Both a particle and a wave?
<span>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</span>
The first one is a transform boundary. The second one is a convergent boundary.
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