Answer:
4. +117,1 kJ/mol
Explanation:
ΔG of a reaction is:
ΔGr = ΔHr - TΔSr <em>(1)</em>
For the reaction:
2 HgO(s) → 2 Hg(l) + O₂(g)
ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)
As ΔHf of Hg(l) and ΔHf O₂(g) are 0:
ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>
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In the same way ΔSr is:
ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)
ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol
ΔSr= 216,8 J/Kmol = <em><u>0,216 kJ/Kmol</u></em>
Thus, ΔGr at 298K is:
ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol
ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>
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I hope it helps!
