They are both equilibrium constants as far as I know.
Kc is in terms of molarity and Kp is in terms of pressure. Also both of them are ratios of respective quantities [ ratio of molarity(s) in Kc and ratio of pressure(s) in Kp], so they should be dimensionless according to dimensional analysis. But in some places I have seen units mentioned along with both Kc and Kp.
It is quite confusing and I don't see any uniformity in this.
Anyone please help.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Try to express $K_\mathrm{c}$ or $K_\mathrm{p}$, say, for $\ce{N2O4 <=> 2NO2}$ and see how "dimensionless" they are. Equilibrium constant can only appear dimensionless when it so happens that the units are cancelled out. $\endgroup$– andselisk ♦Commented Jan 3, 2019 at 4:14
-
$\begingroup$ But in this reaction 2(SO2)[g] + (O2)[g] <=> 2(SO3) [g] a unit would appear in KC. $\endgroup$– GamiraCommented Jan 3, 2019 at 4:25
-
$\begingroup$ ... and how it deviates from/contradicts with what I've written? Also, KC reads as potassium carbide. $\endgroup$– andselisk ♦Commented Jan 3, 2019 at 4:27
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
4
The equilibrium constant $K_p$ is dimensionless as it is defined, ultimately, in terms of the ratio of activities, which are themselves dimensionless. In practice we often use partial pressure instead of activities and then we use the numerical value by effectively dividing each partial pressure by 1 unit of pressure, say 1 atm.
In cases when a mole fraction (or concentration) is used then we find that $K_p=K_x P^{\Delta n}$ where $\Delta n$ is the change in the number of moles, product minus reactant. However, $K_x$ is not a true equilibrium constant since its value must change as $P$ changes to keep $K_p$ constant.
If you are not sure you can check that $K_p$ must be dimensionless from the equation $\Delta G^\mathrm{o}=-RT\ln(K_p)$ as anything inside a function, such as a log, has to be dimensionless.
(Note that as the equilibrium constant $K_p$ is defined in terms of the free energy of the standard states of the gaseous species at 1 atmosphere pressure it is independent of pressure.)
-
1$\begingroup$ >The equilibrium constant Kp is dimensionless as it is defined, ultimately, in terms of the ratio of activity coefficients, which are themselves dimensionless. Excuse me? Won't $K_p$ have dimensions of pressure raised to some power? And what about $K_c$? I'm guessing $[L^{-3}]^a$ ... a is any number ... I think mol is dimensionless so I'm only considering the dimensions of litres. $\endgroup$ Commented Jan 3, 2019 at 14:24
-
3$\begingroup$ @AnuragB. porhyrin is correct. There is a trap: $K_\mathrm{p}$ is indeed dimensionless as it deals not really with the pressures, but fugacity $f$, which, in turn is expressed as $p_i/p°$, where $p_i$ is partial pressure and $p°$ is standard state of pressure so that the units of pressure are cancelled out. Also, no sarcasm this time:) $\endgroup$– andselisk ♦Commented Jan 3, 2019 at 15:06
-
2$\begingroup$ The answer is OK-ish but there is one obvious error, which is now repeated in the comments. Actually, thermodynamic equilibrium constant is defined in terms of activities (fugacities) and not just activity coefficients! Fugacity, for example, is $f=\gamma \frac{p}{p^○}$ where $\gamma$ is an activity coefficient and $p^○$ is a standard pressure. Mind also that the latter is tricky: it was 1 atm in the past, but now IUPAC recommends it to be $10^5$ Pa, or 1 bar in non-SI units. So, you see that it is the activity (fugacity) which is dimensionless. Ask Atkins if you don't believe me. $\endgroup$– voffchCommented Jan 3, 2019 at 19:27
-
$\begingroup$ @voffch, you are correct, I was careless and I have changed my answer to be clearer, I was thinking of $K_x$ when I used 'coefficients'. $\endgroup$ Commented Jan 4, 2019 at 9:09
$\begingroup$
$\endgroup$
Different disciplines and different textbooks treat units of K differently, and you have to be very careful to figure out what their conventions are. Often, the conventions change within a textbook, and often, this is not state explicitly.
Examples where equilibrium constants are clearly treated as dimensionless are Kw = 10e-14, expressions involving ln(K) or -log(K), and once you make the switch from concentrations to activities, or from partial pressure to fugacity.
Examples where equilibrium constant are clearly treated as having dimensions is for the dissociation constant Kd used in biochemistry, or in converting Kc to Kp.
The way you make equilibrium constant dimensionless is to use dimensionless measures of concentrations in their definition. The best way is to use activities because then you get exact results. When that is not possible, you get dimensionless concentrations by dividing the concentration by the respective standard state (given in the same dimension) so that dimensions cancel out. The common standard state for solutes is a concentration of 1 M, whereas the common standard state for solvents is the pure liquid (that's why we usually leave out the solvent from an equilibrium expression). Be aware that unless you are using activities, all your calculations are estimates that become accurate only when working with infinitely diluted solutions.
As an aside, if you want to have a dimensionless version of converting Kp and Kc, you would use (RT M/atm) in the formula instead of (RT) if your agreed-upon standard states for Kc is 1 M and for Kp is 1 atm.
$\begingroup$
$\endgroup$
$K_c$ and and $K_p$ are not dimensionless quantity. Their dimensions change with the reaction you consider.
For example in the reaction you mentioned, clearly $K_c$ should have the units of $\mathrm{\dfrac{L}{mol}}$ which is evident of my statement.
For $K_p$, you can use the formula $K_p=K_c (RT)^{\Delta n_g}$ to find the dimensions. Remember that $R$ and $T$ also have their own dimensions.