How Do Stoichiometric Coefficients Affect Gibbs Free Energy Calculations?

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SUMMARY

The discussion centers on the impact of stoichiometric coefficients on Gibbs Free Energy calculations for the reaction N2(g) + 3H2(g) <--> 2NH3(g) at 298K and 1 bar. The participants clarify that both stoichiometric representations yield valid Gibbs Free Energy values, with the first yielding -32.90 kJ/mol and the second yielding -16.45 kJ/mol. The equilibrium constants differ based on the stoichiometry used, with K = [NH3]^2/[N2][H2]^3 for the first case and K = [NH3]/[N2]^(1/2)[H2]^(3/2) for the second. The necessity of a balanced reaction is emphasized for accurate equilibrium constant reporting.

PREREQUISITES
  • Understanding of Gibbs Free Energy and its relation to chemical reactions
  • Familiarity with stoichiometric coefficients in chemical equations
  • Knowledge of equilibrium constants and their calculations
  • Basic principles of thermodynamics, particularly at standard conditions (298K, 1 bar)
NEXT STEPS
  • Study the derivation of the Gibbs Free Energy equation in chemical thermodynamics
  • Learn how to calculate equilibrium constants from Gibbs Free Energy values
  • Explore the implications of different stoichiometric representations on reaction dynamics
  • Investigate the role of temperature and pressure in Gibbs Free Energy calculations
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Chemistry students, chemical engineers, and anyone involved in thermodynamic calculations and chemical equilibrium analysis.

Puchinita5
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Homework Statement



For the reaction N2(g) + 3H2(g) <---> 2NH3(g)

I am supposed to determine the equilibrium constant at 298K and 1 bar. We have been given a table that states that

for NH3(g):
\Delta_{r}G^{\Theta}= -16.45 kJ/mol

I know that
\Delta_{r}G^{\Theta}= \Sigma v G^{\Theta}_{product} - \Sigma v G^{\Theta}_{reactant}

where v is the stoichiometric coefficient.

But what is confusing me is that you could also write the equation as

1/2 N2(g) + (3/2)(g) <----> NH3(g)

so in the first case, the answer would be

(2*-16.45 kJ/mol) - ((1*0)+(3*0)) = -32.90 kJ/mol

But in the second case, the answer would be

(-16.45 kJ/mol) - (1/2*0 + 3/2 *0) = -16.45 kJ/mol

Since both equations should be valid, what answer is correct? I don't understand which one I should choose because they both look like the should be right.
 
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In the two cases, the expression for the equilibrium constant will be different. In the first case K = [NH3]^2/[N2]

^3 whereas in the second case K = [NH3]/[N2]^(1/2)

^(3/2). This is why you always need to include a balanced reaction along with any equilibrium constant you provide.

 
Hey, I had the same concern. They are both right; it depends on the stoichiometry. The professor would have to specify which one he is looking for like he did last time.
 
Thanks guys! Did he specify last time? I must not have understood him. This was really confusing me, good to know that both are right! I feel better now. :)
 

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