Calculate the expansion work done when 50 g of water is electrolysed under constant pressure at 25C
9 answers
-10KJ
-10KJ
how did u get -10 kj
Please show procedures
I don't no the answer
-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.
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
10287 J
-10287 J