Activity - Equilibrium Expressions

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SUMMARY

The discussion focuses on the concept of activity in equilibrium constant expressions, specifically how Kc and Kp relate to each other. It is established that Kc is not equal to Kp due to differing reference states, with Kc based on a concentration of 1 mol/L and Kp on a pressure of 1 atm or 1 bar. An exception occurs when the sum of the powers in the equilibrium expression's numerator and denominator are equal, resulting in Kc equaling Kp. Additionally, changing the reference pressure affects Kp, altering its expression by a factor of 2Δn, where Δn represents the change in moles of gas.

PREREQUISITES
  • Understanding of chemical equilibrium concepts
  • Familiarity with equilibrium constant expressions (Kc and Kp)
  • Knowledge of reference states in thermodynamics
  • Basic grasp of gas laws and pressure units
NEXT STEPS
  • Study the derivation of equilibrium constant expressions for both Kc and Kp
  • Learn about the impact of changing reference states on equilibrium constants
  • Explore the concept of Δn in chemical reactions and its significance
  • Investigate the implications of non-ideal gas behavior on equilibrium calculations
USEFUL FOR

Chemistry students, chemical engineers, and researchers involved in thermodynamics and reaction kinetics will benefit from this discussion, particularly those focusing on equilibrium expressions and their applications in real-world scenarios.

i_love_science
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When I use the concept of activity to express the equilibrium constant expression, for either equilibrium pressures or concentrations, the units cancel and the value of K has no units (which is how K is customarily reported). But because of the difference in reference states (the reference pressure is 1 atm (or 1 bar) and the reference concentration is 1 mol/L), Kc is not equal to Kp.

Why is there an exception to this principle when the sum of the powers in the numerator and the denominator of the equilibrium expression are the same? Aren't the reference states still different? In this case Kc is equal to Kp. Thank you!
 
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Consider changing the reference pressure, say from 1 bar to 0.5 bar. Each reagent pressure (expressed as a multiple of the reference pressure) will increase by a factor of 2. The expression for Kp will therefore change by a factor 2Δn. If Δn = 0, the value of K is independent of the reference pressure.
 
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Likes   Reactions: etotheipi, i_love_science and jim mcnamara
Thanks, I understand now.
 

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