You try to get
3Fe2O3(s) + CO(g)=>2Fe3O4(s) + CO2(g)
you have (and by the way k stands for kilo and K for °Kelvin so kJ stands for kiloJoules)
1)Fe2O3(s) + 3CO(g)=>2Fe(s) + 3CO2(g) H=-5.7kJ
2)Fe3O4 + CO(g)=>3FeO(s) + CO2(g) H=-4.5kJ
3)Fe(s) + CO2(g)=>FeO(s) + CO(g) H=-0.3kJ
Those equations can be considered as mathematical equations... you can make all sorts of operations on it...multiplications, substractions, sums,
inversions...
If you take equation 1 you should multiply it by 3 to get something closer to your taget equation...eveything must be multiplied by 3 even the output
energy!
1)3Fe2O3(s) + 9CO(g)=>6Fe(s) + 9CO2(g) H=-5.7*3kJ= -17.1 kJ
If you consider equation 2 you noticed that indeed the Fe3O4 is on the wrong side...Then you can invert it... but also the resulting energy and at the
same time multiply everything by a factor 2 to get closer to the desired equation
2)6FeO(s) + 2CO2(g) ==> 2Fe3O4 + 2CO(g) H=+9.0kJ
If you combine those two equations...1 and 2...by summing all, including energy...
1/2)3Fe2O3(s) + 9CO(g) + 6FeO(s) + 2CO2(g) => 2Fe3O4 + 2CO(g) + 6Fe(s) + 9CO2(g) H=-17.1 kJ + 9.0 kJ=-8.1kJ
You now are very close to the desired equation but you have some members that are too much...
like FeO and Fe, by chance those are present in your equation 3...so maybe multiplying that equation by a factor 6 would make things clearer...
3)6Fe(s) + 6CO2(g)=>6FeO(s) + 6CO(g) H=-1.8kJ
By summing this third equation with the mix of the two first...
1/2/3)3Fe2O3(s) + 9CO(g) + 6FeO(s) + 8CO2(g) + 6Fe(s) => 2Fe3O4 + 8CO(g) + 6Fe(s) + 6FeO(s)+ 9CO2(g) H=-8.1kJ-1.8kJ=-9,9kJ
And by simplifying a bit what is present on both sides of the equation...you end up with the desired equation.
1/2/3)3Fe2O3(s) + CO(g) => 2Fe3O4 + CO2(g) H=-8.1kJ-1.8kJ=-9,9kJ
PH Z (PHILOU Zrealone)
"Physic is all what never works; Chemistry is all what stinks and explodes!"-"Life that deadly disease, sexually transmitted."(W.Allen)
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