Calculate ∆g ◦ r for the decomposition of mercury(ii) oxide 2 hgo(s) → 2 hg(ℓ) + o2(g) ∆h◦ f −90.83 − − (kj · mol−1 ) ∆s ◦ m 70.29 76.02 205.14 (j · k −1 · mol−1 ) at 298 k. 1. −117.1 kj · mol−1 2. +246.2 kj · mol−1 3. −64.5 kj · mol−1 4. +117.1 kj · mol−1 5. −246.2 kj · mol−1

4. +117,1 kJ/mol

Explanation:

ΔG of a reaction is:

ΔGr = ΔHr - TΔSr (1)

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) = 181,66 kJ/mol

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 = 0,216 kJ/Kmol

Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ 4. +117,1 kJ/mol

I hope it helps!

the answers would be

the first one and the second one

it is the most commonly accepted one