**Given reaction:** 2 Cu_{2}O(s) + O_{2}(g) → 4 CuO(s)

**Given:** mass of Cu_{2}O(s) = 67.68 g Cu_{2}O(s)

ΔH_{rxn} = – 69.06 kJ

**Calculating the mass of O _{2}(g) that reacts with Cu2O(s):**

$\mathbf{67}\mathbf{.}\mathbf{68}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{Cu}}_{\mathbf{2}}\mathbf{O}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{Cu}}_{\mathbf{2}}\mathbf{O}}}{\mathbf{143}\mathbf{.}\mathbf{1}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{Cu}}_{\mathbf{2}}\mathbf{O}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{O}}_{\mathbf{2}}}{\mathbf{2}\mathbf{}\overline{)\mathbf{mol}\mathbf{}{\mathbf{Cu}}_{\mathbf{2}}\mathbf{O}}}$ **= 0.2365 mol O _{2}**

The oxidation of copper(I) oxide, Cu_{2}O(s), to copper(II) oxide, CuO(s), is an exothermic process, 2 Cu_{2}O(s) + O_{2}(g) → 4 CuO (s). The change in enthalpy upon reaction of 67.68 g of Cu_{2}O(s) is -69.06 kJ.

Calculate the work, w, and the energy change, ΔU_{rxn}, when 67.68 g of Cu_{2}O (s) is oxidized at constant pressure of 1.00 bar and a constant temperature of 25 degree Celcius?

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