Start with the end equation.
FeO(s) + CO(g) ==> Fe(s) + CO2(g)
I want an Fe(s) on the right and CO2(g) on the right. The first equation has that so write that one as is (at least for the time being).
I would reverse equations 2 and 3
and multiply equation 1 by 3, equation 2 by 1, and equation 3 by 2. Then add everything on the left side and everything on the right side, then cancel those molecules/atoms that are on both sides. If I didn't goof I obtained the equation you want by doing that.
Given:
*Fe2O3(s)+3CO(s)-->2Fe(s)+3CO2(g)(delta h is -23kJ)
*3Fe2O3(s)+CO(g)-->2Fe3O4(s)+CO2(g)(delta h is -39kJ)
*Fe3O4(s)+CO(g)-->3FeO(s)+CO2 (delta h is 18kJ)
Desired:FeO(s)+CO(g)-->Fe(s)+CO2(g)
find delta h for the desired reaction.
which reactions do i reverse and then how do i cancel the out the ones i don't need?
5 answers
ok this is what i have gotten:
3Fe2O3(s)+9CO(s)-->6Fe(s)+9CO2(g)
2Fe3O4(s)+CO2(g)-->3Fe2O3(s)+CO(g)
6FeO(s)+2CO2(g)-->2Fe3O4(s)+2CO(g)
but the desired reaction is in a one to one ratio, and also the CO(g) is on the left side in the desired reaction. does that mean i have to switch over again?
3Fe2O3(s)+9CO(s)-->6Fe(s)+9CO2(g)
2Fe3O4(s)+CO2(g)-->3Fe2O3(s)+CO(g)
6FeO(s)+2CO2(g)-->2Fe3O4(s)+2CO(g)
but the desired reaction is in a one to one ratio, and also the CO(g) is on the left side in the desired reaction. does that mean i have to switch over again?
I don't think so. Let's see how it plays out. I'll go through some of it and let you finish.
3Fe2O3 at top left cancels with 3Fe2O3 on right in equation 2.
2Fe3O4 on left in equation 2 cancels with 2Fe3O4 on right of equation 3. Then 9CO in equation 1(left side) cancels with 1 CO on right of equation 2 and 1 CO on right of equation 3 to leave 6 CO on left side. For CO2, we have 2CO2 on left of equation 3 and 1 CO2 on left of equation 2 which subtracts from 9 CO2 on right of equation 1 leaving 6 CO2 on right side. If I've done all that right (you should confirm all of this), that leaves 6FeO + 6CO ==> 6Fe + 6 CO2. So whatever delta H adds up, just divide by 6 to get delta H for the equation you want (1:1:1:1).
3Fe2O3 at top left cancels with 3Fe2O3 on right in equation 2.
2Fe3O4 on left in equation 2 cancels with 2Fe3O4 on right of equation 3. Then 9CO in equation 1(left side) cancels with 1 CO on right of equation 2 and 1 CO on right of equation 3 to leave 6 CO on left side. For CO2, we have 2CO2 on left of equation 3 and 1 CO2 on left of equation 2 which subtracts from 9 CO2 on right of equation 1 leaving 6 CO2 on right side. If I've done all that right (you should confirm all of this), that leaves 6FeO + 6CO ==> 6Fe + 6 CO2. So whatever delta H adds up, just divide by 6 to get delta H for the equation you want (1:1:1:1).
i got -1/3 for delta H. thank you very much, you have been a big help.
You might want to check your work. Since you multiplied equation 1 by 3, did you multiply delta H by 3? And you multiplied equation 3 by 2, did you multiply delta H by 2? If those are delta Hs for the reaction as shown, then the multiplications are in order and that will change the final answer.
3 answers