Why is $K_c$ not affected by change in pressure? I know the mathematical explanation, but I don't really understand the reason when only looking at $K_c$. The explanation I know is with the reference to $K_p$, but is it possible to explain without $K_p$?

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    $\begingroup$ Kc and Kp are.constants for a reaction. They dont get affected by changing volume or pressure. They tell you the value of Q at the equilibrium condition. $\endgroup$ Commented Apr 9, 2016 at 12:18
  • $\begingroup$ I agree. This sounds like K vs. Q confusion. When the pressure, volume, concentrations, whatever change, then Q changes. However, only one value of Q defines equilibrium. The system responds to reestablish equilibrium to that $Q=K$ once again. $\endgroup$
    – Ben Norris
    Commented Apr 10, 2016 at 10:43

4 Answers 4


Equilibrium is reached when the forward and reverse reaction rates are equal. These rates are determined by the frequency of successful collisions, which is turn is determined by concentration (seen in the rate laws), temperature, activation energy, and how often molecules collide with the correct orientation (together these last three determine the rate constant, $k$, as shown in the Arrhenius equation).

The equilibrium constant, $K_c$ is the ratio of the rate constants, so only variables that affect the rate constants can affect $K_c$. Pressure doesn't show in any of these relationships.




$$k=A\mathrm e^{-\frac{E_\mathrm a}{RT}}$$

It may also be useful to think about different ways pressure can be changed.

  1. The volume of the reaction can be changed. This will affect concentrations of reactants and products, and the reaction will likely have to shift left or right to reestablish equilibrium. If volume were decreased, the rates of the forward and reverse reaction will each increase due to higher concentration. If the exponents in the rate laws are different, the amount of increase for each will be different. Although the rates change, the rate constants do not, so $K_c$ doesn't change either.

  2. The temperature can be changed. This will affect the rate constants of the forward and reverse reactions according to the Arrhenius equation. An increase in temperature will increase $k_\text{forward}$ and $k_\text{reverse}$, but by different amounts because their activation energies are rarely equal. This changes $K_c$, which is the ratio of the two rate constants.

  3. More gas can be added without changing the volume of the container.

    a. If the gas is one of the reactants or products, this would affect one of the concentrations, and the reaction will have to shift to reestablish equilibrium. Again, this changes one of the rates, but does not affect the rate constants, so $K_c$ is unaffected.

    b. If the gas is inert and doesn't react with anything involved in the equilibrium, the concentration of each reactant is unchanged. Neither the rates nor the rate constants are affected, so the system should still be at equilibrium and $K_c$ is unaffected.


$K_p$ is defined in terms of free energies of the gaseous substances in their standard states at 1 atm pressure so must be independent of pressure, i.e. is a constant. With concentrations we use $c_i=n_i/V$ for species $i$ and for example, if $K_c=c_\ce{A}^ac_\ce{B}^b/(c_\ce{C}^cc_\ce{D}^d)$ then for an ideal/perfect gas $K_c=K_p(RT)^{-\Delta n}$ where $\Delta n=a+b-c-d$ and so is independent of pressure just as is $K_p.$

So the explanation is that $K_p$ is defined that way and so $K_c$ is defined similarly as it measures the same equilibrium just expressed with different quantities, i.e. concentration instead of pressure.

What is confusing is that a typical expression could be $K_p=4a^2p/(1-a^2)$ where $p$ is total pressure and $a$ degree of dissociation. If $p$ changes $K_p$ does not but instead $a$ must to keep $K_p$ constant.


TLDR: While $K_p$ is pressure-independent for ideal gases, it is pressure-dependent for real gases. To get a pressure-independent constant for real gases, we need to substitute the fugacities for the pressures, giving us $K_f$. This is not merely definitional. The pressure-dependence of $K_p$ for non-ideal reacting systems has real-world consequences for the pressure-dependence of the relative amounts of reactants and products at equilibrium.

As porphyrin and Buck Thorn nicely explained, once you establish that $K_p$ is independent of pressure then, assuming an ideal system, it necessarily follows that $K_c$ is indepedent of pressure as well.

But I'd like to take a closer look at the pressure-independence of $K_p$. The reason $K_p$ is pressure-independent is not directly because it is defined as such. Rather, the reason $K_p$ is pressure-independent is because it's assumed that the gases in the equilibrium expression behave ideally. That's the fundamental stipulation; the pressure-independence is a consequence of this.

[I.e., an assumption of ideality is needed not only to obtain pressure-indepedence for $K_c$, but also to obtain pressure-independence for $K_p$.]

But then you might ask: What if the gases don't behave ideally? Will $K_p$ be independent of pressure? Here the answer is no. Consider the following generic reaction:

$$\ce{aA(g) + bB(g)<=>cC(g) + dD(g)}$$


$$K_p=\frac{p_C^c p_D^d}{p_A^a p_B^b}$$

Now consider that the fugacity, $f$, is related to the pressure through the fugacity coefficient, $\gamma (T,p)$: $f = \gamma (T,p)p$. Note that $\gamma$ is both pressure- and temperature-dependent*.

[*This is stated explicitly in many physical chemistry textbooks, e.g. see bottom p. 276 of Thomas Engel and Philip Reid, Thermodynamics, Statistical Thermodynamics, & Kinetics (3rd Edition), Pearson, 2012.

For more on fugacity, see my answer at: Fugacity vs compressibility (note that there I use $\varphi$ instead of $\gamma$ as the fugacity coefficient, since there I was following Wikipedia's notation, while here I'm following that in Engel & Reid).]

Substituting $p_i = \frac{f_i}{\gamma_i (T,p)}$ into our expression for $K_p$, we have:

$$K_p=\frac{\Big(\frac{f_C}{\gamma_C (T,p)}\Big)^c \Big(\frac{f_D}{\gamma_D (T,p)}\Big)^d }{\Big(\frac{f_A}{\gamma_A (T,p)}\Big)^a \Big(\frac{f_B}{\gamma_B (T,p)}\Big)^b} = \frac{f_C^c f_D^d}{f_A^a f_B^b} \frac{\Big(\frac{1}{\gamma_C (T,p)}\Big)^c \Big(\frac{1}{\gamma_D (T,p)}\Big)^d }{\Big(\frac{1}{\gamma_A (T,p)}\Big)^a \Big(\frac{1}{\gamma_B (T,p)}\Big)^b} = \frac{f_C^c f_D^d}{f_A^a f_B^b} \frac{\gamma_A (T,p)^a \gamma_B (T,p)^b}{\gamma_C (T,p)^c \gamma_D (T,p)^d}$$


$$K_f=\frac{f_C^c f_D^d}{f_A^a f_B^b}$$

We have:

$$K_p = K_f \frac{\gamma_A (T,p)^a \gamma_B (T,p)^b}{\gamma_C (T,p)^c \gamma_D (T,p)^d}$$

Since the $\gamma_i$'s are pressure-dependent, we can see that $K_p$ will be pressure-dependent as well, if the reaction involves at least one real gas. [Note: $K_f$ is pressure-independent. Thus there is no compensating pressure-dependence from $K_f$ that cancels out the pressure-dependence of the $\gamma_i$'s.]

To simply illustrate the real-world physical consequences of this, consider this reaction:


We can write:

$$K_x = \frac{x_B^b}{x_A^a}=K_p \Big(\frac{p}{p^{\circ}}\Big)^{a-b}$$

Since $a=b=1$, this reduces to:

$$K_x = \frac{x_B}{x_A}=K_p$$

But we've already established that, while $K_p$ is pressure-independent for ideal gases, it is pressure-dependent for real gases. Hence $K_x$ will be as well.

Thus, at equilibrium, the relative mole fractions of A and B will be independent of pressure if A and B are ideal. But they will vary with pressure if A and B are non-ideal. This directly illustrates the consequences of varying the pressure of non-ideal reacting systems.

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    $\begingroup$ Actually, I think you are sort of wrong (but I'm not sure, it's subtle, as your answer points out): it's true that ideality is important, but it is baked into $K_p$s definition through the expression $\Delta G^\circ=-RT\ln K_p$. Both $K_p$ and $\Delta G^\circ$ together define ideal behavior relative to an ideal standard state. Once those definitions are in place, the substances are allowed to behave nonideally, but we use fugacities (for instance) instead and continue using $K_p$. $\endgroup$
    – Buck Thorn
    Commented Feb 18, 2021 at 10:09
  • $\begingroup$ I.e., instead of starting with $\mu(T,p) = \mu^o(T) + RT ln \frac{p}{p^o}$, we start with $\mu(T,p) = \mu^o(T) + RT ln \frac{f}{f^o}$ (see eqns. 7.13 and 7.14, p 175, Engel & Reid 3rd Ed.) $\endgroup$
    – theorist
    Commented Feb 18, 2021 at 10:49
  • $\begingroup$ Well, I saw your point about the importance of ideality in your answer, but now your comment seems to play down that importance, so I am a little confused about how you see things. To determine a fugacity you need a measure of how far you are from ideality, so presumably you have defined an ideal state for the system (for gases, allowing the pressure to go to zero). If you did things right, you obtained $\Delta G^\circ$ by extrapolating to this ideal state. $\endgroup$
    – Buck Thorn
    Commented Feb 18, 2021 at 10:51
  • $\begingroup$ I'll have to sit down and think some more about this, obviously. Thanks for your replies! $\endgroup$
    – Buck Thorn
    Commented Feb 18, 2021 at 11:37
  • $\begingroup$ @BuckThorn I edited my answer to more explicitly demonstrate why $K_p$ is pressure-dependent for real gases. $\endgroup$
    – theorist
    Commented Dec 5, 2021 at 2:48

Elaborating on @porphyrin's answer, for an ideal system $K_c$ does not depend on pressure because if the temperature is constant then the concentration of a component depends only on its partial pressure $p_i$ and not on the total pressure P:


In other words, at constant T a number concentration can be used as a proxy for the partial pressure.

On the other hand the mole fraction $\chi_i$ of a component depends on both the partial and the total pressure:


Therefore the equilibrium constant based on mole fraction is a function of pressure. Since % yields are based on mole fractions, varying the pressure is a potentially important way of controlling the yield of a reaction, as Le Chatelier's Principle reminds us.

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    $\begingroup$ You might line to add that $K_x=K_PP^{-\Delta n}$ so if $\Delta n \ne 0$ the mole fraction at equilibrium will depend on total pressure $P$ even though $K_P$ does not. $\endgroup$
    – porphyrin
    Commented Feb 18, 2021 at 9:00
  • $\begingroup$ @porphyrin Thanks, I left that out as an optional exercise for the reader. Also, I thought of leaving a comment under your answer but figured there was space for another perspective. $\endgroup$
    – Buck Thorn
    Commented Feb 18, 2021 at 9:58

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