# 6.4: Interpreting Equilibrium Constants Values

Learning Objectives

• How to figure out in which direction a reaction will go (i.e. towards making product, or more reactant)
• Calculating equilibrium concentrations. This may involve knowing equilibrium values for some of the reactants and products and determining the concentration of an unknown.
• Alternatively, we may be provided with the starting concentrations of reactants and products and may be asked to find the equilibrium concentrations

Your ability to interpret the numerical value of a quantity in terms of what it means in a practical sense is an essential part of developing a working understanding of Chemistry. This is particularly the case for equilibrium constants, whose values span the entire range of the positive numbers:

• If  Keq is a large number (Keq > 1), it means that, at equilibrium, the concentration of the products is large. In this case, the reaction favors the formation of products.

• If  Keq is a small number (Keq<1), it means that, at equilibrium, the concentration of the reactants is large. In this case, the reaction favors the formation of reactants.

• If  Keq ≈ 1, it means that, at equilibrium, there are significant amounts of both reactants and products.

As an equilibrium constant approaches the limits of infinity, the reaction can be increasingly characterized as a one-way process; we say it is “complete” or “irreversible”.

Kinetically Hindered Reactions

Although it is by no means a general rule, it frequently happens that reactions having very large equilibrium constants are kinetically hindered, often to the extent that the reaction essentially does not take place.

The examples in the following table are intended to show that numbers (values of Keq), no matter how dull they may look, do have practical consequences!

Table $$\PageIndex{1}$$: Examples of Reversible Reactions
Reaction
Keq
remarks
$$N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$$ $$5 \times 10^{–31}$$ at 25°C,
0.0013 at 2100°C
These two very different values of K illustrate very nicely why reducing combustion-chamber temperatures in automobile engines is environmentally friendly.
$$3 H_{2(g)} + N_{2(g)} \rightleftharpoons 2 NH_{3(g)}$$ $$7 \times 10^5$$ at 25°C,
56 at 1300°C
See the discussion of this reaction in the section on the Haber process.
$$H_{2(g)} \rightleftharpoons 2 H_{(g)}$$ $$10^{–36}$$ at 25°C,
$$6 \times 10^{–5}$$ at 5000°
Dissociation of any stable molecule into its atoms is endothermic. This means that all molecules will decompose at sufficiently high temperatures.
$$H_2O_{(g)} \rightleftharpoons H_{2(g)} + ½ O_{2(g)}$$ $$8 \times 10^{–41}$$ at 25°C You won’t find water a very good source of oxygen gas at ordinary temperatures!
$$CH_3COOH_{(l)} \rightleftharpoons 2 H_2O_{(l)} + 2 C_{(s)}$$
$$K_c = 10^{13}$$ at 25°C This tells us that acetic acid has a great tendency to decompose to carbon, but nobody has ever found graphite (or diamonds!) forming in a bottle of vinegar. A good example of a super kinetically-hindered reaction!

## Do Equilibrium Constants have Units?

The equilibrium expression for the synthesis of ammonia

$\ce{ 3 H2(g) + N2(g) -> 2 NH3(g)} \label{15.4.1}$

$K_eq = \dfrac{[NH_3]^2}{[N_2] [H_2]^3} \label{15.4.3}$

If concentrations are expressed in mol/L, then  $$K_eq$$ would be expressed in mol–2 L2. And yet Keq is often represented as being dimensionless. The reason for this is that, due to the fact that units for the equilibrium depend on the stoichiometric coefficients for each reaction, the Keq units are not consistent. Besides, in carrying out calculations with Keq, the use of units will make things more complicated. Therefore,  there is rarely any real need to show the units.

Strictly speaking, we should be consistent in the use of concentrations and always use the same units for reactants and products, that is usually molarity (M) or mol/L.

## Summary

The magnitude of the equilibrium constant, $$K$$eq, indicates the extent to which a reaction will proceed:

• If $$K$$eq is a large number, it means that the equilibrium concentration of the products is large. In this case, the reaction as written will proceed to the right (resulting in an increase in the concentration of products)
• If $$K$$eq is a small number, it means that the equilibrium concentration of the reactants is large. In this case, the reaction as written will proceed to the left (resulting in an increase in the concentration of reactants)

Knowing the value of the equilibrium constant, $$K$$, will allow us to determine: (1) the direction a reaction will proceed to achieve equilibrium and (2) the ratios of the concentrations of reactants and products when equilibrium is reached