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,

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.

Thank you.

Show work please.

Answer

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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)

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

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