[AmbitiousStudent]

I had some problems with the question below. How can I check up if the the driving force is because HCl and SO2 are gases? I cannot find any values for SO2 when it is in liquid phase.
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According to the the litterature, the driving force in the reaction below is because HCl and SO2 is gases. Check this up with calculate the reaction enthalpy and reaction entropy at 298K. If you need more data will you have to check it up in a table work

HCOOH+ClSOCl -> HCOCl+HCl+SO2

ClS(O)Cl (aq): Delta Hf -785 (kJ/mol) Sm 220 (J/K/mol)
HC(O)Cl(aq) :Delta Hf -376.56 (kJ/mol) Sm 246.52 (J/K/mol)
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I started to check up values for HCOOH, SO2 and HCl

HCOOH (aq): Delta Hf -424,72 (kJ/mol) Sm 128,95 (J/K/mol)
HCl(q) :Delta Hf -92,31 (kJ/mol) Sm 186,91 (J/K/mol)
SO2(q) :Delta Hf -296,83 (kJ/mol) Sm 248,22 (J/K/mol)
HCl(aq) :Delta Hf -167,16 (kJ/mol) Sm 56,5 (J/K/mol)
SO2(aq) : No values

Delta Hf= -376,56-92,31-296,83-(-424,72-785)= 444,02 KJ/mol

Delta S= 246,52+186,91+248,22-128,95-220=332,7 J/K/moll

When delta H and delta S are positive the reaction is entropy driven if SO2 and HCl are in gas phase.

BUT how can I check up if the the driving force is because HCl and SO2 are gases? I cannot find any values for SO2 when it is in liquid phase

[MrTeo]

Just a small piece of advice: try to use (l) for the liquid phase and (g) and (s) for gaseous and solid phases, as (aq) is usually referred to an aqueous solution of the compound considered.

When ∆H and ∆S are positive the reaction is entropy driven if SO₂ and HCl are in gas phase.

Ok. I suppose you found all the values in the tables provided so I'll trust your work. 
How do you expect the enthalpy and entropy terms to change when you have liquid products?

[AmbitiousStudent]

Just a small piece of advice: try to use (l) for the liquid phase and (g) and (s) for gaseous and solid phases, as (aq) is usually referred to an aqueous solution of the compound considered.

When ∆H and ∆S are positive the reaction is entropy driven if SO₂ and HCl are in gas phase.

Ok. I suppose you found all the values in the tables provided so I'll trust your work. 
How do you expect the enthalpy and entropy terms to change when you have liquid products?

You are right about l instead of aq, thanks.

When it goes from gas phase to liguid phase will the value for Hf and S decrease.  When the value for HCl and SO2 decrase will the value for delta Hf and delta S decrease since they are on the product side of the reaction.

But I do not really know how I can check it up if the driving force is that SO2 and HCl is in gas phase since I do not have any values on SO2(l). I do not think that the differnce between SO2(g) and SO2(l) is that big that either delta S delta Hf or delta G will go from positive to negative.

[Corribus]

Any time you go from a liquid to a gas the entropy is going to increase (and vice-versa).  I don't know off the top of my head of any exceptions to that.  What really determines the spontaneity (Gibbs energy) then is the enthalpic term: is the mutual attraction between molecules strong enough to overcome the enthalpic driving force for the phase change from liquid to gas? Naturally this depends a lot on the temperature.

[AmbitiousStudent]

I agree with you but can you really state that the driving force in this reaction is that HCl and SO2 is in gas phase since both the entalpi and the entropy are increasing when it goes to gas phase? Or should I state that the statement is false (i.e that is not because HCl and SO2 is in gas phase) ?

[Corribus]

It's a false dilemma: it's never just one thing or the other.  The reaction is spontaneous at the temperature in question because the entropic term exceeds the enthalpic term in magnitude.  It is appropriate to say that at this temperature, the reaction is entropically driven.  Most likely the largest contribution to the entropic favorability is the fact that you're forming a lot of gas.  So I think it would be ok to say that the driving force is primarily the formation of two gasses.  But in reality there's always lots of things which contribute to the favorability (or lack thereof) of a reaction, and it's not always possible to identify or quantify every one of them.

[AmbitiousStudent]

It's a false dilemma: it's never just one thing or the other.  The reaction is spontaneous at the temperature in question because the entropic term exceeds the enthalpic term in magnitude.  It is appropriate to say that at this temperature, the reaction is entropically driven.  Most likely the largest contribution to the entropic favorability is the fact that you're forming a lot of gas.  So I think it would be ok to say that the driving force is primarily the formation of two gasses.  But in reality there's always lots of things which contribute to the favorability (or lack thereof) of a reaction, and it's not always possible to identify or quantify every one of them.

But is the reaction really spontaneous since delta G is positive according to my calculation ?

Delta G = 444kJ-298*332.7

[Corribus]

Sorry, yes, I didn't actually do any calculation. You are right, the Gibbs energy is positive which means the reaction is no spontaneous at the temperature in question.  Which means that although the entropy is positive, there's not enough heat energy available to make the reaction spontaneous.

Do keep in mind that this does not mean any product will be formed.  Equilibrium will always be reached.  A positive Gibbs energy just means that equilibrium will favor reactants.