Gibbs Energy and Reaction Quotient

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SUMMARY

The discussion centers on the relationship between Gibbs energy and the reaction quotient (Q) in chemical reactions. When pure reactants A and B are mixed to form product C, the initial concentration of C is zero, leading to a reaction quotient Q of zero. This results in a Gibbs energy change (ΔG) that is theoretically undefined at the start of the reaction, as ΔG is calculated using the equation ΔG = ΔG(standard) + RTln(Q). However, it is acknowledged that the Gibbs energy can be considered infinitely negative, indicating a strong driving force for the formation of C.

PREREQUISITES
  • Understanding of Gibbs energy and its significance in thermodynamics
  • Familiarity with the reaction quotient (Q) and its calculation
  • Basic knowledge of chemical reaction dynamics
  • Concept of standard Gibbs energy (ΔG(standard))
NEXT STEPS
  • Study the implications of Gibbs energy in chemical equilibrium
  • Learn about the calculation and significance of reaction quotients in various reactions
  • Explore the concept of standard Gibbs energy and its applications in predicting reaction spontaneity
  • Investigate the role of temperature and concentration in Gibbs energy changes
USEFUL FOR

Chemistry students, chemical engineers, and researchers interested in thermodynamics and reaction kinetics will benefit from this discussion.

llabesab16
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I have been putting some thought into understanding Gibbs energy but I can't quite figure one thing out. Here is my dilemma:

Say that someone wants to react A with B to form C, and they mix pure A with pure B. At the moment the reaction starts, there is no C in the mixture (is this correct?). If the reaction quotient is [C]/[A], then shouldn't the reaction quotient equal zero at the beginning of the reaction? If that is the case, then wouldn't the change in Gibbs energy be undefined at the beginning of the reaction if deltaG = deltaG(standard) + RTln(Q)?

Thanks for your help!
 
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Sure. You could also think of it as being infinitely negative, so the formation of C will be very favorable.
 

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