Thermochemistry and the formation of pure oxygen

[Zaros]

Homework Statement

Pure oxygen was first obtained in the laboratory by the French scientist Antoine Lavoisier in the 18th century by decomposing red mercury(II) oxide, HgO, to give silver mercury liquid and gaseous oxygen:
2HgO(s) \Leftrightarrow 2Hg(l) + O₂

(a) From values of \Delta _f H\circ and S\circ , calculate \Delta_rxnH\circ and \Delta_rxnS\circ for this reaction.

(b) From your values of \Delta_rxnH\circ and \Delta_rxnS\circ, calculate the temperature at which this reaction will become product-favoured.

(c) What is the equilibrium constant, K_p, for this reaction at the temperature where it becomes product-favoured?

(d) At 200\circC, below what pressure of oxygen gas will this reaction become product favoured?

\Delta_fH\circ(Hg) = 60.83 kJ/mol
\Delta_fH\circ(O₂) = 0 kJ/mol
\Delta_fH\circ(HgO) = -91 kJ/mol

S\circ(Hg) = 76 J/(k.mol)
S\circ(O₂) = 205 J/(k.mol)
S\circ(HgO) = 70 J/(k.mol)

Homework Equations

\Delta_rxnH\circ = \Sigman\Delta _f H\circ(products) - \Sigmam\Delta _f H\circ(reactants)

\DeltaG = \DeltaH -T\DeltaS

The Attempt at a Solution

a) \Delta_rxnH\circ = (2*60.83 + 0) - (2*(-91))
= 303.66 kJ/mol

\Delta_rxnS\circ = (2*76 + 205) - (2*70)
= 217 J/(mol.k)

b) T = (303.66 - \DeltaG)/(217*10^-3)

for products to be favoured \DeltaG <0

T> 1.40*10³K

This was as far as I was able to get as was unsure how I would go about calculating the equilibrium constant. Also I'm unsure if what I have done so far is correct.

 

[Zaros]

can anyone help?

 

[Zaros]

I have managed to solve part (c) not sure if its correct though.

c) \DeltaG\circ= -2*-59
=118kJ/mol
=118,000 J/mol

(-\DeltaG\circ)/(RT) = (-118000)/(8.314*1.4*10³) = -10.1
k=e^-10.1=3.96*10^-5

Im still having trouble with part (d) though as I am unsure what equation to be using. I've pulled the information out of the question:
T=200\circC = 473k
P(O₂)=?
\DeltaG<0

Input would be appreciated even if its just saying if what i have done so far is correct.
Thanks
Zaros

 

[pugdog]

Heya Zaros,

Looks like we are doing the same assignment. I am not sure how to do fonts on this forum, only just joined, so I hope you understand what I am saying:

Your H value of (Hg) = 60.83 kJ/mo is wrong... make sure you use the (l) not the (g) Hg value. So your H of the reaction should be (0 + 0) - (2 * -91) = 182 kJ/mo.

Otherwise that'll throw out a lot of the assignment!

Am interested in any other answers you get for the rest of teh questions though.

 

[Zaros]

ahh thanks for that. Good job you picked up on it. Ill get back to you once I've sorted it out

 

[Zaros]

Okay iv gone back through all my working replacing the values and the new answers I got were:
a) \Delta_rxnH\circ=182kJ/mol

b) T>8.39*10²k

c) k=e^-16.9=4.50*10^-8

I've only written up the values that have changed as everything else will have remained the same as my original posts. Still haven't figured out part (d) though. I'm thinking that its on something that is missing from my textbook (or in another section) as it doesn't mention K_p at all just normal k so I am going to do some alternate research for it in the extra material I have in hopes of finding out. If you get any further or need help somewhere just ask. Also for all the symbols etc that i have been using click on the \Sigma symbol and an extra menu will appear which will automatically insert the symbol that you click on. This is part of a hand in assignment for UNE's CHEM110 unit are you by any chance doing the same unit?

 

[pugdog]

Thanks for the font tips :)

Yep, I am doing the same unit :) Have done all the assignment as well but was a bit unsure of this last question too. Will have more of a look later today adn tell you what I come up with. :)

 

[pugdog]

Ok, I've come up with:

6 (a) 182 kJ

(b) T > 8.31 * 10² K

(c) K = 3.83 x 10^-8

Ive realized the reason we are slightly off is because you have used the SI data book S\circ and I used the back of the textbook. I might change mine just b/c I need to use the SI book for HgO, keep it all the same. We are both using the same formulas though so hopefully that's a good sign.

Im stuck on d, am going to show this to a chem pHd student who is a buddy, will let you know what he says if you like.

 

[Zaros]

Ok yes that's good idea to keep them all the same. Hopefully your buddy will be able to help as all my searching has just come up with using differentiation with some of the equations which seems way ahead of what we have been doing so far in this unit. I could probably do it myself using this method as I have previously done some of a physics degree which went into much more detail with differentiating then any of the maths in MATH120. But I still think that It would be the wrong method as they would not expect us to be able to do this yet.
Thanks