Questions: Copper(I) oxide can be oxidized to copper(II) oxide: Cu2O(s) + 1/2 O2(g) → 2 CuO(s) ΔHrxno = -146.0 kJ Given ΔHfo of Cu2O(s) = -146.0 kJ/mol, find ΔHfo of CuO(s). Be sure your answer has the correct number of significant figures. kJ/mol 18

Determine the enthalpy of formation for \(\mathrm{CuO}(s)\).

Understand the given reaction and data

The reaction given is:

\[ \mathrm{Cu}_{2} \mathrm{O}(\mathrm{~s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CuO}(\mathrm{~s}) \]

The enthalpy change for this reaction is \(\Delta H_{\mathrm{rxn}}^{\mathrm{o}} = -146.0 \, \mathrm{kJ}\).

The enthalpy of formation for \(\mathrm{Cu}_{2} \mathrm{O}(s)\) is given as \(\Delta H_{f}^{\circ} = -146.0 \, \mathrm{kJ/mol}\).

Apply Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction. We can use this to find the enthalpy of formation for \(\mathrm{CuO}(s)\).

The enthalpy change for the reaction can be expressed in terms of the enthalpies of formation:

\[ \Delta H_{\mathrm{rxn}}^{\circ} = \left[2 \times \Delta H_{f}^{\circ}(\mathrm{CuO})\right] - \Delta H_{f}^{\circ}(\mathrm{Cu}_{2}\mathrm{O}) - \left[\frac{1}{2} \times \Delta H_{f}^{\circ}(\mathrm{O}_{2})\right] \]

Since \(\Delta H_{f}^{\circ}(\mathrm{O}_{2}) = 0\) (as it is a standard state element), the equation simplifies to:

\[ -146.0 = \left[2 \times \Delta H_{f}^{\circ}(\mathrm{CuO})\right] - (-146.0) \]

Solve for \(\Delta H_{f}^{\circ}(\mathrm{CuO})\)

Rearrange the equation to solve for \(\Delta H_{f}^{\circ}(\mathrm{CuO})\):

\[ -146.0 = 2 \times \Delta H_{f}^{\circ}(\mathrm{CuO}) + 146.0 \]

Subtract 146.0 from both sides:

\[ -292.0 = 2 \times \Delta H_{f}^{\circ}(\mathrm{CuO}) \]

Divide by 2:

\[ \Delta H_{f}^{\circ}(\mathrm{CuO}) = -146.0 \, \mathrm{kJ/mol} \]

\(\boxed{-146.0 \, \mathrm{kJ/mol}}\)

The enthalpy of formation for \(\mathrm{CuO}(s)\) is \(\boxed{-146.0 \, \mathrm{kJ/mol}}\).