Question

The oxidation of copper(I) oxide, Cu2O(s), to copper(II) oxide, CuO(s), is an exothermic process.

2Cu₂O(s)+O₂(g) ----> 4CuO(s)

The change in enthalpy upon reaction of 74.84 g of Cu2O(s) is -76.36 kJ. Calculate the work, w, and energy change, ?Urxn, when 74.84 g of Cu2O(s) is oxidized at a constant pressure of 1.00 bar and a constant temperature of 25 °C.

Answer 1

2Cu₂O (s) + O₂ (g) 4CuO (s)

Molar mass of Cu₂O = 143.1 gmol-1

Therefore,

Number of moles of Cu₂O reacted = 74.84 g / 143.1 g mol^-1 = 0.522 mol

From the balanced equation 1 mole of O₂ reacts with 2 mol of Cu₂O

Therefore,

Number of moles of O₂ reacted = 0.522 / 2 = 0.261 mol

Change in number of moles (n) of the components in the gaseous stat

n = n_{products (g)} - n_reactants(g) = 0 - 0.261 = -0.261

We have, Work, w = - nRT = - (-0.261) x 8.314 x 298 = 646.64 J = 0.646 KJ

w = 0.646 KJ

We have, enthalpy change of the reaction at constant pressure,

H = U + PV

Therefore,

U = H – PV

U = -76.36 - 0.646 = -77 KJ

U = -77 kJ