※ 引述《dum54 (Mr.D)》之銘言：
: Consider the reaction as following:(C2H2:26.04 amu)
: C2H2(g) + 5/2 O2(g) ---> H2O(g) + 2CO2(g) ΔH = -1255.5 kJ
: 1.Please calculate the PV work for the reaction of 13.0g of acetylene at
: atmospheric pressure if the volume change is -13.0L
既然在大氣壓底下那就是1atm=101300pa 13L=0.013m^3
W=-PΔV=101300*0.013=1316.9J
: 2.Please use the data described above to calculate the heat(q) for the
: reaction of 13.0g of acetylene at atmospheric pressure if the volume change
: is -13.0L.
ΔH=Qp 所以應該是-1255.5kJ/1mol → -627.8kJ
: 3.Please use the data described above to calculate the value of ΔE for the
: reaction of 13.0g of acetylene at atmospheric pressure if the volume change
: is -13.0L.
: 謝謝各位~
-627.8k=ΔE+Δ(PV)=ΔE+PΔV ΔE=-626.4kJ
老實說非常不確定....
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