The oxidation of copper(I) oxide, Cu2O(s), to copper(II) oxide, CuO(s), is an exothermic process,
2 CuO2(s) + 02(g) ---> 4 CuO (s)
The change in enthalpy upon reaction of 67.68 g of CuO (s) is -69.06 kJ
Calculate the work, w, and the energy change, delta Urxn, when 67.68 g of Cu2O (s) is oxidized at constant pressure of 1.00 bar and a constant temperature of 25 degree celcius?
(the two answer must be in kJ)
I can't get the right answer for some reason.
Show work please.
The balanced chemical reaction is :
2Cu2O (s) + O2 (g) -------------> 4CuO (s)
given that 67.68 g of Cu2O is reacted i.e.
number of moles of Cu2O reacted = 67.68 g / 143.1 g mol-1 = 0.473 mol
The stoichiometry of the balanced equation tells that each 2 mol of Cu2O reacts with 1 mole of O2
number of moles of O2 reacted = 0.473 / 2 = 0.24 mol
change in number of moles (Δn) of the compound in the gaseous state in the reaction is :
Δn = nproducts (g) - nreactants(g)
= 0 - 0.24
and work done in chemical reaction given by
w = - ΔnRT
= - (-0.24 mol) * 8.314 J K-1 mol-1 * 298 K
= 594.62 J
~= 0.594 kJ which is actually expansion (compression) work (-PΔV)
positive sign indicates that work is done on the system.
Also, ΔH = ΔU + Δ(PV)
at constant pressure,
ΔH = ΔU + PΔV
=> ΔU = ΔH - PΔV
= -69.06 - 0.59
~= -69.65 kJ