Dentista08

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Given: Fe(s) + H2O(g) à FeO(s) + H2(g)∆H = -24.7 kJ
2 FeO(s) + 1/2O2(g) à Fe2O3(s)∆H = -289.2 kJ
H2(g) + 1/2O2(g) à H2O(g)∆H = -241.8 kJ

Calculate the enthalpy change, ∆H for the following reaction:
2Fe(s) + 3H2O(g) à Fe2O3(s) + 3H2(g)

isnt it products- reactants?
im getting [-289.2 + 3(241.8)] - [2(24.7)+3(-241.8)]

the correct answer is -96.8kJ

i dont get thier explanation :uhno:
 
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MTD52

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Given: Fe(s) + H2O(g) à FeO(s) + H2(g)∆H = -24.7 kJ
2 FeO(s) + 1/2O2(g) à Fe2O3(s)∆H = -289.2 kJ
H2(g) + 1/2O2(g) à H2O(g)∆H = -241.8 kJ

Calculate the enthalpy change, ∆H for the following reaction:
2Fe(s) + 3H2O(g) à Fe2O3(s) + 3H2(g)

isnt it products- reactants?
im getting [-289.2 + 3(241.8)] - [2(24.7)+3(-241.8)]

the correct answer is -96.8kJ

i dont get thier explanation :uhno:


You have to balance the equations so that you can get to the reaction they are asking about. In other words, you multiply reactions or switch them around so when you "add" them all up it equals what they're asking for. It's sort of hard to explain and I feel like what I'm saying isn't making sense lol, so try to just follow what I did:

first reaction multiply by 2 (including ∆H) :
2*(Fe(s) + H2O(g) --> FeO(s) + H2(g); ∆H = -24.7 kJ)
which equals
2Fe(s) + 2H2O(g) --> 2FeO(s) + 2H2(g); ∆H = -49.4 kJ

2nd reaction leave alone:
2 FeO(s) + 1/2O2(g) --> Fe2O3(s); ∆H = -289.2 kJ

3rd reaction switch around (makes the sign of ∆H opposite of what it was):
H2O(g) --> H2(g) + 1/2O2(g); ∆H = 241.8 kJ

If you put those reactions one on top of another and add the products and reactants, canceling out what is equal on both sides (1/2O2 and 2FeO), you are left with the reaction they are asking for:
2Fe(s) + 3H2O(g) --> Fe2O3(s) + 3H2(g)

Then just add the three enthaplies --> -49.4 + (-289.2) + 241.8 = -96.8kJ
 

Dentista08

looking for a practice!
10+ Year Member
5+ Year Member
Jul 25, 2008
228
0
Boston, MA
Status (Visible)
  1. Dental Student
  2. Dentist
You have to balance the equations so that you can get to the reaction they are asking about. In other words, you multiply reactions or switch them around so when you "add" them all up it equals what they're asking for. It's sort of hard to explain and I feel like what I'm saying isn't making sense lol, so try to just follow what I did:

first reaction multiply by 2 (including ∆H) :
2*(Fe(s) + H2O(g) --> FeO(s) + H2(g); ∆H = -24.7 kJ)
which equals
2Fe(s) + 2H2O(g) --> 2FeO(s) + 2H2(g); ∆H = -49.4 kJ

2nd reaction leave alone:
2 FeO(s) + 1/2O2(g) --> Fe2O3(s); ∆H = -289.2 kJ

3rd reaction switch around (makes the sign of ∆H opposite of what it was):
H2O(g) --> H2(g) + 1/2O2(g); ∆H = 241.8 kJ

If you put those reactions one on top of another and add the products and reactants, canceling out what is equal on both sides (1/2O2 and 2FeO), you are left with the reaction they are asking for:
2Fe(s) + 3H2O(g) --> Fe2O3(s) + 3H2(g)

Then just add the three enthaplies --> -49.4 + (-289.2) + 241.8 = -96.8kJ
now i get it! thank you!!!
 
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