Question

Repeat the procedure from Part C for the other three oxides,

Oxide	ΔH∘f (kJ/mol)
K2O(s)	−363.2
CaO(s)	−635.1
TiO2(s)	−938.7

then rank all four oxides according to their enthalpy of reduction.

Rank from greatest (most positive) to least enthalpy of reduction. To rank items as equivalent, overlap them.

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Reset Help CaO{\rm CaO} TiO2{\rm TiO_2} K2O{\rm K_2O} V2O5{\rm V_2O_5} Least enthalpy Greatest enthalpy

Answer 1

The formation of each of these oxides is due to addition of oxygen and increase in the oxidation state of the metals

Ca(s) + 1/2O2(g) -> CaO(s)

Ti(s) + O2(g) -> TiO2.(s)

2K(s) + 1/2O2(g) -> K2O(s)

2V(s) + 5/2O2(g) -> V2O5(s)

delta Hf is the enthalpy change of formation of the substance.

More negative delta Hf value means that the product formed is more stable. So, more stable oxide will mean the metal it contain can be easily oxidised and thus a better reducing agent. (Substances which can get easily oxidised are the best reducing agents)

So, V2O5 has deltaHf = -1550.590 kJ/mol (wikipedia) has the largest negative delta Hf value among the given oxides and thus its the best reducting agent.

followed by TiO2 with delta Hf = −938.7 kJ/mol then CaO with delta Hf = −635.1 kJ/mole and the weakest reducing agent K2O with delta Hf = −363.2 kJ/mole

So, the order from positive to negative enthalpy of reduction will be reverse of the order explained above

So, the order will be K2O , CaO , TiO2 , V2O5