The equilibrium constant

A reaction with a negative free energy change should run until a reactant is gone, and almost none of them do, which means the criterion of the previous lesson is incomplete rather than wrong.

The previous lesson established that a change at constant temperature and pressure goes when $Δ G < 0$ , where $Δ G = Δ H - T Δ S$ . Take that literally and nothing should ever stop halfway. Yet acetic acid in water is barely ionised, calcium carbonate heated in a sealed vessel reaches a fixed carbon dioxide pressure and stays there, and a reactor full of nitrogen and hydrogen makes some ammonia and then makes no more. This lesson repairs the criterion, and the repair turns out to be the most useful single equation in chemistry.

Why the free energy stops falling

The quantity that must decrease is the free energy of the whole mixture, and the fault in the naive picture is treating that as a straight line between pure reactants and pure products. It is not, because a mixture has an entropy that neither pure state has.

Follow the extent of reaction $ξ$ from the first lesson. At $ξ = 0$ the vessel holds pure reactants; at $ξ_{\max}$ pure products. If mixing contributed nothing, $G$ would run linearly between the two ends and its minimum would be at whichever end is lower. But at every intermediate composition both reactants and products are present, and the entropy of mixing them is positive, which pulls $G$ down in the middle and away from both ends. The mixing term is largest at intermediate compositions and falls to zero at each pure limit, with an infinite slope there, and that infinite slope is decisive: however unfavourable a reaction, the first trace of product always lowers $G$ .

So $G$ plotted against $ξ$ is a curve with a minimum strictly inside the range, and the system slides downhill to that minimum and stops. Equilibrium is the bottom of that curve, where

{(\frac{\partial G}{\partial ξ})}_{T, p} = 0

The derivative on the left is what $Δ_{r} G$ actually means: the slope of the free energy with respect to extent, in joules per mole of reaction, at the composition the mixture currently has. A reaction with $Δ_{r} G^{ominus} = - 32$ kJ mol⁻¹ does not have that slope throughout; it has it at one particular reference composition, and the slope climbs toward zero as products build up.

This also explains why no reaction ever goes fully to completion. Complete conversion is a pure state, where the mixing term has infinite slope in the reverse direction, so the last trace of reactant never disappears. Sometimes what remains is one molecule in $10^{60}$ , which is completion for any practical purpose, but the position of equilibrium is always strictly inside.

Activity, and the composition dependence

To make this quantitative we need how the free energy of one substance depends on how much of it is there. For an ideal gas at constant temperature the answer follows from the work of isothermal compression, and comes out as a logarithm. Writing the molar free energy of a species as its chemical potential $μ$ ,

μ_{i} = μ_{i}^{ominus} + R T \ln a_{i}

where $a_{i}$ is the activity, a dimensionless measure of how much of the substance is present relative to its standard state. For a gas it is the partial pressure divided by $10^{5}$ Pa; for a dissolved solute it is the concentration divided by $1$ mol dm⁻³; and for a pure solid or pure liquid it is $1$ , because a pure substance is already in its standard state. That last rule is why solids never appear in an equilibrium expression: adding more solid does not change its activity, and a lump of calcium carbonate has the same chemical potential whether there is a gram of it or a tonne.

These are the ideal forms. Real gases at high pressure and real ions in solution deviate, and the honest statement is $a_{i} = γ_{i} c_{i} / c^{ominus}$ with an activity coefficient $γ_{i}$ that approaches $1$ at infinite dilution. For a $0.1$ mol dm⁻³ solution of a simple salt, $γ$ is already around $0.78$ , so calculations that ignore it are good to a few per cent at best in ionic systems, and better than that in dilute or gaseous ones. Everything below uses $γ = 1$ , and it is worth remembering that this is where the error lives.

The reaction quotient and the exact relation

Now assemble the mixture. The slope of $G$ with respect to $ξ$ is the stoichiometric sum of the chemical potentials, $Δ_{r} G = \sum ν_{i} μ_{i}$ , so substituting the expression above and collecting the logarithms gives

Δ_{r} G = Δ_{r} G^{ominus} + R T \ln Q

where the reaction quotient $Q$ is the product of the activities each raised to its stoichiometric number, products on top and reactants underneath. For $+ 3 \to 2 N$ it is $Q = a (N)^{2} / [a () a ()^{3}]$ .

Two things follow at once. First, $Δ_{r} G^{ominus}$ is simply $Δ_{r} G$ when every activity is $1$ , which is what "standard conditions" means and why a standard free energy change never on its own tells you what a real flask will do. Second, at equilibrium the slope is zero, and the value $Q$ has taken there is by definition the equilibrium constant $K$ :

Δ_{r} G^{ominus} = - R T \ln K

This is the bridge the whole course has been building toward. On the left are tabulated enthalpies and entropies measured in a calorimeter; on the right is the composition a reactor will actually reach. $K$ is dimensionless, because every activity is, and it depends on temperature alone: not on pressure, not on the starting amounts, not on whether a catalyst is present.

The exponential makes it a violent amplifier. At $298$ K, $R T = 2.479$ kJ mol⁻¹, so every $5.7$ kJ mol⁻¹ of free energy is a factor of ten in $K$ .

Example. Ammonia synthesis has $Δ_{r} G^{ominus} = - 32.5$ kJ mol⁻¹ at $298.15$ K. Find $K$ . Then do the same for the combustion of methane, $Δ_{r} G^{ominus} = - 818.0$ kJ mol⁻¹.

For ammonia, $\ln K = 32500 / (8.314 \times 298.15) = 13.11$ , so $K = 5.0 \times 10^{5}$ . Equilibrium lies well over toward ammonia, which is a fact about the room temperature reaction that no ammonia plant has ever been able to use. For methane, $\ln K = 818000 / 2479 = 330$ , so $K = 2 \times 10^{143}$ . There is no meaningful sense in which any methane remains, and equally no sense in which a bottle of methane and air is in danger of reacting by itself.

Now you. Dinitrogen tetroxide dissociates as $(g) \to 2 N(g)$ , with $Δ_{f} G^{ominus} = 97.9$ kJ mol⁻¹ for and $51.3$ for $N$ . Find $Δ_{r} G^{ominus}$ and $K$ at $298.15$ K.

Answer

$Δ_{r} G^{ominus} = 2 (51.3) - 97.9 = + 4.7$ kJ mol⁻¹. Then $\ln K = - 4700 / 2479 = - 1.896$ , so $K = 0.150$ . Positive but small, so at standard pressure the mixture is mostly the dimer with an appreciable amount of the brown monomer, which is what the tube looks like.

Computing an equilibrium composition

With $K$ in hand the composition follows from a bookkeeping table: initial amounts, change in terms of one unknown, equilibrium amounts. The unknown is the extent of reaction in whatever units the constant uses, and the equation to solve is $K$ set equal to the quotient of those equilibrium values.

Example. $1.00$ mol of is placed in a $10.0$ dm³ vessel at $298.15$ K, where $K = 0.150$ . Find the equilibrium partial pressures and the degree of dissociation.

The initial pressure is $n R T / V = 1.00 \times 0.083145 \times 298.15 / 10.0 = 2.479$ bar, using $R$ in bar dm³ mol⁻¹ K⁻¹. Let $x$ bar of dissociate. Then at equilibrium $p () = 2.479 - x$ and $p (N) = 2 x$ , and since the standard pressure is $1$ bar the activities are these numbers. So

K = \frac{(2 x)^{2}}{2.479 - x} = 0.150

which rearranges to $4 x^{2} + 0.150 x - 0.3719 = 0$ and gives $x = 0.287$ bar. The equilibrium pressures are $p (N) = 0.573$ bar and $p () = 2.192$ bar, and the degree of dissociation is $0.287 / 2.479 = 11.6$ per cent. Checking, ${0.573}^{2} / 2.192 = 0.150$ .

Now you. Repeat with the same $1.00$ mol in a $40.0$ dm³ vessel, so the initial pressure is $0.620$ bar. Find the degree of dissociation and compare.

Answer

Now $4 x^{2} + 0.150 x - 0.0930 = 0$ , giving $x = 0.1348$ bar, so $p (N) = 0.270$ and $p () = 0.485$ bar. The degree of dissociation is $0.1348 / 0.620 = 21.8$ per cent, nearly double. $K$ has not changed at all; the composition has, because diluting a reaction that makes more molecules pushes it forward.

That last observation is the whole of the next lesson in miniature. The constant stayed constant and the answer moved.

Where the constant is easier than it looks

Two shortcuts save a great deal of algebra. When $K$ is very small, the change $x$ is small compared with the initial amounts, so the denominators can be left at their starting values, giving an explicit answer rather than a quadratic. The rule of thumb is that this is safe when $x$ comes out below five per cent of the smallest initial value, and the discipline is to compute $x$ and then check that, because when the approximation fails it fails silently. Weak acids in the ninth lesson are where this matters most.

The second is that a constant can be assembled from others. Reverse a reaction and $K$ becomes $1 / K$ ; double the coefficients and it becomes $K^{2}$ ; add two reactions and their constants multiply. All three follow from $Δ_{r} G^{ominus} = - R T \ln K$ and the fact that free energies add while logarithms turn addition into multiplication.

One conversion is worth stating explicitly. If a gas phase constant is written in concentrations rather than pressures, the two differ by the ideal gas relation $p = c R T$ applied to each species, so $K_{p} = K_{c} (R T / p^{ominus})^{Δ n}$ where $Δ n$ is the change in moles of gas. When $Δ n = 0$ , as in the hydrogen and iodine reaction below, the two are numerically the same and the distinction can be ignored.

Q against K says which way

The reaction quotient is defined for any composition, not just the equilibrium one, and comparing it with $K$ answers the practical question directly. From $Δ_{r} G = R T \ln (Q / K)$ , obtained by substituting the definition of $K$ into the earlier expression, the sign of $Δ_{r} G$ is the sign of $\ln (Q / K)$ . If $Q < K$ there is too little product and the reaction runs forward; if $Q > K$ it runs backward; if $Q = K$ nothing happens on average, though both directions continue at the molecular level.

Example. Hydrogen and iodine react as $+ \to 2 HI$ with $K = 54.3$ at $698$ K. A vessel holds all three at $0.0100$ , $0.0100$ and $0.0200$ mol dm⁻³. Which way does it go, and where does it end if it starts instead from $1.00$ mol dm⁻³ each of hydrogen and iodine?

$Q = {0.0200}^{2} / (0.0100 \times 0.0100) = 4.0$ , well below $54.3$ , so the mixture makes more hydrogen iodide. Starting from $1.00$ each, let $x$ react: at equilibrium the reactants are $1.00 - x$ each and the product is $2 x$ , so $K = 4 x^{2} / (1 - x)^{2}$ . Since both sides are perfect squares, take the square root: $2 x / (1 - x) = \sqrt{54.3} = 7.369$ , giving $x = 0.787$ . The equilibrium concentrations are $0.213$ mol dm⁻³ for each reactant and $1.573$ for hydrogen iodide.

Now you. For the same reaction and constant, a vessel holds at $0.0500$ , at $0.0200$ and $HI$ at $0.400$ mol dm⁻³. Which way does the reaction run?

Answer

$Q = {0.400}^{2} / (0.0500 \times 0.0200) = 0.160 / 0.00100 = 160$ , which exceeds $54.3$ , so $Δ_{r} G > 0$ in the forward direction and hydrogen iodide decomposes until $Q$ falls to $54.3$ .

What moves it, and what does not

Everything in this lesson holds the temperature fixed, because $K$ is a function of temperature and of nothing else. Change the amounts and $Q$ moves while $K$ stays put, and the system responds by returning $Q$ to $K$ . Change the temperature and $K$ itself moves, which is a different kind of change and needs a different equation.

That distinction, together with the case of compressing a gas mixture, where the amounts do not change but the activities do, is the content of the next lesson. Applied to the synthesis of ammonia the three answers pull against each other, and the resolution is one of the more consequential compromises in industrial history.