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Author: Subject: Enthalpy question
yoyoils
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sad.gif posted on 19-8-2011 at 17:05
Enthalpy question


Use the thermochemical equations shown below to determine
the enthalpy for the reaction: 3Fe2O3(s) + CO(g)=>2Fe3O4(s) + CO2(g)


Fe2O3(s) + 3CO(g)=>2Fe(s) + 3CO2(g) H=-5.7KJ

Fe3O4 + CO(g)=>3FeO(s) + CO2(g) H=-4.5KJ

Fe(s) + CO2(g)=>FeO(s) + CO(g) H=-0.3KJ

The rules I'm trying to follow are
Reactant must appear on reactant side, and product belongs on product side then multiple or divide following by summing up of the enthalpy but the thing I don't understand is equation #2 how CO & Fe3O4 are both on the same side so I'm not sure how to change it. Thank you in advance for your time! I'm in the middle of working this out as I post this.




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PHILOU Zrealone
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[*] posted on 4-9-2011 at 13:57


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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