Chemical Equilibrium Constant Calculation

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SUMMARY

The discussion centers on calculating the equilibrium constant (Keq) for the reaction C(s) + O2(g) → CO2(g) at 25°C, given the reaction quotient (Q = 1.0 x 10^42). The standard Gibbs free energy change (dG°) values provided are C(s) = 0, O2(g) = 0, and CO2(g) = -394.36 kJ/mol. The correct formula to use is dG°rxn = -R*T*ln(Keq), leading to the calculation of Keq as 1.3 x 10^69, which was initially miscalculated. The user expressed confusion regarding the relationship between Q and Keq.

PREREQUISITES
  • Understanding of Gibbs free energy and its relation to chemical reactions
  • Familiarity with the ideal gas law and standard conditions
  • Knowledge of logarithmic functions and their application in chemistry
  • Proficiency in using the equation dG°rxn = -R*T*ln(Keq)
NEXT STEPS
  • Study the relationship between reaction quotient (Q) and equilibrium constant (Keq)
  • Learn about the implications of Gibbs free energy in chemical thermodynamics
  • Explore the derivation and application of the van 't Hoff equation
  • Investigate the effects of temperature on equilibrium constants
USEFUL FOR

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

hotwheelharry
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Homework Statement



If Q = 1.0 1042 for the reaction C(s) + O2(g) → CO2(g) at 25°C, calculate Keq.

dGof (Kj/mol) of...
C(s)=0
O2(g)=0
CO2(g)=-394.36

Homework Equations


dGorxn = -R*T*ln(K)

The Attempt at a Solution



-394.36*1000 = -8.314472(298)ln(K)
K=1.3e69

This is the wrong answer. I'm lost as to what to do. Please help.
I might be using the wrong equation for the situation. The only information I was given was in the problem statement.
 
Last edited:
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What is Q? How is it related to Keq?
 

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