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Zaros
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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) [tex]\Leftrightarrow[/tex] 2Hg(l) + O2
(a) From values of [tex]\Delta[/tex] f H[tex]\circ[/tex] and S[tex]\circ[/tex] , calculate [tex]\Delta[/tex]rxnH[tex]\circ[/tex] and [tex]\Delta[/tex]rxnS[tex]\circ[/tex] for this reaction.
(b) From your values of [tex]\Delta[/tex]rxnH[tex]\circ[/tex] and [tex]\Delta[/tex]rxnS[tex]\circ[/tex], calculate the temperature at which this reaction will become product-favoured.
(c) What is the equilibrium constant, Kp, for this reaction at the temperature where it becomes product-favoured?
(d) At 200[tex]\circ[/tex]C, below what pressure of oxygen gas will this reaction become product favoured?
[tex]\Delta[/tex]fH[tex]\circ[/tex](Hg) = 60.83 kJ/mol
[tex]\Delta[/tex]fH[tex]\circ[/tex](O2) = 0 kJ/mol
[tex]\Delta[/tex]fH[tex]\circ[/tex](HgO) = -91 kJ/mol
S[tex]\circ[/tex](Hg) = 76 J/(k.mol)
S[tex]\circ[/tex](O2) = 205 J/(k.mol)
S[tex]\circ[/tex](HgO) = 70 J/(k.mol)
Homework Equations
[tex]\Delta[/tex]rxnH[tex]\circ[/tex] = [tex]\Sigma[/tex]n[tex]\Delta[/tex] f H[tex]\circ[/tex](products) - [tex]\Sigma[/tex]m[tex]\Delta[/tex] f H[tex]\circ[/tex](reactants)
[tex]\Delta[/tex]G = [tex]\Delta[/tex]H -T[tex]\Delta[/tex]S
The Attempt at a Solution
a) [tex]\Delta[/tex]rxnH[tex]\circ[/tex] = (2*60.83 + 0) - (2*(-91))
= 303.66 kJ/mol
[tex]\Delta[/tex]rxnS[tex]\circ[/tex] = (2*76 + 205) - (2*70)
= 217 J/(mol.k)
b) T = (303.66 - [tex]\Delta[/tex]G)/(217*10-3)
for products to be favoured [tex]\Delta[/tex]G <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.