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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. $\endgroup$ Apr 9 '16 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
    Apr 10 '16 at 10:43
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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.

$$K_c=\frac{k_\text{forward}}{k_\text{reverse}}=\frac{[\text{products}]^y}{[\text{reactants}]^x}$$

$$\text{rate}_\text{forward}=k_\text{forward}[\text{reactants}]^x$$

$$\text{rate}_\text{reverse}=k_\text{reverse}[\text{products}]^y$$

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

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.

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$K_p$ 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.

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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:

$c_i=\frac{p_i}{RT}$

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:

$\chi_i=\frac{p_i}{P}$

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
    Feb 18 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
    Feb 18 at 9:58
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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 is 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 $K_p$ vary with pressure? Here you run into a definitional issue. Consider the following reaction:

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

Then the generic equilibrium expression is:

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

If those partial pressures are actual partial pressures, and if those gases behave non-ideally, then $K$ will not be pressure-independent. To make this explicit, we can substitute the symbol $f$, for fugacity, for the partial pressures, as follows, and call the resulting expression $K_f$:

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

Thus, in sum, the following expression is pressure-independent only if it is stipulated that the gases behave ideally:

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

If the gases don't behave ideally, and one changes the total pressure and then enters the actual partial pressures of those gases into this expression, the value of this expression will be found to change with total pressure.

My concern with saying that $K_p$ is pressure-independent because it is defined as such is that this could mislead students into thinking if you put actual pressures into the equilibrium expression, and change the total pressure, then $K_p$, by definition, can't change. This is false.

To address this, some use the term $K_p$ where it is understood that the expression only contains ideal gases (in which case $K_p$ actually is pressure-independent), and use the term $K_f$ where it is understood that real gases are involved.

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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
    Feb 18 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
    Feb 18 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
    Feb 18 at 10:51
  • $\begingroup$ Yes, you're right, my comment was confusing, so I deleted it, and $f^o=p^o$, so $\Delta G^o$ is defined relative to a standard state. But that's the case both for $K_p$ and $K_f$, yet $K_f$ is not p-independent. I.e., you seem to be arguing that $K_p$'s p-independence is already baked into the defintion of $\Delta G^o$, so no separate stipulation of ideality is necessary. I say no, because (as you yourself pointed out), $K_f$ is also based on that ideal standard state, yet it is not p-independent. $\endgroup$
    – theorist
    Feb 18 at 11:06
  • $\begingroup$ I.e., the specification of the ideal standard state is not sufficient to get p-independence. You need to explicity also specify ideal gas behavior over all pressures you are examining, which is exactly my point: it's only that specific stipulation that gets you p-independence. p-independence cannot be obtained merely from the defintion of $K_p$ as based on an ideal std. state $\endgroup$
    – theorist
    Feb 18 at 11:08
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Consider the equilibrium: A+2B (reversible reaction arrow) 2C.

If [C] and [B] are 2 moldm^-3 while [A] is 1 moldm-3 originally: Kc = [C]^2/([A][B]^2) = 2^2/(12^2) = 0.8 mol^-1dm^3

  1. If pressure increases and nothing happens (which cannot happen in actual): [A], [B] and [C] increase by say, 3/2 times because an increase in pressure can only be due to an increase in the number of particles per unit volume unless there is a change in temperature, which is considered another factor as it results in a change in the value of Kc. The new Kc value will then be: Kc = 3^2/(1.5*3^2) = 0.0.667 (3.s.f.), which is lower than the actual Kc value. Hence, this is not possible as a pressure change in the system cannot result in a change in the Kc value, according to Le Chatelier's principle.
  2. If pressure increases and and Le Chatelier’s principle is applied (which happens in actual): [A], [B] and [C] increase by say, 3/2 times because an increase in pressure can only be due to an increase in the number of particles per unit volume unless there is a change in temperature, which is considered another factor as it results in a change in the value of Kc. However, unlike scenario 1, Le Chatelier’s principle is applied. When Le Chatelier’s principle is not applied, Kc value would decrease as the numerator ([C]^2) would increase by a smaller proportion than the denominator ([A]+[B]^2). However, when it is applied, some A and B is converted into C, thereby increasing the proportion by which ([C]^2) increases and decreasing the proportion by which ([A]+[B]^2) increases. Hence, according to Le Chatelier’s principle, the Kc value will increase back to it’s normal value at 0.8 mol^-1dm^3 from 0.783 mol^-1dm^3 when pressure of the system increases.
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  • $\begingroup$ I had edited this answer because of the poor usage of symbols, but my edit wasn't accepted? Is this a glitch? $\endgroup$
    – user104162
    Feb 17 at 12:22
  • $\begingroup$ @DestinyFindsYou Your edit was rejected because of improper use of MathJax which introduced more problems than fixes. Note that on Chemistry.SE we have mhchem for typesetting chemical expressions and physical units. Please visit this page, this page and this one. $\endgroup$
    – andselisk
    Feb 19 at 8:13
  • $\begingroup$ @Rabbit Hold on, are you suggesting that for the equilibrium $$\ce{A + B <=> 2 C}$$ $$K_c = \frac{[\ce{C}]^2}{[\ce{A}]+[\ce{B}]^2}?$$ Are you sure about the denominator? $\endgroup$
    – andselisk
    Feb 19 at 8:15
  • $\begingroup$ I have now edited my answer so that it matches the equilibrium equation I mentioned. $\endgroup$
    – Rabbit
    Feb 19 at 14:56
  • $\begingroup$ @Rabbit What is "B^2" supposed to mean in a chemical reaction? Still, if two here is the stoichiometric coefficient and the reaction is $$\ce{A + 2 B <=> 2 C},$$ you got $K_c$ wrong. There is no sum in the denominator. And please check out links in my comment to Destiny regarding the formatting so that we can understand each other better. $\endgroup$
    – andselisk
    Feb 19 at 16:02

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