What effect does increasing the pressure and temperature have on the equilibrium constant, kc

​The equilibrium position of a reaction may be changed by:•Adding or removing a reactant or product•Changing the pressure by changing the volume (equilibria involving gases)•Dilution (for equilibria in solution)

•Changing the temperature.​

•Le Chatelier’s Principle is very useful in determining how the position of equilibrium can be changed to ensure more product is formed.•Le Chatelier’s Principle states that if a system is at equilibrium and the temperature, pressure or concentrations of the species are changed, the reaction will proceed in such a direction as to oppose this change.

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Changes in concentration and changes in pressure for gaseous systems will alter the equilibrium position but not the value of the equilibrium constant, K.

The addition of a reactant moves the position of equilibrium forward, producing more product. The value of the equilibrium constant, K, is unchanged.

An increase in pressure for a gaseous system moves the position of equilibrium to the side of the equation that has fewer moles of substance, and hence lower pressure. The value of the equilibrium constant, K, is unchanged.
Eg: For the reaction:
        2SO2(g) + O2(g)   <----->        2SO3(g)
           3 Particles                          2 Particles
The reaction goes in the forward direction (a net forward reaction) to form 2 particles.
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When solution is diluted the reaction will go in the direction that will form the greater amount of particles.

Eg.      Fe^3+(aq) + SCN^-(aq)    <----->          Fe(SCN)^2+ (aq)

​The direction of the equilibrium will change for an increase in temperature depending on whether the reaction is endothermic or exothermic.
Increasing temperature will result in:•A net back reaction for an exothermic reaction•A net forward reaction for an endothermic reaction.

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The addition of a catalyst does not change the position of equilibrium or the value of the equilibrium constant ,K.  Equilibrium is simply reached more quickly.
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​The addition of an inert gas at constant volume does not change the position of equilibrium or the value of the equilibrium constant, K. The total pressure is increased.
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​→ indicates a net forward reaction← indicates a net back reaction•Add reactants: →•Add Products: ←•Increase Pressure: direction of the smaller amount of particles•Dilution: direction of larger amount of particles•Increase Temperature:•Exothermic Reaction ←•Endothermic Reaction →•Catalyst: Increased rate but no change in equilibrium.

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•There are two things we would like to do in industry to make processes more viable: increase the RATE and YIELD of the reaction.
•We can do this by altering five main variables:

Image courtesy of  http://en.wikipedia.org/wiki/Image:Lechatelier.jpg, Public Domain, https://commons.wikimedia.org/w/index.php?curid=428795

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Increasing the temperature of an endothermic reaction would favour the forward reaction, thereby increasing the concentration of the products compared to reactants, which in hand increases the value of Kc, the equilibrium constant. Increasing pressure would not change the value of Kc, as it is independent of pressure, however it would shift the position of equilibrium towards the side with the least moles of gas.

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:

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

Then:

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

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:

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

Substituting

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

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:

$$\ce{A(g)<=>B(g)}$$

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.