Question

The enthalpy changes, Δ, for three reactions are given.

Answer 1

Using Hess's law, the heat of formation for CaO(s) using the reaction shown is; -637 kJ/mol

Definition;

Hess's law states that the standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction can be divided, while each occurs at the same temperature.

The reaction which represents the formation of CaO(s) and whose heat of formation is to be deter mined is;

• Ca(s). + 1/2O2(g) ⟶ CaO(s)

Using Hess's law;

The reactions given are;

1. H2(g). + 1/2O2(g) ⟶ H2O(l). ΔH=−286 kJ/mol
2. Ca(s). + 2H+(aq) ⟶ Ca2+(aq). + H2(g). ΔH=−544 kJ/mol
3. CaO(s). + 2H+(aq) ⟶ Ca2+(aq)+H2O(l). ΔH=−193 kJ/mol.

To arrive at the target equation; we must reverse equation 3; so that we have;

Ca2+(aq)+H2O(l). ⟶. CaO(s). + 2H+(aq) ΔH= 193 kJ/mol.

By the aggregation of equations 1 and 2 to the modified equation 3; the net equation is;

Ca(s). + 1/2O2(g) ⟶ CaO(s)

Ultimately, the heat of formation for CaO(s) using the reaction is;

• ΔH = −286 kJ/mol − 544 kJ/mol + 193 kJ/mol

• ΔH = -637 kJ/mol.

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