Suppose the gases involved in a reaction behave as real gases with significant intermolecular attractions.

I just learned that $$K_c(RT)^{\Delta n_r} = K$$

Would the value of Kc calculated from the equation above be too high, too low, or the same as the actual value?

I thought that it would be the same, but I don't think that's right because deviation from ideal behavior should deviate from equilibrium.


1 Answer 1


I believe that the real $K_c$ would be higher. This is because in ideal gases london dispersion forces, hydrogen bonding, etc. is ignored. These would increase the number of collisions between particles since they would be attracted to each other more. $$\ce k=zp*e^{-E_a/RT}$$ where k= rate constant, z= frequency of collisions, p= fractions of collisions with proper orientation to cause a reaction, $E_a$= activation energy, T= temperature. So Since z would increase, the K would increase. Hence the $K_c$ would increase.

  • $\begingroup$ Wow, that was actually a really helpful answer. I do want some confirmation from other users as well but your answer actually really helped me out. Thank you! $\endgroup$
    – phi2k
    Jan 1, 2017 at 19:17
  • $\begingroup$ The first time I read this I was confused. Are you saying that $K_c$ would be higher for all ideal gases, or simply saying that $K_c$ would be higher for gases that have significant intermolecular attractions? (I.e., would $K_c$ be lower for gases that have intermolecular repulsions?) $\endgroup$
    – Curt F.
    Jan 2, 2017 at 2:50
  • $\begingroup$ Well it would be higher for all gases. Ideal gases don't technically exist. All gases have some intermolecular forces thought it may be small in certain cases. The difference between the true $K_c$ and "fake ideal" $K_c$ of a gas would increase with stronger intermolecular interactions. Since the question specifically asked for gasses with significant intermolecular attractions I aimed it towards attraction not repulsion. Other than heat, the intermolecular attractions are much larger than the repulsion in all gases. And heat is summed up in T in the equation which is assumed constant. $\endgroup$ Jan 2, 2017 at 3:08

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