Thermodynamics Question

[thegenius]

There is a reaction whose (delta H) = 873 kJ/mol and the (delta S) = 3 J/mol K.

What would make this reaction occur spontaneously?

A. the temperature is sufficiently high
B. the correct catalyst is found
C. the temperature is sufficiently low
D. it will never occur spontaneously because (delta G) is positive.

OK, I say the best answer is B. However I'm wrong because you can have a temperature that would make (delta G) negative, and that is

873,000/3 = 291,000 K (notice the units above.)

So this begs the question, I thought that temperature is basically unachievable. I suppose if you were in the middle of a star you could reach that temperature - but I said that B was the answer. Why would B not be a good answer with this one?

---

Members don't see this ad.

 

[Medikit]

Why would B not be a good answer with this one?

Catalysts do not effect the spontaneity of the reaction. They simply change the rate.

The question asks what would make the reaction spontaneous, not if it is a reasonable reaction for doing in the lab.

 

[DieselPetrolGrl]

i would pick A
loosk like a direct application of dg=-dh-tds
and with the h so positive and the s positive too i would like a big ole T to offset that and get a nice big neg # for the "-TdS" portion - big enough to offset that +873 dH value
um yeah thats how i would do it ...why it inst B...all a catalist can do is lower activation energy - raise the rate - it cant fool with the spontianaiaiiaiaty of a reaction

PS ~no i refuse to spell and yes i use the term d as delta ie dH=delta H and ds=delta s..yadda yadda yadda~

 

[DieselPetrolGrl]

Catalysts do not effect the spontaneity of the reaction. They simply change the rate.

yeah what the poster with the cool av said!

 

[Medikit]

yeah what the poster with the cool av said!

--> To the person with the cool hair.

 

[thegenius]

Catalysts do not effect the spontaneity of the reaction. They simply change the rate.

The question asks what would make the reaction spontaneous, not if it is a reasonable reaction for doing in the lab.

Thanks for the response. I'll definitely make note of that.

 

[Fusion]

I concur, the answer is A. Refer to the Gibbs free energy eqn. Despite the fact that the enthalpy is so high, a high enough temperature multiplied by the value for the entropy (which is positive) will result in a product whose absolute value is larger than the enthalpy value. Subtracting this product from the enthalpy will result in a negative value for delta G, indicating an exothermic rxn.

Also, catalysts affect kinetics, not thermodynamics. In other words, they speed things along but do not affect the overall energy gained by or released from the reactants. Catalysts do lower the activation energy, but this is not considered a thermodynamic effect.

Hope this makes sense.

 

[Assembler]

The answer is A.

It's spontaneous if the change in Gibb's free energy is less than zero. Obviously, it's nonspontaneous (endergonic) if delta G is > 0, and at equilibrium if the change is equal to zero.

Delta (G) = Delta (H) - T*Delta(S)

H is large, S is small...it seems to be a positive number so far. Maximize T, then H-T*S would be in negative territory - then it becomes spontaneous.

Gibb's free energy can be affected by various conditions, but one thing that is always constant is the standard free energy change (Go')

If I remember correctly:
Delta (Go') = -RTln(Keq)
Delta (G) = Delta (Go') + RTln(Keq)

 

[willthatsall]

Yeah, make sure you don't over think the question. I had a question on one test and it showed a reaction with positive entropy said, "what can you say is likely true about this reaction at something like 1500 K" I think. And I was thinking, at a temperature that high, it is going to be spontaneous. Of course that was wrong, there is no way to tell. So just stick with gibbs free energy equation in that situation, it won't lead you astray.

 

[DieselPetrolGrl]

Yeah, make sure you don't over think the question. I had a question on one test and it showed a reaction with positive entropy said, "what can you say is likely true about this reaction at something like 1500 K" I think. And I was thinking, at a temperature that high, it is going to be spontaneous. Of course that was wrong, there is no way to tell. So just stick with gibbs free energy equation in that situation, it won't lead you astray.

i remember that kaplan Q
i picked the answer bc i was liek WHOAAA thats def high enough
i blame the Q that was such a whack-trap

 

[Groove]

Just to reiterate what everyone said:

Don't make it more complicated than it has to be.

We know that Delta G = Delta H - T * Delta S

We also know that G must be negative for the reaction or process to be spontaneous. You have a large H and a small S, therefore in order to make this number negative, we need to make sure that (T * Delta S) is sufficiently larger than H. The only way to do this is to have a large Temperature. Hence, the answer is A.

A catalyst is only going to affect the activation energy and the rate of the reaction. One tip I'd suggest with one of these problems is that if the answer can be solved mathematically by one of the formula's, then always go with that answer because that is the underlying proof of the concept. Who cares if it could be actually carried out in a lab? The math speaks for itself, that it IS possible theoretically.

Remember, catalyst will change rate and activation energy, but will never affect Delta G or equilibrium constants.

 

[Assembler]

Pretty much what everyone said. A catalyst just lowers the activation energy barrier, hence stabilizing the transition state intermediate. It increases the rate of the reaction, doesn't affect delta G, is never consumed, and only leads to equilibrium faster.

 

[willthatsall]

i remember that kaplan Q
i picked the answer bc i was liek WHOAAA thats def high enough
i blame the Q that was such a whack-trap

Yeah, I knew it but I still picked the wrong answer. That happens at least once per test when you are just like, "Ok, I know this is stupid, but I'm going with this answer even though it's probably wrong" and then you think about how stupid you are later.

 

[thegenius]

Pretty much what everyone said. A catalyst just lowers the activation energy barrier, hence stabilizing the transition state intermediate. It increases the rate of the reaction, doesn't affect delta G, is never consumed, and only leads to equilibrium faster.

Well I certainly appreciate everyone's response. I got more than I suspected - but that is good.

Kaplan suggests studying with people, and I do think it's a good idea. You end up hearing the same theme in different ways - and you learn a little as a result.

Like WillThatsAll, I sometimes overthink questions. Had I remembered that catalysts only affects kinetics and not thermodynamics - then I would have probably picked A because B is wrong! I was thinking 287,000 K, that is ridiculous.

Ok, this thread can die.