How to use Hess's Law to calculate enthalpy for the reaction?

How to use Hess's Law to calculate enthalpy for the reaction?

• Given the thermochemical equations H2 (g) + Br2 (g) => 2HBr (g) deltaH = -72 kJ H2 (g) => 2H (g) deltaH = 436 kJ Br2(g) => 2Br (g) deltaH = 224 kJ calculate the enthalphy change for H(g) + Br (g) => HBr (g) also if you could also solve this one (if you can!) In the process of isolating iron from its ores, carbon monoxide reacts with Iron (III) Oxide as described by the equation Fe2O3 (s) + 3CO (g) => 2 Fe (s) + 3 CO2 (g) delta H = -24.8 kJ The enthalpy change for the combustion of carbon monoxide is 2CO (g) + O2 (g) => 2 CO2 (g) deltaH= -566 kJ Use the preceding themochemical equations to calculate the enthalphy change for 4Fe (s) + 3 O2 (g) => 2 Fe2O3 (s) thank you so much! if you could answer one or the other or both it would be GREATLY appreciated! :)

• Answer:

  we have to calculate delta H for : H + Br ------> HBr let, H2 + Br2 -----> 2HBr ......delta H = -72 kj....(1) H2 ------> 2H ...delta H = 436 kj......(2) Br2 -------> 2Br ......delta H = 224 kj ......(3) subtracting (2) from (1)... H2 + Br2 - H2 ------> 2HBr - 2H .....delta H = -72-436 = -508 kj Br2 + 2H -----> 2HBr ....delta H = -508 kj ....(4) subtracting (3) from (4)... Br2 + 2H - Br2 ------> 2HBr - 2Br ......deltaH = -508 - 224 = -732 kj 2H + 2Br ------> 2HBr .....delta H = -732 kj dividing it by 2... H + Br ------> HBr ....delta H = -732/2 = -366 kj so the delta H for the reaction is -366 kj --------------------------------------… let, Fe2O3 + 3CO -----> 2Fe + 3CO2 ...delta H = -24.8 kj....(1) 2CO + O2 ------> 2CO2 .....delta H = -566 kj....(2) multiplying (1) by 2... 2Fe2O3 + 6CO ------> 4Fe + 6CO2 ...delta H = -24.8 X 2 = -49.6 kj ....(3) multiplying (2) by 3... 6CO + 3O2 ------> 6CO2 .....delta H = -566 X 3 = -1698 kj.....(4) subtracting (3) from (4)... 6CO + 3O2 - ( 2Fe2O3 + 6CO) ------> 6CO2 - ( 4Fe + 6CO2) ...delta H = -1698 - -49.6 = -1698 + 49.6 = -1648.4 kj 6CO + 3O2 - 2Fe2O3 - 6CO -----> 6CO2 -4Fe - 6CO2 ...delta H = -1648.4 kj 3O2 - 2Fe2O3 -----> -4Fe...delta H = -1648.4 kj 4Fe + 3O2 ------> 2Fe2O3 ....delta H = -1648.4 kj