Gibb's free energy (very simple one)

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SUMMARY

The discussion centers on calculating the equilibrium constant (K) for the formation of nitric oxide at 25°C using Gibbs free energy (G). The user calculated G as 1.73x105 J/mol and applied the equation ln K = -G/RT, resulting in ln K = -69.78 and K = 4.95x10-31. However, the user received an error from the submission system, prompting inquiries about potential unit discrepancies, particularly regarding the values for R and S. The CRC Handbook of Chemistry and Physics was recommended as a resource for unit conversions.

PREREQUISITES
  • Understanding of Gibbs free energy and its significance in chemical reactions
  • Familiarity with the equation ln K = -G/RT
  • Knowledge of unit conversions, particularly between J/mol and kJ/mol
  • Basic proficiency in thermodynamics and equilibrium concepts
NEXT STEPS
  • Review the CRC Handbook of Chemistry and Physics for unit conversions and constants
  • Learn about the significance of Gibbs free energy in chemical thermodynamics
  • Study the relationship between enthalpy (H), entropy (S), and Gibbs free energy (G)
  • Practice calculating equilibrium constants using various values of G, H, and S
USEFUL FOR

Chemistry students, educators, and professionals involved in thermodynamics and chemical equilibrium calculations will benefit from this discussion.

pari786
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Homework Statement



Calculate the equilibrium constant for the reaction forming nitric oxide at room temperature, 25C. G= 1.73x10^5 J/mol, H = 180.5KJ/mol an S = 24.8J/(mol.K)

Homework Equations



ln K = -G/RT

The Attempt at a Solution



I subsiuted ln K = (-1.73x10^5) / (8.3145x298.15)
ln K = -69.78
e^-69.78
I'm getting K as 4.95*10^-31

but the computer (where I submit online assignment ) is saying its wrong ... I've only one single attempt left ..can anyone see where I'm wrong.
Thanks and immediate help please!
 
Last edited:
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Please show some more work. In particular, how did you obtain the value for G? What units did you use for R and S? Example: H is in KJ/mol and S is in J/molK. When I mess up these calculations, it is usually related to units.

BTW, for a very useful compilation of R in various units, consult the CRC Handbook of Chemistry and Physics. In the 84th edition (2003-2004), a very complete table is printed on page 1-54.
 

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