Asked by DC
                help - confusion :(
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.
Here's my line of reasoning.. but it's been marked wrong. Thank you in advance.
converting 75.3g of Cu2O to moles of CuO, I got 1.0525mol CuO
W=-delta(n)RT = -(1.0525mol)(8.3144J/molK)(25+273.15)= -2.609 kJ
energy change = -2.609kj+ - 76.83kJ = -79.439kJ
            
            
        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.
Here's my line of reasoning.. but it's been marked wrong. Thank you in advance.
converting 75.3g of Cu2O to moles of CuO, I got 1.0525mol CuO
W=-delta(n)RT = -(1.0525mol)(8.3144J/molK)(25+273.15)= -2.609 kJ
energy change = -2.609kj+ - 76.83kJ = -79.439kJ
Answers
                    Answered by
            bobpursley
            
    the heat of reaction is 76.83kJ per mole of Cu2O, not CuO.
Then, work=moles*Cu2O*76.83kJ
and that is it. Your delta(n)RT is off the thinking mark.
    
Then, work=moles*Cu2O*76.83kJ
and that is it. Your delta(n)RT is off the thinking mark.
                    Answered by
            DrBob222
            
    You appear to be calculating the change in volume from mols Cu2O ==> 2CuO. But each is a solid and they don't count do they? Isn't the change in volume the change due to the gas which is oxygen? Shouldn't you calculate how much oxygen was there initially and how much is there at the end (zero of course) and p*delta V then is the work done by the surroundings on the system. The mols O2 you can calculate from mols Cu2O given in the problem. 
    
                    Answered by
            DC
            
    ok I recalculated and got the w=.006523kJ and the energy change to be -76.82kJ... The energy change is right but the w is wrong for some reason... 
I calculated w as w=-delta(n)RT = -(0-.26312molO2)(8.314452x10^-2 Lbarr/molK)(25+273.15) = 6.5226J=.006523kJ...
    
I calculated w as w=-delta(n)RT = -(0-.26312molO2)(8.314452x10^-2 Lbarr/molK)(25+273.15) = 6.5226J=.006523kJ...
                    Answered by
            HELP!!!!!!!!
            
    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 70.64 g Cu2O(s) is −72.08 kJ .
Calculate the work, 𝑤 , and energy change, Δ𝑈rxn , when 70.64 g Cu2O(s) is oxidized at a constant pressure of 1.00 bar and a constant temperature of 25∘ C .
Note that Δ𝐸rxn is sometimes used as the symbol for energy change instead of Δ𝑈rxn .
im confused
    
2Cu2O(s)+O2(g)⟶4CuO(s)
The change in enthalpy upon reaction of 70.64 g Cu2O(s) is −72.08 kJ .
Calculate the work, 𝑤 , and energy change, Δ𝑈rxn , when 70.64 g Cu2O(s) is oxidized at a constant pressure of 1.00 bar and a constant temperature of 25∘ C .
Note that Δ𝐸rxn is sometimes used as the symbol for energy change instead of Δ𝑈rxn .
im confused
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