GC Question

[doc1986]

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An unknown compound dissolves in H2O spontaneously. If the temperature of the water decreases, which is true?

a) deltaG > 0, deltaH > 0, deltaS > 0
b) deltaG > 0, deltaH < 0, deltaS > 0
c) deltaG < 0, deltaH > 0, deltaS < 0
d) deltaG < 0, deltaH < 0, deltaS > 0
e) deltaG < 0, deltaH > 0, deltaS > 0

Tell me what you think and I will tell you the correct answer and why I disagree with it.

 

[DoctaWalnuts]

Spontaneous G is negative G<0
Water loses heat = endothermic H > 0
Solid --> liquid (dissolves) more entropy S > 0

when I first started on these, I was confused when they said, "the water loses heat"...the reaction takes heat from the water (surroundings) and thus the reaction requires energy and is endothermic.

I might be wrong because..I'm still learning as well. 🙂

 

[doc1986]

Whoops I read question wrong the first time. I just looked at the temperature of water decreasing part and thought water->solid (like freezing water to ice cube) and reasoned that entropy was decreasing as the molecules were becoming closer packed. I forgot to read that a compound dissolves (which is actually solid->liquid) and therefore increased entropy. I understood the other 2 variables.

 

[ttran9229]

I concur. My answer is also E, but it does depend on what the unknown compound is. One can assume it's a solid, but it can also be a gas. Given the problem statement and the answer choices, it would be "safe" to assume that the compound is transitioning from a solid into an aqueous form since not many gas substances will spontaneously dissolve in water.

 

[GiTsticker]

For this question, you need only know that:
1) The reaction is spontaneous
2) The reaction is endothermic

G = H - TS

If G is negative and H is positive, then S must be positive. Therefore it doesn't matter whether it was a gas, solid, liquid, aqueous or anything.

 

[ttran9229]

For this question, you need only know that:
1) The reaction is spontaneous
2) The reaction is endothermic

G = H - TS

If G is negative and H is positive, then S must be positive. Therefore it doesn't matter whether it was a gas, solid, liquid, aqueous or anything.

Yes, this is a very good way of looking at the problem.

 

[drpharMD]

Uh, I believe the answer is D. Losing heat is an EXOTHERMIC PROCESS.

 

[Kneecoal]

this is a destroyer question, right? i think i got it wrong too because i took the "temperature decreases" to mean the outside was getting colder aka endothermic. buttt apparently they mean it's giving off heat, and that's why the temp decreases.

i think the question could have been worded better. unless the point was they were trying to trick you.

 

[GiTsticker]

Uh, I believe the answer is D. Losing heat is an EXOTHERMIC PROCESS.

The temperature decreased. That means it is absorbing heat. If by "it" you mean the environment, then yes it is losing heat, but the environment is irrelevant in this context. Heat from the environment is being absorbed by the water which is taking on a new, higher energy form by dissolving some substance. Note that this isn't usually the case as dissolution is usually an exothermic process resulting in a more stable solution (which explains the higher boiling points etc.). In any case, colder=endothermic warmer=exothermic with reference to the reaction. With reference to the environment....just don't worry about the environment.

 

[GiTsticker]

i took the "temperature decreases" to mean the outside was getting colder aka endothermic.

You were right in your conclusion that it was endothermic, but there is a better way of thinking about it. Imagine heat not as temperature, but as something that changes from one system to another. In this case, heat from the environment is changed from heat into intermolecular energy in the solution. The result is that temperature goes down and the energy of the solution goes up. Therefore energy in the form of heat is moving INTO (endo) the reaction. The only way this could happen spontaneously is if a gain in the product of temperature and entropy occurred that was greater in magnitude than the increase in enthalpy.

 

[Locdawgg]

For this question, you need only know that:
1) The reaction is spontaneous
2) The reaction is endothermic

G = H - TS

If G is negative and H is positive, then S must be positive. Therefore it doesn't matter whether it was a gas, solid, liquid, aqueous or anything.

If the reaction is spontaneous, then Delta G = negative. If the reaction is endothermic, then Delta H = positive. If Delta S "must be" positive that would mean Delta S x negative T = negative value. Since temperature is decreased we would assume T is a small number?? Wouldn't that make Delta G = positive?

 

[emitpeels]

Going from a solid to a liquid increase entropy. Since the reaction cools down, it must be endothermic, which is +delta H.