# How does the value of the equilibrium constant give information about the extent of reaction at equilibrium?

Pretty much everywhere I've read, the numerical value of the equilibrium constant gives you information about the relative concentrations of products and reactants in the equilibrium mixture; the larger the value.

However, for reactions with an unequal number of moles on the reactant and product side, the equilibrium constant has units in terms of concentration or pressure, allowing me to make the numerical value arbitrarily large or small by choosing the appropriate units. Does the numerical value give no qualitative information about the position of the equilibrium in this case? Or are there a specific set of units that I'm supposed to use?

I am aware of a thermodynamic equilibrium constant that uses activities and is thus dimensionless, but equilibrium constants are only defined in terms of molar concentrations or partial pressures at my level

• For a gaseous mixture, the equilibrium composition fits at the same time the value expressed in partial pressures and the value expressed in molar concentrations. I see no big deal. Jul 1 at 5:31
• There is also an equation that relates the two which is $\ce{K_p = k_c(RT)^{\Delta n}}$
– M.L
Jul 1 at 5:35
• I am not asking about the relation between them. Kc and Kp for many reactions have units (instead of being unitless) so their numerical values can be altered arbitrarily by choosing appropriate units. In this case, how does their magnitude give any information about the position of equilibrium? Jul 1 at 6:24
• Does the following post answer your question: chemistry.stackexchange.com/questions/117636/… ? Jul 1 at 7:04
• @BuckThorn I know that the thermodynamic equilibrium constant is dimensionless but Kp and Kc, which do have units, are used much more often and used to predict the positon of the equilibrium Jul 1 at 7:07

Yes you can't compare extent of reactions directly based on the numerical value of the equilibrium constant.

A very good example of this is the comparison of solubility of different salts based on their solubility products.

Let's take the example of calcium and silver carbonates:

$$\begin{array}{|c|c|c|} \hline \text{Salt} & K_\text{sp} & \text{Solubility} \\ \hline \ce{CaCO3} & \pu{8.7 \times 10^{-9} M^2} & \pu{9.3 \times 10^{-5} M}\\ \hline \ce{Ag(CO3)2} & \pu{6.2 \times 10^{-12} M^3} & \pu{1.15 \times 10^{-4} M}\\ \hline \end{array}$$

As can be observed the solubility of silver carbonate is more than that of calcium carbonate even though it has a smaller solubility product.

This contradiction occurs due to the difference in the units of the solubility product. This all sums back to the elementary teaching, "Don't compare things with different units."

We can mathematically compare only the concentration of products and not their equilibrium constants if they have different units.

Directly comparing equilibrium constants is not a problem, in case both the constants have the same dimensions and the comparison is done in the same units. For example you can't compare two equilibrium constants if one of them is in pascals and other in bars.

So if you want to compare extent of two reactions either calculate the final concentration/pressure and compare them or if their equilibrium constants have the same dimensions than compare them in the same units.

The equilibrium constant has to be dimensionless otherwise expressions such as $$\Delta G^0=-RT\ln(K)$$ are not possible. In a reaction such as

$$\ce{A_2 <=> 2A}$$

if $$\alpha$$ is the degree of dissociation and $$1-\alpha$$ of $$A_2$$ reacts and $$2\alpha$$ of A is produced using the partial pressures $$(P_A=2\alpha P/(1+\alpha); P_{A_2}=(1-\alpha)P/(1+\alpha))$$ and as $$K_p=P_A^2/P{A_2}$$ it follows that

$$K_p=\frac{4\alpha^2P}{1-\alpha^2}$$

where $$P$$ is total pressure divided by 1 atm (or 1 bar) and is dimensionless, and so the degree of dissociation is dependent on the equilibrium constant as

$$\alpha =\sqrt{\frac{K_p}{4P+K_p}}$$

which is what you are asking about. This ($$\alpha$$) has to be the same irrespective of any units used.

• $\Delta G^0 = 1.$ Jul 1 at 7:27