Set933/10012/12/2015 0533 PM Gr Print CalauatorPorlodic Table on 5 of 16 Mapdt T

Question

Set933/10012/12/2015 0533 PM Gr Print CalauatorPorlodic Table on 5 of 16 Mapdt The oxidation of coppert) axide, CuO(s), to coppor/ll) oxide, CuO(s), is an exothermic process. The change in enthalpy upon reaction of 18.03 g of CuO(s) is-18.40 kJ. Calculate the work, w, and energy change, AUn, when 18.03 g of CuO(s) is oxidized at a constant pressure of 1 CO bar and a constant temperature of 25 c. Number Number Note that n is sometimes used kJas the symbol for energy change instead of A Hint o@ e ()| anything

Explanation / Answer

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 18.03 g of Cu2O(s) is -52.80 kJ. Calculate the work, w, (in KJ) and energy change, Urxn, when 51.75 g of Cu2O(s) is oxidized at a constant pressure of 1.00 bar and a constant temperature of 25°C.

1. So first step I did was use the 18.03 g of Cu2O to calculate the moles which was .2580 mol Cu2O.

2. Next I used the equation PV=nRT substituting all my values to find the volume of Cu2O which was
6.3955 L Cu2O.

3. The equation for work is w=-PdeltaV
I did the same process of using the equation PV=nRT for O2 and achieving a volume of 3.1978 L
O2.

4. I subtracted 4.4819 L O2 and 8.9639 L Cu2O and got -3.1976 L

5. From here I converted the -4.482 L to -.0031976 m^3

6. I finally plugged that value into the equation w=-PdeltaV where my P was 1.00 bar but I converted it
to 10^5Pa so altogether it was:
w=-(10^5 Pa)(-.0031976 m^3)=+319.76 J
since my answer had to be in KJ I converted that value to +.3197 KJ

In this case delta U = -w

= -.3197

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