Question on the reduction of FeO by CO

  • #1
Some numbers below: these are the standard molar enthalpies and entropies at 298K that I got from a website.
In the reaction FeO + CO = Fe + CO2, the deltas are:
delta H = (0+ -393) - (-266.5+ -110.5) = -16.3 KJ/mol.
delta S = (27.2+213.8) - (197.9+54) = -10.9 J/mol/K
Now at 1200 K,
Go = -16.3 + (1200)(-10.9) = -3.23
I have ignored the effect of temperature on enthalpy as both the reactants and the products are at the same temperature.
The issue with this is that it suggests that the reaction is spontaneous at 1200K. Therefore, in G = Go + RT ln Q, ln Q is >0 and therefore pCO2/pCO>1. However, we know the opposite to be the case in real life where pCO2/pCO = 1/2.3 in the blast furnace
Where am I going wrong??

Hf (KJ/mol K)Sf (J/mol K)
FeOs
-266.5​
54​
COg
-110.5​
197.9​
CO2g
-393.3​
213.8​
Fes
0​
27.2​
 

Answers and Replies

  • #2
Bystander
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I have ignored the effect of temperature on enthalpy as both the reactants and the products are at the same temperature.
Where am I going wrong??
 
  • #3
Thanks - I took a second look at the calcs, and it appears that I was dividing the Cp in KJ/Kg by the molar mass instead of multiplying it. With that adjustment, the temperature adjustment of the enthalpy is no longer negligible. I am still getting a number different than 2.3, but at least Ln(Q) is negative.
 
  • #4
mjc123
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Therefore, in G = Go + RT ln Q, ln Q is >0
What are you taking as G and G0 here?
Are you assuming that the blast furnace is at equilibrium? Is that true? Do they add extra CO to force the reaction to the right?
 
  • #5
dG (note: should have used the deltas in the original post) = 0 in the limiting case. I'm trying to find the corresponding value of ln Q = pCO2/pCO

dGo = dH-TdS; where for every reactant and product, H = Ho + Cp(T-298K). It's a bit complicated because the Cp itself changes with temperature.

The blast furnace has many zones in it, corresponding to different temperatures, as the iron ore travels down from the top and the CO comes up from the bottom. The CO is not injected externally. The ratio of CO2 to CO comes about from thermodynamic balance as the CO travels up from the bottom and reduces the iron ore in the layers above.
 

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