[Win,odd Dhamnekar]

Compute ΔH°,ΔS°,ΔG° at 25°C for the following reactions:-

1)Combustion of Methane

2)Formation of gaseous HCl from its elements.

3)Formation of H₃SiCl from its elements [ We will need to use bond energies and estimate ΔS°]

My answer:-

[mjc123]

First, S is the symbol for entropy, not "spontaneity".

You have made some mistakes in calculation.

For reaction 1, you have not multiplied the enthalpy of formation of water by the stoichiometric coefficient of 2. The entropy calculation looks correct, but ΔG will be wrong because ΔH is wrong.

For reaction 2, you have calculated ΔH for 1 mole of HCl, but ΔS for 2 moles. For the reaction as you have written it, the reaction enthalpy (or entropy, or Gibbs energy) is twice the molar enthalpy (etc.) of formation of HCl, but the wording of the question suggests you want it per mole of HCl. As ΔH and ΔS are inconsistent, ΔG is of course wrong, and you have made a sign mistake.

For reaction 3, where did you get your numbers? The question says you need to use bond energies and estimate ΔS.

[Win,odd Dhamnekar]

Thanks for pointing out my mistakes in the computations of ΔH°_f,ΔS°_rxn,ΔG_rxn° which otherwise  would remain unnoticed.
 I have rectified my mistakes in reaction 1 and 2 as follows.

1)ΔH°_rxn and ΔG°_rxn for reaction 1, are  -890.5 kJ/mol and -818.12 kJ/mol respectively.

2)ΔH°_rxn and ΔG°_rxn for reaction 2, are -92.3 kJ/mol and -95.28kJ/mol respectively.

3) I have taken the values of ΔH°_f,ΔS° of Chlorosilane(H₃SiCl) for reaction 3 from(https://webbook.nist.gov/cgi/cbook.cgi?Formula=H3SiCl&NoIon=on&Units=SI&cTG=on&cTR=on

 Corrected ΔS°_rxn is 250.76 J/mol K + 223.1 kJ/mol K - 18.8 J/mol K - 168.9 J/mol K=285.56 J/mol K The third reaction is non-spontaneous at low temperature.

Corrected ΔG°_rxn for reaction 3 is 359.76 kJ/mol - 298 K*(285.56 J/mol K)= 274.75 kJ/mol

Note:ΔH°_f for HCl(aq) is -167.2 kJ/mol

  I hope the above computations are now errorless.

[mjc123]

The calculations now look correct. For completeness you should state the correct ΔS for reaction 2.

On a point of notation, for reaction 2 as you have written it, ΔH_rxn = 2ΔH_f(HCl). It is ΔH_f(HCl) that you are asked for. In that case, it would be better to write the reaction as
1/2 H₂ + 1/2 Cl₂  HCl
for which ΔH_rxn = ΔH_f(HCl).

I have just noticed that question 3 asks for the formation of H₃SiCl from its elements, so the reaction you have written is the wrong one.

[Win,odd Dhamnekar]

 Thanks again for pointing out my mistake.

So, the corrected ΔS°_rxn for reaction 2, is 186.9 J/mol K - 165.2 J/mol K - 114.2 J/mol K= -93 J/mol K

So, the correct equation for the reaction 3, is 3H + Si + Cl H₃SiCl

ΔH°_rxn for above reaction is -141.84 kJ/mol -654.0 kJ/mol -121.3kJ/mol=-917.14 kJ/mol. https://webbook.nist.gov/cgi/cbook.cgi?ID=C13465786&Units=SI&Mask=1#Thermo-Gas   

ΔS°_rxn for above reaction is 250.76 J/mol K - 344.1 J/mol K -18.8 J/mol K - 165.2 J/mol K =-277.34 J/mol K

ΔG°_rxn for the above reaction is -917.14 kJ/mol - 298.15 K*(-277.34 J/mol K)=-834.45 kJ/mol

 Third reaction is spontaneous at low temperature.

[mjc123]

No, that is completely wrong. The correct ΔS is simply half the value you calculated before, i.e. 10 J/mol/K. You have calculated ΔS for the reaction
H + Cl  HCl
This is NOT the same as
1/2 H₂ + 1/2 Cl₂ HCl
Do you understand the difference?
Likewise, the equation for formation of H₃SiCl from its elements in their standard states is
3/2 H₂ + 1/2 Cl₂ + Si  H₃SiCl
not what you have written.

[Win,odd Dhamnekar]

Thanks for pointing out my mistakes.

So for reaction 2,Δ_rH° ,Δ_rS°, Δ_rG° are -92.3 kJ/mol, 10 J /mol K, -95.28 kJ/mol respectively.

Likewise, for reaction 3, Δ_rH°, Δ_rS°, Δ_rG° are -141.84 kJ/mol,  -75.64 J/mol K, -119.3 kJ/mol respectively.

Second reaction is spontaneous at all temperatures, while first and third reactions are spontaneous at low temperature.

[mjc123]

That looks correct now.
Nevertheless, I initially thought you understood the chemistry, but had just made some careless errors in the calculations. Now I'm not so sure. Can you confirm that you understand the difference between
1/2 H₂ + 1/2 Cl₂  HCl
and H + Cl  HCl
and why the first is correct and the second is wrong (for the formation of HCl from its elements)?

[Win,odd Dhamnekar]

There are seven elements that naturally occur as homonuclear diatomic molecules in their gaseous states: hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine.
The only chemical elements that form stable homonuclear diatomic molecules at standard temperature and pressure (STP) (or typical laboratory conditions of 1 bar and 25°C) are the gases hydrogen (H₂), nitrogen (N₂), oxygen (O₂), fluorine (F₂), and chlorine (Cl₂).

Now, Let us see our first equation which is redox reaction.

[tex]\frac12H_2 + \frac12Cl_2 \rightarrow H^+Cl^-\tag{1}[/tex]

In (1), H₂ is reducing agent and Cl₂ is oxidizing agent.

Now,look at the second reaction.
[tex] H^+ Cl^- \rightarrow H^+Cl^- \tag{2}[/tex]  No reaction takes place in (2).

That's why reaction equation (1) is correct and reaction equation (2) is wrong.

[mjc123]

Yes, these elements are diatomic gases in their standard states. So these diatomic gases must be the starting materials when you talk about "enthalpy of formation from elements".
Note that gaseous HCl is a covalent molecule, not ionised, so it is incorrect to write it as H⁺Cl^-.

You appear to have misread my second equation. I did not write the nonsensical
H⁺Cl^-  H⁺Cl^-
but H + Cl  HCl
i.e. a free H atom reacts with a free Cl atom to give a HCl molecule.
My point was that hydrogen exists in its standard state as H₂ molecules, not free H atoms, likewise chlorine as Cl₂ not Cl, and this must be reflected in the equation you write.

[Win,odd Dhamnekar]

 I made a typographical error while writing equation (2), It should be
[tex] H^+ + Cl^- \rightarrow HCl(g)\tag{2}[/tex]

No reaction takes place in (2)

Is this correct now?

[mjc123]

No. I meant uncharged atoms, not ions. I wrote that because in Reply #4 you used entropy values for H(g) and Cl(g), as I could tell by comparing with tables, so I assumed that was the reaction you were thinking of. If you actually meant H⁺(g) and Cl^-(g), you were reading your tables wrongly.

[Win,odd Dhamnekar]

 Thanks for finding out my mistake.
 Does free atoms of Hydrogen, Nitrogen, Oxygen, Fluorine, Chlorine, Bromine, Iodine exist? or we have to separate it from their homonuclear, diatomic molecule?

[Win,odd Dhamnekar]

  Yes, Atoms of Hydrogen exists in the nature in the form of cation, anion and neutral form.