True or False

[SmurfinUSA]

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True or False:

When T is increased, Keq always increases.

why?

ty~~~~~~

 

[SmurfinUSA]

sorry for replying on my own post, but I'm thinking that the answer is false, because Keq depends on if the rxn is exo or endothermic.

R -> P + heat (exothermic)

R + heat -> P (endothermic)

R= reactants
P= products

So, if T is increased for both reactions, rxn 1 will not be favored, and rxn 2 will be favored. Does this mean Keq1 is decreased and Keq2 is increased?

Here, I treat heat as a product or a reactant, but I also know that products do not change the equilibrium constant, though they do change the final concentrations.

But we know that Keq is changed by Temperature!!! ahhhh!

 

[Rabolisk]

It has to be false, by logic.

Assume the statement to be true. Then whenever T is increased, K is increased for any reaction. But all equilibrium reactions can go both ways theoretically. The equilibrium constant for the reverse reaction, K' is by definition, 1/K. When T is increased, K is increased by assumption, which leads to K' being decreased. But K' is an equilibrium constant, leading to a contradiction. Hence the statement is not true.

 

[SmurfinUSA]

thanks rab

Is my explanation valid?

 

[ilovemcat]

sorry for replying on my own post, but I'm thinking that the answer is false, because Keq depends on if the rxn is exo or endothermic.

R -> P + heat (exothermic)

R + heat -> P (endothermic)

R= reactants
P= products

So, if T is increased for both reactions, rxn 1 will not be favored, and rxn 2 will be favored. Does this mean Keq1 is decreased and Keq2 is increased?

Here, I treat heat as a product or a reactant, but I also know that products do not change the equilibrium constant, though they do change the final concentrations.

But we know that Keq is changed by Temperature!!! ahhhh!

It seems as though you're on the right track, but I'm confused what you meant by the last part. First note that a change in temperature will change Keq. How a change in temperature (addition of heat) affects the reaction you're considering depends whether the reaction is endothermic or exothermic.

Exothermic Reactions favor Products, because Keq > 1
Endothermic Reactions favor Reactants, because Keq < 1

Keq = [Products]/[Reactants]

An endothermic reaction (heat is a reactant) will proceed in the forward direction with the addition of (more) heat.
More products are formed to use up heat. Keq will increase.

An exothermic reaction (heat is a product) will proceed in the reverse direction with the addition of (more) heat.
More reactants are formed to use up heat. Keq will decrease.

Like Rabolisk said, the statement is false.

 

[SmurfinUSA]

Thanks for the clarification guys,

One more..

T/F:

There exists a reaction where with the addition of heat, no change in Keq will be present.

 

[ilovemcat]

Thanks for the clarification guys,

One more..

T/F:

There exists a reaction where with the addition of heat, no change in Keq will be present.

That's impossible. Maybe Rabolisk could correct me if I'm wrong, but from my understanding temperature always changes Keq.

 

[Rabolisk]

Here, I treat heat as a product or a reactant, but I also know that products do not change the equilibrium constant, though they do change the final concentrations.

What he means is that the addition of a product or a reactant does not change K, although it does change the equilibrium concentrations of the species. When one treats heat as a product/reactant, the difference is that it does change the K.

 

[BerkReviewTeach]

Thanks for the clarification guys,

One more..

T/F:

There exists a reaction where with the addition of heat, no change in Keq will be present.

If the reactant and product are of equal energy and equal entropy, then in theory it could be true. Consider the ring flip process for cyclohexane. The two conformers are identical in energy and entropy, so no matter how much you heat the system or cool the system, Keq will always be 1. While this isn't exactly a reaction, the ring flip process can be described as an equilibrium process and assigned a Keq. You will always have equal parts of the two ring structures, although the reality is that you can't distinguish one from the other, so this is purely an exercise in thought and not practice.

 

[ilovemcat]

If the reactant and product are of equal energy and equal entropy, then in theory it could be true. Consider the ring flip process for cyclohexane. The two conformers are identical in energy and entropy, so no matter how much you heat the system or cool the system, Keq will always be 1. While this isn't exactly a reaction, the ring flip process can be described as an equilibrium process and assigned a Keq. You will always have equal parts of the two ring structures, although the reality is that you can't distinguish one from the other, so this is purely an exercise in thought and not practice.

That's interesting. By the way, I read somewhere that at room temperature, there's an energy barrier of about 11 kcal/mol between two chair conformations of cyclohexane. Not sure how I remember that or if it's even important to know, lol. Regardless though, I understand the point you were making 😛

 

[kupa]

If the addition of heat is not enough to break the chemical bond of the compound, there will be no change in the concentration of the products and reactants.. (referring to a simple decomposition process)

 

[justhanging]

That's interesting. By the way, I read somewhere that at room temperature, there's an energy barrier of about 11 kcal/mol between two chair conformations of cyclohexane. Not sure how I remember that or if it's even important to know, lol. Regardless though, I understand the point you were making 😛

The energy barrier is there because it has to go through the boat conformation and there is an activation energy associated with changing to the boat conformation. As far as the original problem I always think of Keq as the ratio of the forward rxn k and the backward rxn k. Adding heat will increase the k's for both reaction but it will have a larger effect on one, depending on exothermic or endo. That is why catalysts don't affect Keq because they affect the forward and the reverse rxn by the same amount.