Promena entalpije

pmf93's picture

Prema datim termohemijskim jednacinama i toplotnim kapacitetima izracunati promenu entalpije stvaranja N2O na   298 K i 423 K:

C(s) + N2O(g) = CO(g) + N2(g)               deltaH(298) = -192.9 kJ/mol

C(s) + O2(g) = CO2(g)                            delta H(298) = -393.5kJ/mol

CO(g) + 1/2O2(g) = CO2(g)                     delta H(298) = -283.0 kJ/mol

C(N2)= (27.0+ 0.006T) J/mol*K

C(O2)= (25.6*0.014T) J/mol*K

C(N2O)= (27.2*0.044T) J/mol*K

milicaradenkovic's picture

CO(g)+N2(g)-------------------->C(s)+N2O(g)                           deltaH1(298)=192.9 kJ/mol

C(s)+O2(g)------------------>CO2(g)                                        deltaH2(298)= -393.5 kJ/mol

CO2(g)-------------------------->CO(g)+1/2O2(g)                          deltaH3(298)= 283.0 kJ/mol

CO(g)+N2(g)+C(s)+O2(g)+CO2(g)------------------->N2O(g)+C(s)+CO2(g)+CO(g)+1/2O2(g)

N2(g)+1/2O2(g)------------------>N2O(g)

deltaH(298)=deltaH1(298)+deltaH2(298)+deltaH3(298)=192.9-393.5+283.0=82.4 kJ/mol

Ja nemam masu da bi nasla promenu entalpije.