Increase in entropy

[justin7w2011]

What results in a greater increase in entropy an increase in mols of gas or a phase change? I thought I remembered chad saying mols prevailed but I had a top score question that the correct answer was the phase change over the greater change in mols of gas.

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[jwan14]

Intuitively I would say phase change. Molecules going from liquid to gas would be much more disordered than gas to more gas

 

[justin7w2011]

I went to chad's to try and look and I found this in a solution from one of chad's quizzes "The single largest indicator for

S is a change in the number of moles of gas." Not sure how I should approach those questions in the future now.

 

[DocBella]

2 I Cl (g) --> I2 (g) + Cl2 (g)

Can anyone please tell me how delta S (change in entropy) is negative for this "non-spontaneous" rxn? (mentioned in the question stem)

I understand as a rule of thumb that if a rxn is non-spontaneous at any temperature, delta H is positive while delta S is negative to yield a positive delta G. I just don't see how in this particular reaction there's a decrease in randomness.... in other words, how is 2 I Cl (g) more random/disordered than I2 (g) and Cl (g) ? If the number of molecules are increasing while the phase is the same, then how does that translate to a negative delta S?

 

[TheEpicFruitCake]

2 I Cl (g) --> I2 (g) + Cl2 (g)

Can anyone please tell me how delta S (change in entropy) is negative for this "non-spontaneous" rxn? (mentioned in the question stem)

I understand as a rule of thumb that if a rxn is non-spontaneous at any temperature, delta H is positive while delta S is negative to yield a positive delta G. I just don't see how in this particular reaction there's a decrease in randomness.... in other words, how is 2 I Cl (g) more random/disordered than I2 (g) and Cl (g) ? If the number of molecules are increasing while the phase is the same, then how does that translate to a negative delta S?

because you go from 3 atoms to 2 atoms aka less entropy

 

[DocBella]

because you go from 3 atoms to 2 atoms aka less entropy

3 atoms?

 

[TheEpicFruitCake]

3 atoms?

lol molecules. the point is 3 > 2

also you don't give us the state that 2I is in.

 

[DocBella]

lol molecules. the point is 3 > 2

also you don't give us the state that 2I is in.

I got that you meant molecules.... i don't see how there are 3 molecules on the left.... ? (Sorry if I'm missing something) And the question stem didn't say anything about the state of the reaction, it simply gave this rxn and said it's NOT spontaneous, and from that we had to choose answer choices which had combinations of -/+ delta H and -/+ delta S.

Thanks for the help!

 

[DocBella]

I got that you meant molecules.... i don't see how there are 3 molecules on the left.... ? (Sorry if I'm missing something) And the question stem didn't say anything about the state of the reaction, it simply gave this rxn and said it's NOT spontaneous, and from that we had to choose answer choices which had combinations of -/+ delta H and -/+ delta S.

Thanks for the help!

Isn't the 2ICl saying that there are 2 molecules of Iodine monochloride? Isn't 2 a coefficient for the entire molecule as a whole and not pertaining to Iodine alone? =/

 

[TheEpicFruitCake]

I got that you meant molecules.... i don't see how there are 3 molecules on the left.... ? (Sorry if I'm missing something) And the question stem didn't say anything about the state of the reaction, it simply gave this rxn and said it's NOT spontaneous, and from that we had to choose answer choices which had combinations of -/+ delta H and -/+ delta S.

Thanks for the help!

Yeah well... i've never seen ICl so I assumed I and Cl seperately seeing as you added a space inbetween. My bad. But only because you say entropy is decreasing. I would say this is because they are turning into pure gasses This is just a guess, its actually very close to zero, but yeah. Also, this question is making my raise my eyebrow lol. I looked up Iodine and its boiling point is 370K. It would make much more sense if it was ICl(aq)
dG = dH-TdS. Not Spontaneous will be +/- for all temps. +/+ for low temps (or since entropy is practically 0 its nearly infinite temps.

But let's not hijack this dudes thread. PM me if you don't get it

 

[DocBella]

Yeah well... i've never seen ICl so I assumed I and Cl seperately seeing as you added a space inbetween. My bad. But only because you say entropy is decreasing. I would say this is because they are turning into pure gasses This is just a guess, its actually very close to zero, but yeah. Also, this question is making my raise my eyebrow lol. I looked up Iodine and its boiling point is 370K. It would make much more sense if it was ICl(aq)
dG = dH-TdS. Not Spontaneous will be +/- for all temps. +/+ for low temps (or since entropy is practically 0 its nearly infinite temps.

But let's not hijack this dudes thread. PM me if you don't get it

HAHA we aren't hijacking it, we're adding to it! loll his question was very similar and related to this topic, so maybe it will clear more things for him/her and anyone else who will search this same topic

But yeh, this question confused the crap out of me too, hopefully someone will come across it and a light will go off bc my eyebrows have BEEN raised on this question! lol

 

[aquaniff]

Sorry first time replying one second!

 

[aquaniff]

DeltaG=deltaH-TdeltaS (law of thermodynamics)
to use short hand the equation is dG=dH-TdS

T is in Kelvin and must always be positive.

If the reaction is spontaneous, then dG must be negative (another thermo law)

In order for dG to be negative dH-tdS<=0
dH<=tdS

Now for this question,

For the reaction to be nonspontaneous (dG>=0):
dH>=tdS
t is always positive and will scale dS up. The only way to ensure that this statement is true (that the reaction is non-spontaneous-- dG<0) is for dS to be negative.

If this does not make sense plug in number for dH, t and dS and see which ones will make the statement true.

When dS is negative, there is more order and less chaos.

The moles are there to throw you off. Remember for spontaneous and nonspontaneous reactions with gasses, temperature and enthalpy (dH) play a role as well.

Hope this helps! let me know if anything is confusing

 

[TheEpicFruitCake]

It definitely doesn't go from 2 to 4 moles

Please google "diatomic"

and yes, you are correct for everything else

 

[aquaniff]

the moles stay the same. my fault

 

[aquaniff]

@TheEpicFruitCake does the rest make sense? I forgot to count the compound as 4 moles instead of 2 (distribute the 2 to each chemical)

 

[aquaniff]

It definitely doesn't go from 2 to 4 moles

Please google "diatomic"

and yes, you are correct for everything else

I am familiar with diatomic, but would you agree that the total moles stay the same for this reaction?

 

[DocBella]

I thought that the only way entropy would change is when there's a phase change (or an increase or decrease in the molar amount too?) idk, maybe there's a typo in this problem.... (stupid Kaplan, I've found plenty of typos and incorrect/opposite explanations in their material) but everything that you said, aquaniff, makes sense. And I am aware of thermodynamics laws but I just can't apply it to this reaction. Maybe it's just me, but i hope to God that this kinda rxn doesn't come on the real thing cuz imma break the computer! lol Thanks for the explanation!

Thanks TheEpicFruitCake! my thread hijack partner