1. Oblicz entalpię reakcji Fe2O3(s) + 2Al(s) = 2Fe(s) + Al2O3 mając dane 4Fe(s) + 3O2(g) = 2Fe2O3(s) delta h = -1644 kJ 4Al(s) + 3O2(g) = 2Al2O3(s) delta h = -3340kj

4Fe(s) + 3O2(g) = 2Fe2O3(s) delta H1= -1644 kJ 2Fe2O3(s)= 4Fe(s) + 3O2(g)  delta H1=  1644 kJ Fe2O3(s)= Fe(s) + 3/2O2(g)  delta H1=  822 kJ 4Al(s) + 3O2(g) = 2Al2O3(s) delta H2 = -3340kj 2Al(s) + 3/2O2(g) = Al2O3(s) delta H2 = -1670kj delta H = delta H1+delta H2 = 822-1670=(-848kJ)

ΔH1= -1644kJ ΔH2= -3340kJ ΔHx= ? *(-1/2)/ 4Fe(s) + 3O2(g) --------> 2Fe2O3(s) / ΔH1 *(1/2)/ 4Al(s) + 3O2(g) ---------> 2Al2O3(s) / ΔH2 _______________________________ Fe2O3(s) + 2Al(s) -------> 2Fe(s) + Al2O3 / ΔHx   ΔHx= 1/2ΔH2 + (-1/2)ΔH1 ΔHx= 1/2*(-3340) + (-1/2)* (-1644) ΔHx= -1670 +822 ΔHx= -848kJ