Asked by Dana
Calculate the expansion work done when 50 g of water is electrolysed under constant pressure at 25C
Answers
Answered by
frank
-10KJ
Answered by
SALISU SANI
-10KJ
Answered by
mau
how did u get -10 kj
Answered by
fahad
Please show procedures
Answered by
mukarakate tendai
I don't no the answer
Answered by
Victor
-10kJ because you have to consider the work due the H2 (-6,7kJ) and the work due the O2 produced (-3.3kJ) in the reaction.
Answered by
Andre Tjahjono
Here is the answer
2H2O -------> O2 + 2H2
2.7 mol 1,38 mol 2,7 mol
mol H2O = 50g/18 g/mol = 2.7 mol
for general eq -> w = - Pex.delta V
Pex= external pressure
delta V = Vfinal (V2) - V initial(V1)
neglect the intial volume because the final volume (after the production of gas) is so much larger and delta V = Vf- Vi=Vf=nRT/Pex, then : W = -(n.R.T)
wO2 = -(nxRxT)
wO2 = 1,38 mol x 8,3145 J.K^-1.mol^-1 x 298 =-3,4 Kj
same for calcualting O2 then wH2 = -6,8 kJ
Wtotal = WO2 + WH2 = -3,4 Kj + (-6,8 Kj) = 10,2 Kj or 10 Kj
2H2O -------> O2 + 2H2
2.7 mol 1,38 mol 2,7 mol
mol H2O = 50g/18 g/mol = 2.7 mol
for general eq -> w = - Pex.delta V
Pex= external pressure
delta V = Vfinal (V2) - V initial(V1)
neglect the intial volume because the final volume (after the production of gas) is so much larger and delta V = Vf- Vi=Vf=nRT/Pex, then : W = -(n.R.T)
wO2 = -(nxRxT)
wO2 = 1,38 mol x 8,3145 J.K^-1.mol^-1 x 298 =-3,4 Kj
same for calcualting O2 then wH2 = -6,8 kJ
Wtotal = WO2 + WH2 = -3,4 Kj + (-6,8 Kj) = 10,2 Kj or 10 Kj
Answered by
bigun
10287 J
Answered by
bigun
-10287 J
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