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April 17, 2014

April 17, 2014

Posted by **DC** on Sunday, November 3, 2013 at 1:38pm.

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

2 Cu2O + O2 --> 4CuO

The change in enthalpy upon reaction of 75.30 g of Cu2O(s) is -76.83 kJ. Calculate the work, w, and energy change, ΔUrxn, when 75.30 g of Cu2O(s) is oxidized at a constant pressure of 1.00 bar and a constant temperature of 25°C.

Additional Details

Useful equations

Ideal Gas: PV = nRT

Work: w = –PΔV

Energy Change: ΔUrxn = ΔHrxn – PΔV

Thank you in advance! :)

- Chemistry -
**DrBob222**, Sunday, November 3, 2013 at 8:08pmIf you had shown your work perhaps could have found the error. When you did pdV did you leave that in L*atm or did you convert that to kJ? Why not wshow your work and let us look at it?

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