Calculate $\Delta$ G_r $\degree$for the decomposition of mercury(II)oxide at 298 K. 2HgO(s)$\rightarrow$ 2Hg(l)+ O₂(g) ------------ $\Delta$ H_f $\degree$,kJ.mol ^- ¹ -90.83 S_m $\degree$,J.K ^- ¹.mol ^- ¹ 70.29 76.02 205.14

Question

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To solve this problem, we need to use the enthalpy of formation values for each compound involved in the reaction. Using the given data, we have:

HgO (s) -> Hg (l) + 1/2 O2 (g): ΔHf = +90.83 kJ/mol
Hg (l): ΔHf = 0 kJ/mol
O2 (g): ΔHf = 0 kJ/mol

Using Hess's Law, we can calculate the ΔH for the reaction:

2HgO (s) -> 2Hg (l) + O2 (g)

ΔH = 2(-90.83 kJ/mol) + 0 kJ/mol - 0 kJ/mol
ΔH = -181.66 kJ/mol

Now, we can use the ΔH value to calculate the ΔHf for the unknown compound as follows:

ΔHf = ΔHf(products) - ΔHf(reactants)
ΔHf = 2(0 kJ/mol) + OS - (-181.66 kJ/mol)
ΔHf = OS + 181.66 kJ/mol

We can see that the ΔHf for the unknown compound is positive, which indicates that it is an endothermic reaction (i.e. heat is absorbed). Therefore, we can eliminate choices A, B, and E, which have positive values.

Now, we need to choose between choices C and D. Since the reaction is endothermic, we expect the value to be positive. Choice D is the only one that has a positive value (+117.1 kJ/mol), so the answer is D.

Calculate Δ\DeltaΔ Gr °\degree° For the Decomposition of Mercury(II)oxide at 298 K →\rightarrow→

Calculate $\Delta$ G_r $\degree$ For the Decomposition of Mercury(II)oxide at 298 K $\rightarrow$