What affects reactant/product ratios?

[confusedliberal]

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TBR chemistry question:

All of the following affect the reactant/product ratios of the equilibrium mixture EXCEPT:

A. increasing the volume of the reaction vessel
B. heating the reaction vessel
C. cooling the reaction vessel
D. adding reactant to the reaction vessel

Obviously B and C are incorrect. I know that pressure does not change equilibrium so A is the right choice.

But what about D? Why would adding reactant change the product and concentration ratios? Wouldn't the equilibrium constant thus change if that happens? I thought only temperature is the only thing that can change the equilibrium constant. Help?

 

[Doodl3s]

pressure MAY change the equilibrium constant if you're looking at gases...

EDIT: yea, just did a quick google.. if you go from 2 moles to 4 moles... increasing volume increases the "4 mole side" of the ratio... So considering A MIGHT change the equilibrium mixture ratios, and the fact that D NEVER should... I would actually be D... does the book say "A"?

 

[confusedliberal]

The correct answer is A. The solution says "adding moles of either reactant or product will shift the equilibrium to the opposite side". Still confused with this, anyone else can help?

 

[phEight]

Can you tell me what page this is on?

 

[phEight]

Maybe we're supposed to view this question as it asking what happens INSTANTLY after you add reactant into the vessel. Since equilibrium is already established, when reactant is initially added, the product/reactant ratios will be different. Imagine a reaction that occured very slowly, you would have to wait a long time for equilibrium to be re-established. The question isn't asking if the equilibrium constant changes, but whether the ratio changes.

 

[Doodl3s]

TBR chemistry question:

All of the following affect the reactant/product ratios of the equilibrium mixture EXCEPT:

Im sorry phEight, but that sounds an awful lot like the ratios AT equilibrium.... this is terribly worded for sure... horrible question.

 

[sv3]

i would have thought D for sure.
Pressure can change equilibrium as stated by an above poster. Sure its only for gases that pressure has this effect but the question doesn't state what form the compounds are in.
I see no reason why the answer wouldn't be D...even if you only consider the circumstances right after the change (the change of increasing volume or adding reactant)

very odd......

 

[sv3]

Maybe we're supposed to view this question as it asking what happens INSTANTLY after you add reactant into the vessel. Since equilibrium is already established, when reactant is initially added, the product/reactant ratios will be different. Imagine a reaction that occured very slowly, you would have to wait a long time for equilibrium to be re-established. The question isn't asking if the equilibrium constant changes, but whether the ratio changes.

Wouldnt changing the pressure also have this effect?
also, i think when the prep books or whatever say only Temp affects Keq, this is assuming some things stay constant, one being pressure.....

 

[Dr Gerrard]

Yeah, I would think the answer would be D.

Isn't Keq the reactant to produce ratio?

This stays constant at a given temperature.

Wouldn't change in pressure change the equilibrium constant, Kc?

Change in pressure would keep Kp constant, but I didn't think it would keep Kc constant.

 

[phEight]

Im sorry phEight, but that sounds an awful lot like the ratios AT equilibrium.... this is terribly worded for sure... horrible question.

Could very well be... I see the question stating there is some mixture at equilibrium, and then the answer choice says "Adding reactant to this mixture", so that makes me think you're adding reactant to something already at equilibrium. I just donno how else to make sense of it. In fact I'm not even sure what I'm talking about at all really

I remember answering this question and I got it right, and I hadn't even really thought about it... it just made sense to me that increasing volume won't do anything, it seems that's the BEST answer. D seems to have it's intricacies, but that isn't the BEST answer in this case. Maybe it's just practice in choosing the best answer

What if you look at this from the perspective of K and Q. When K = Q, the system is at equilibrium. When you add reactant, equilibrium is disturbed... K remains the same but Q is different. However, according to whether Q is less than or greater than K, the reaction is will proceed in the forward or reverse direction until equilibrium is reestablished. Q is also a measure of ratios of products and reactants, no?