# The oxidation of copper(i) oxide cu2o

The oxidation of copper(I) oxide,
Cu2O(s), to copper(II) oxide, CuO(s), is an exothermic
process,
The oxidation of copper(I) ,
Cu2O(s), to copper(II) oxide, CuO(s), is an exothermic
process,
2 CuO2(s) + 02(g) —> 4 CuO (s)
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
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?
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)
(the two answer must be in kJ)

I can’t get the right answer for some reason.
I can’t get the right answer for some reason.
Thank you.
Thank you.

Also Read :   Correctly identify this tissue type and then label the features of the tissue

classmAte Date Page Mo 143-) PAo eik ニー587-18 /Tou
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
therefore;
number of moles of O2 reacted = 0.473 / 2 = 0.24
mol
Now,
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
= -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)