Question

References t6-3 hess's law for the reaction: agi(s) + br2(g) → agbr(s) + i2(s), δh° = –54.0 kj δhf° for agbr(s) = –100.4 kj/mol δhf° for br2(g) = +30.9 kj/mol the value of δhf° for agi(s) is:

Answer 1

Answer: 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