What does it mean Keq=1?

Question

I’m struggling with a chemistry question related to equilibrium in a reaction. I don’t understand the meaning of $K_\text{eq} = 1$. After some research I found that when $K_\text{eq}$ is 1 it means that the concentration of products and reactants is the same. But this is not true! If my reaction is of the type $\ce{A <=> 2B}$ only if their concentration is $\pu{1 M}$ this is correct, but it is only a special case and not the rule. So what does really mean $K_\text{eq} = 1$? I know in this case neither products nor reactants are favourites, but why? Also the fact that the standard free Gibbs energy is 0 is not clear to me. Maybe it means all the energy it is used for the reaction itself and no energy is left to be converted into work? If possible I’ll appreciate a numerical example.

Answer 1

After some research I found that when Keq is 1 it means that the concentration of products and reactants is the same. [...] If my reaction is of the type $\ce{A <=> 2B}$ only if their concentration is 1M this is correct, but it is only a special case and not the rule.

IUPAC's Green Book discerns two types of equilibria constants $K$. Either the one which fits $K = - \exp(-\Delta_rG/-RT)$, or the ones which are derived from a basis of pressure, concentration, or molality. For the later $K$ is a quotient of e.g., corresponding products of concentrations to the power of the corresponding stoichiometry factors.

Let's assume your reaction $\ce{A <=> 2B}$ in diluted solution is balanced, then the constant is defined as

$$K_c = \frac{[\ce{B}]^2} {[\ce{A}]}$$

In this case, at chemical equilibrium, rearrangement of the mathematical equation leads to the general forms of

$$[\ce{B}] = \sqrt{K_c \cdot [\ce{A}]}$$

and

$$[\ce{A}] = \frac{[\ce{B}]^2} {K_c}$$

A detail frequently forgotten by beginners not familiar (yet) with dimensional analysis: the unit of $K$. If $K$ was derived from $\Delta_rG$, the constant is unit less. If derived from pressure, concentration, or molality, the sum of the stoichiometric coefficients has to be taken into account to yield a constant in unit of either $\pu{Pa}^{\sum \nu_\text{B}}$ (for pressure), $\pu{(mol m^{-3})}^{\sum \nu_\text{B}}$ (for molar concentration), or $\pu{(mol kg^{-1})}^{\sum \nu_\text{B}}$ (for molality).*

Only for a reaction $\ce{A <=> B}$ the units cancel out each other. For your above reaction of $\ce{A <=> 2B}$, however

$$K = \frac{(\pu{1 mol/L})^2} {\pu{1 mol/L}} = % % \frac{\pu{1^2 mol^2/L^2}} {\pu{1 mol/L}} = % % \pu{1 mol/L} $$

* Cohen, E. R. et al. (eds.) Quantities, Units and Symbols in Physical Chemistry, IUPAC Green Book, 3rd edition / 2007; section 2.11 chemical thermodynamics, p. 58.

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• If the hypothetical reaction were balanced $\ce{3A + 2B <=> 5C + 4D}$

  $$K_c = \frac {[\ce{C}]^5 \cdot [\ce{D}]^4} {[\ce{A}]^3 \cdot [\ce{B}]^2}$$

  or more general, $\ce{\alpha A + \beta B + \ldots <=> \sigma S + \tau T}$,

  $$K_c = % % \frac {[\ce{S}]^\sigma \cdot [\ce{T}]^\tau \ldots} {% [\ce{A}]^\alpha \cdot [\ce{B}]^\beta \ldots}$$

• $K_c$ (derived from concentrations in $\pu{mol/L}$), and $K_p$ (derived from partial pressures of gaseous equilibria) are approximations for diluted reactants. The higher the concentration of molecules, however, the more they can hinder each other during a reaction; at some point, thermodynamic activities are the better descriptions of the reagents.

• Note, capital italic $K$ is about the thermodynamic equilibrium constant. Lower case italic $k$ typically is used to describe the kinetic rates of reactions. This is convention of nomenclature compiled in IUPAC's Green Book (see resources page). At high school level, the freely available 4 page summary (link to pdf) likely provides enough information for you.

Answer 2

The equilibrium constant $ K=1$ has no more significance than that when $K=10$ or $K=0.1$ etc. You will know that the equilibrium shifts as the temperature changes, so consider the gas phase equilibrium

$$\ce{PCl_5(g) = PCl_3(g) + Cl_2(g) }$$

which has $K=0.230$ at $472$ K and at a slightly higher temperature $524$ K, $K=2.69$ and clearly $K=1$ is to be found somewhere in between these two temperatures. The Gibbs free energies at the two temperatures are $\Delta G^\text{o}=+5.77$ kJ/mol and $-4.31$ kJ/mol. (In 'thermodynamic speak' this means going from non-spontaneous to spontaneous reaction as the temperature increases and although it may seem so it is actually not very revealing).

In terms of the equation expressed as partial pressures (as appropriate for a gas phase reaction)

$$\displaystyle K=\frac{p_{PCl_3}p_{PCl_2} }{p_{PCl_5}}\equiv\frac{p_{PCl_3}^2}{p_{PCl_5}}$$

as $K$ is a constant at a given temperature we only ever know the ratio of the species. In this example $p_{PCl_5}K =p_{Pcl_3}^2$ or $p_{PCl_5} =p_{Pcl_3}^2$ when $K=1$.

It is necessary to point out at this point that using partial pressure in the gas phase or concentrations in solution is not technically correct, although very practical. We should use activities or equivalently fugacities which are dimensionless. To get around this we must divide each partial pressure or concentration by 1 unit, say 1 atmosphere or 1 mol/dm$^3$.

Using thermodynamics we know that the change in standard reaction free energy is related to the equilibrium constant as $\Delta G^\text{o}=-RT\ln(K)$, which trivially means that $K$ has to be dimensionless (see above), and when $K=1$ then $\Delta G^\text{o}=0$ which using $\Delta G = \Delta H -T\Delta S$ means that only at this temperature does $\Delta H^\text{o}=T\Delta S^\text{o}$, in other words an exact balance between heat change and entropy change. However, although it may seem that it is, this has no particular significance, other that it is the tipping point one way or the other in the reaction.

Answer 3

Andrea, you seem to have asked a deep question without a simple answer; I will Try. The problem is in carefully figuring activities. If there is strict stoichiometry and the coefficients = 1, the activities are equal. In this case there is no change in the number of moles in the reaction. Should a component be augmented activities will change and no longer be equal. A reaction that could fit this description is a simple esterification: ester + water = acid + alcohol.

The simplest example I can think of is an ice-water equilibrium. Pure ice and pure water each have mole fractions of 1.. At equilibrium the Keq = 1/1 = 1. This means that as long as both water and ice are present the temperature is constant [Gibbs phase rule]. Lowering the water activity by adding a solute lowers the melting point [ice melts to remove heat] until the activity of the ice equals that of the water. Keq is still 1 but the activities are lower. [An interesting thought! Should a solute raise the chemical potential of liquid water would the freezing point be increased?] A change in pressure is similar, the activities change, temperature adjusts until the activities are equal.

I am trying to give a rigorous explanation for a more complex system but am having trouble expressing activities in a manner that makes sense, so I am posting this partial answer to put this question back into the queue and give it its due.