Chemical Equilibrim question

[mohad]

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I feel like I'm being dumb here because I don't understand why I got this question wrong, but here it is. I would like some of your opinions on it.

The ratio of products to reactants at equilibrium for Reaction I will change with a change in the:

Reaction 1:
ATP + glutamate + NH3

ADP + phosphate + glutamine

A.concentration of ATP.
B.

G°
C. pressure
D. Temperature

The answer is temperature apparently, but I thought this was a basic Le'Chatalier's principle question so I picked A. The explanation given is "The only way to change a reaction's equilibrium constant is to change the temperature." and I don't even understand where they brought up the idea of an equilibrium constant since it was never mentioned in the question. Could someone explain why the answer is D please?

 

[BilalL]

I feel like I'm being dumb here because I don't understand why I got this question wrong, but here it is. I would like some of your opinions on it.

The ratio of products to reactants at equilibrium for Reaction I will change with a change in the:

Reaction 1:
ATP + glutamate + NH3

ADP + phosphate + glutamine

A.concentration of ATP.
B.

G°
C. pressure
D. Temperature

The answer is temperature apparently, but I thought this was a basic Le'Chatalier's principle question so I picked A. The explanation given is "The only way to change a reaction's equilibrium constant is to change the temperature." and I don't even understand where they brought up the idea of an equilibrium constant since it was never mentioned in the question. Could someone explain why the answer is D please?

A reaction at equilibrium is entirely dependent on the Keq which is different at different temperatures. The Keq is not dependent on delta G and very little with pressure.

These reactions proceed to equilibrium so if you add ATP the ratio will remain the same at equilibrium.

Le Chatliers principle in this sense just would refer to an increase or decrease on the rate of achieving equilibrium.

 

[TheRealAngeleno]

The equilibrium constant Keq = [Products] / [Reactants] . The only way to change the Keq costant is by temperature.

 

[mohad]

OH CRAP! Now I get what you're saying. I just thought that as soon as you add ATP, the reaction would shift right, making more product. But ultimately, at the new equilibrium, the ratio of products to reactants will be the same. now I get what you guys are saying by the ratio will change when the temperature changes. Thanks a bunch for clearing that up!

 

[MShopes]

I feel like I'm being dumb here because I don't understand why I got this question wrong, but here it is. I would like some of your opinions on it.

The ratio of products to reactants at equilibrium for Reaction I will change with a change in the:

Reaction 1:
ATP + glutamate + NH3

ADP + phosphate + glutamine

A.concentration of ATP.
B.

G°
C. pressure
D. Temperature

The answer is temperature apparently, but I thought this was a basic Le'Chatalier's principle question so I picked A. The explanation given is "The only way to change a reaction's equilibrium constant is to change the temperature." and I don't even understand where they brought up the idea of an equilibrium constant since it was never mentioned in the question. Could someone explain why the answer is D please?

It is the law of mass action!!!!! At equilibrium, the ratio of product concentrations to reactant concentration is constant at constant temperature. Therefore, changing this ratio would lead to change in temperature whether decreasing or increasing.

And you are not dump, you probably just forgot or thought too deep like we all do sometimes. By more practice, you will remember better 🙂

 

[MShopes]

I feel like I'm being dumb here because I don't understand why I got this question wrong, but here it is. I would like some of your opinions on it.

The ratio of products to reactants at equilibrium for Reaction I will change with a change in the:

Reaction 1:
ATP + glutamate + NH3

ADP + phosphate + glutamine

A.concentration of ATP.
B.

G°
C. pressure
D. Temperature

The answer is temperature apparently, but I thought this was a basic Le'Chatalier's principle question so I picked A. The explanation given is "The only way to change a reaction's equilibrium constant is to change the temperature." and I don't even understand where they brought up the idea of an equilibrium constant since it was never mentioned in the question. Could someone explain why the answer is D please?

By the way, the answer would be A if it said: An increase in the products (reaction shifts to the right) would occure with an increase in? Then the answer would be A because apparantely increase in a reactant concentration would shift the reaction to the right and therefore the product will increase. However, since this is equilibrium, the ratio stays the same. That is what the system actually is trying to do. When it gets more of a substance in one side, it sends it to the other side right away to balance both sides (equilibrium). That is what Le'Chatalier's principle is actually. It states that when a constraint is applied to a system in equilbirum, the equilibrium will shift in a direction to encounter that constraint.

Hint: As long as the reaction does not include any state symbols (s, l, g) (solid, liquid, gas), then you can ignore the choice of pressure right away because pressure would depend on the presence of gaseous products or reactants but since we don't know the states, then we can ignore the pressure. And any ways the pressure does not change the equilibrium. It can only tell you in which direction the reaction will shift if a pressure is applied to it.

Also, Free energy change (Delta G) or Standard free energy change would be useless without the entropy change (Delta S) and the Enthalpy change (Delta H) so if you don't see these two, you can also eliminate Delta G right away. Now you eliminated two choices in 2 seconds. And again these factors doesn't change the equilibrium. They only tell you about shifting of reaction.

The details can go farther more but that will be too complicated. Just know that only thing that can affect the equilibrium is temperature!
It just depends on how fast can you shoot these concepts in your head into the question and solve it quickly. With practice, like I said, it will be faster!

 

[Pre-Med Village]

Adding reagent does cause A to form B, but this produces a negative free energy change not by shifting the equilibrium constant but by shift the current state of the system off of equilibrium. Le Chatelier's would tell you that, but the question is oddly stated, with a cloaked version of the equilibrium constant on the prompt - ratio of products to reagents, etc - so even though the reaction is no longer at equilibrium, the equilibrium constant doesn't change.

I don't really understand how to eliminate standard free energy change, except that this doesn't make sense, the standard free energy is a physical presupposition, so I guess that answer is eliminated versus temperature as being 'more worse' or something. My point is that if you could alter the standard free energy change of a reaction in a Godlike way, you would change the equilibrium constant, which directly depends on the standard free energy change at a give temperature.

Changing the temperature, however, does change the standard free energy change and thus the equilibrium constant, so that is the best answer. Changing the temperature alters the entropic impetus of heat flow between the system to the environment so that a new equilibrium occurs with a new equilibrium constant where microscopic heat flows are reversible. Le Chatelier's also predicts this change, but it is because increasing temperature favors the endothermic direction because heat flows out aren't as spontaneous at high temperature. Le Chatelier's tells you that changing the concentrations or changing temperature will alter the direction of a reaction but the interior logic in terms of the relationship of concentrations, equilibrium constant, and temperature is a little different in either case.

 

[liveoak]

how does temp change it?

it depends on deltaH right?

 

[MT Headed]

Yes, it depends on deltaH.

If the reaction is endothermic, heat is a sneaky little hidden reactant. If the reaction is exothermic, heat is a sneaky little hidden product.

As a reactant or product, it does shift the amounts of other reactants and products according to LeChat, in order to achieve the new equilibrium concentrations.

Since adding or removing heat will adjust the concentrations, and heat itself does not show up in the K=[products]/[reactants] equation, the net effect of a change in temperature is to change the value of K itself.

There are other formulas for K that do include temperature, like K=e^(-dG/RT), but temp is not in the K=[products]/[reactants] formula.

 

[Pre-Med Village]

There's a simple answer, which is just to give a formula, or a deeper answer.

Here's the formula -

or a bit better for my eyes -

I think the best way to get at the underlying scientific principle is to imagine the state of equilibrium much like I think Gibbs did when he was able to apply Carnot's work in general chemistry.

Think about what a system is like at equilibrium. We learn that the system is at equilibrium, so delta G is 0. What does this mean

delta G is 0, then delta H minus T delta S equals 0

uh huh

divide both sides by delta T to give a different perspective

delta H divided by T equals delta S at equilibrium. This is the meaning that the free energy change is 0

In other words, the equilibrium state describes the particular state of the system in which microscopic heat flows would be reversible, just as likely to flow from the system to the environment as the converse, so there is no direction in which change can occur to increase the entropy of the universe.

For example, when an exothermic reaction is releasing heat out into the surroundings, you can understand conceptually that it is more likely to go in that direction than for the heat to coalesce, enter the reaction vessel, and reverse the reaction, like wood unburning itself. Equilibrium ultimately has to do with entropy. The free energy depends on the comparison of the intrinsic entropy change versus the entropy change due to heat flows. At equilibrium the two are in balance so heat is just as likely to flow one way as the other.

When you change the temperature, however, the heat flowing out into the surroundings doesn't produce as big an entropy change in the environment. delta S equals Q divided by T at constant temperature, so when the temperature is greater, heat flowing out of the system doesn't cause as great an increase in the number of new microstate ensembles which are now possible, due to the heat flow, using the way statistical mechanics has of describing this.

These are pretty advanced ideas. For a lot of folks, Le Chatelier's is enough for the MCAT, except for the few really advanced types of questions, but the underlying reason is that higher temperature decreases the stakes of heat flow for entropy change. The reverse of the exothermic direction in terms of microscopic heat flows becomes proportionally more likely, so the equilibrium constant changes.

how does temp change it?

it depends on deltaH right?