Energikonvertering ved Solid Oxide Electrolyser Cells

A.2. UDREGNINGER AF ∆H I EES
Løsninger
∆gco2 = 203202[k J/kmol ] ∆gh2o = 203949[k J/kmol ]
∆hco2 = 282221[k J/kmol ] ∆hh2o = 248549[k J/kmol ]
E0;co2 = −1,053[V ] E0;h2o = −1,057[V ]
Etn;co2 = 1,463[V ] Etn;h2o = 1,288[V ]
F = 96485[Coul omb/mole] gpr oduct s;co2 = −423094[k J/kmol ]
gpr oduct s;h2o = −236911[k J/kmol ] gr eact ant s;co2 = −626296[k J/kmol ]
gr eact ant s;h2o = −440860[k J/kmol ] hpr oduct s;co2 = −71187[k J/kmol ]
hpr oduct s;h2o = 37934[k J/kmol ] hr eact ant s;co2 = −353407[k J/kmol ]
hr eact ant s;h2o = −210615[k J/kmol ] P = 5[MPa]
spr oduct s;co2 = 313,3[k J/kmol -K ] spr oduct s;h2o = 244,7[k J/kmol -K ]
sr eact ant s;co2 = 243[k J/kmol -K ] sr eact ant s;h2o = 205[k J/kmol -K ]
T = 1123[K ]
VII