling the coefficients would yield an entirely different reaction and therefore would require an entirely new equilibrium constant as a result.

 

[DNA]

I can't think of a time when you would want to have 2's in front of each reactant and product. To my knowledge, the lowest common denominator is best. why is this important? can you post the webpage? edit: I mean, I understand you're trying to figure out when Keq changes. If it means anything, I think Keq changes when pressure changes too. This is because of a change in temperature though. I suppose you could relate any of this to PV=nRT if they say temp changes?

That's an easy way to confuse yourself. Keep it simple. Only temperature effects Keq. If a reaction took place in a sealed (constant volume) container, then increasing pressure would indirectly increase temperature. An increase in temperature would then result in an alternative Keq value. It's the temperature itself changing the Keq, not the pressure.

To be honest, the whole PV=nRT is irrelevant. The question is simply about the proper way to calculate a Keq expression provided the coefficients for a given reaction is changing. It's not asking about what factors can effect a given Keq value.

 

[stevvo111]

By the way, you are right in realizing that temperature is the only thing that effects the equilibrium constant. But you need to realize here that doubling the coefficients would yield an entirely different reaction and therefore would require an entirely new equilibrium constant as a result.

Sweet, thanks dude! I understand now!