BR Chem Ch 8, Passage 1, Q2

[barondankness]

For an exothermic reaction, if ^S is positive, then as the temperature is increased, what is observed?

Answer: D. The Keq-to-Q ratio increases, while ^G decreases.

I understand the rationale given for the answer. Because ^G = ^H - T^S, increasing temp will decrease ^G. And from ^G = RT*ln(Q/K), as ^G decreases, K/Q increases.

However, my confusion comes from the fact that the reaction is also exothermic, and in past examples, if you add heat to an exothermic reaction, the Keq will decrease as more reactant will be formed. Assuming that Q is not changing (otherwise the Keq-to-Q ratio would just become 1 over time), this rational would lead me to belive that the Keq-to-Q ratio decreases. Can anybody bridge this gap in my understanding?

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[sazerac]

Nobody wants to touch this one yet huh? It's a profound problem here and I am curious what the answer really is. Here are some pontifications:

(1) The original problem is unchanged if the reaction is exothermic or endothermic. It doesn't seem to matter here.

(2) What does a negative delta G mean for a written chemical reaction? It means that it will run in the forward direction (maybe slowly, maybe quickly) and release free energy. This does not seem to take into account product and reactant concentrations though. Strange. It's not like we are assuming all reactants and products are at standard conditions (one molar, etc) because that would be delta G zero, which is a different value and a different concept. If the reaction increases entropy (delta S positive) and is exothermic (delta H negative) then delta G is always going to be negative under all conditions, using up all reactants and producing only products. But that seems a little bogus... a reaction that NEVER runs backwards???!?

(3) What does a negative delta G mean when you take concentrations products and reactants into account? It means, for the given concentrations, the reaction will forwards until we have used up enough more reactants and created enough products so that we are now at equilibrium.

Honestly, item #2 sounds like a little white lie we tell chemistry students when they are first starting out, and item #3 sounds like the reality of what is really going on in the world. I'm curious what some other thermo-heads have to say about this one.

 

[baltimoreman]

I am not that far into TBR, but I have some background in thermo and will offer my opinion.

First, what sazerac says in (3) is correct.

Next, the equilibrium reactant and product concentrations are not dependent on the starting concentrations. You can start with any concentrations you want, and if you wait long enough, the system will reach equilibrium.

I think it's easier to look at the endothermic/exothermic and temperature change part a little differently. It is true that if you add heat to an exothermic reaction, Keq will decrease, and the equilibrium will shift towards the reactant side of the reaction. The problem doesn't say that heat is added. Only that the temperature is increased. Imagine doing a new experiment at a higher temperature instead of adding heat to the currently-running experiment. That would give you the right answer. The wording of the question is a little confusing.

 

[baltimoreman]

I am not that far into TBR, but I have some background in thermo and will offer my opinion.

First, what sazerac says in (3) is correct.

Next, the equilibrium reactant and product concentrations are not dependent on the starting concentrations. You can start with any concentrations you want, and if you wait long enough, the system will reach equilibrium.

I think it's easier to look at the endothermic/exothermic and temperature change part a little differently. It is true that if you add heat to an exothermic reaction, Keq will decrease, and the equilibrium will shift towards the reactant side of the reaction. The problem doesn't say that heat is added. Only that the temperature is increased. Imagine doing a new experiment at a higher temperature instead of adding heat to the currently-running experiment. That would give you the right answer. The wording of the question is a little confusing.

 

[barondankness]

I am not that far into TBR, but I have some background in thermo and will offer my opinion.

First, what sazerac says in (3) is correct.

Next, the equilibrium reactant and product concentrations are not dependent on the starting concentrations. You can start with any concentrations you want, and if you wait long enough, the system will reach equilibrium.

I think it's easier to look at the endothermic/exothermic and temperature change part a little differently. It is true that if you add heat to an exothermic reaction, Keq will decrease, and the equilibrium will shift towards the reactant side of the reaction. The problem doesn't say that heat is added. Only that the temperature is increased. Imagine doing a new experiment at a higher temperature instead of adding heat to the currently-running experiment. That would give you the right answer. The wording of the question is a little confusing.

I am not sure I buy that reasoning. Ch 3 of TBR gen chem seems to contradict this, table 3.5 on pg 180 specifically. I have also read that the only way to change Keq for a given reaction is to change temperature. Why does it make a difference to the Keq value if the temperature is changed at equilibrium or with a new reaction?

Maybe I am overlooking the Q in the Keq-to-Q ratio. I do not think the question explains Q to change as Keq changes, because if so, the Keq-to-Q ratio would just become equal to 1 over time.

 

[barondankness]

double post

 

[baltimoreman]

I did some more reading, and I realized that my previous answer was incorrect. I don't know how to explain this.