Lesson 2 · 25 min

The equilibrium constant and the reaction quotient

Whatever mixture you start from, [HI]²/([H₂][I₂]) ends up near 54. Write K, tell Kc from Kp, deal with solids and pure liquids, and predict the direction of change by comparing Q with K.

In this lesson

Write the equilibrium-constant expression from a balanced equation, heterogeneous equilibria included
Convert between Kc and Kp with $K_p = K_c(RT)^{\Delta n}$
Judge from the size of K whether an equilibrium mixture is mostly reactants or products
Calculate the reaction quotient Q and predict the direction of change by comparing it with K

In 1864 the Norwegians Guldberg and Waage stated the law of mass action: at a given temperature, at equilibrium, “the product of product concentrations divided by the product of reactant concentrations” — each raised to its coefficient — is a constant. The table shows four starting points for $\ce{H2} + \ce{I2} \mathrel{\ce{<=>}} \ce{2HI}$ at 430 °C. Their equilibrium compositions are completely different, yet the last column always comes out the same.

Four starting points for H₂ + I₂ ⇌ 2HI at 430 °C (mol/L; computed by the exact ICE solver with Kc = 54.3)
Start H₂, I₂, HI	Equilibrium H₂, I₂, HI	$[\ce{HI}]/([\ce{H2}][\ce{I2}])$	$[\ce{HI}]^2/([\ce{H2}][\ce{I2}])$
0.500, 0.500, 0	0.107, 0.107, 0.786	69.0	54.3
0.200, 0.600, 0	0.00676, 0.407, 0.386	141	54.3
0, 0, 1.000	0.107, 0.107, 0.786	69.0	54.3
0.100, 0.100, 0.500	0.0747, 0.0747, 0.551	98.6	54.3

Source: Kc = 54.3 at 430 °C (standard general-chemistry value).

Predict first

What does the table show? At equilibrium, which quantity does not depend on the starting mixture?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

[\ce{HI}]

[\ce{HI}]/([\ce{H2}][\ce{I2}])

[\ce{HI}]^2/([\ce{H2}][\ce{I2}])

The sum of the three concentrations

Writing the equilibrium constant

a\,\mathrm{A} + b\,\mathrm{B} \rightleftharpoons c\,\mathrm{C} + d\,\mathrm{D}\qquad K_c = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b},\qquad K_p = \frac{p_\mathrm{C}^{\,c}\,p_\mathrm{D}^{\,d}}{p_\mathrm{A}^{\,a}\,p_\mathrm{B}^{\,b}}

Concentrations are relative to 1 mol/L and pressures to 1 bar, so strictly K has no units. Pure solids and pure liquids (including the solvent of a dilute solution) have activity 1 and are left out.

$[\mathrm{C}]$: equilibrium concentration of C, mol/L
$p_\mathrm{C}$: equilibrium partial pressure of C, bar
$a, b, c, d$: coefficients in the balanced equation

In a heterogeneous equilibrium the solids drop out of K. For limestone decomposing, $\ce{CaCO3(s)} \mathrel{\ce{<=>}} \ce{CaO(s)} + \ce{CO2(g)}$ , the equilibrium constant is simply $K_p = p_{\ce{CO2}}$ : about 0.22 bar at 800 °C as measured; the bench below includes heat capacities and gets 0.23 bar (extrapolating the 298 K values of ΔH° and ΔS° in a straight line would give 0.44 bar, about twice too high; the Gibbs energy lesson explains why). Whether the kiln holds one lump or a tonne of limestone, the equilibrium $\ce{CO2}$ pressure is the same. Likewise K for $\ce{H2O(l)} \mathrel{\ce{<=>}} \ce{H2O(g)}$ is just the vapour pressure: 0.0317 bar at 25 °C.

Kc and Kp: for ideal gases $p = cRT$ , so substituting for every partial pressure gives

K_p = K_c\,(RT)^{\Delta n},\qquad R = 0.08314\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}

where $\Delta n$ is the coefficients of gaseous products minus those of gaseous reactants. For $\ce{H2} + \ce{I2} \mathrel{\ce{<=>}} \ce{2HI}$ , $\Delta n = 0$ and $K_p = K_c$ ; for $\ce{N2O4} \mathrel{\ce{<=>}} \ce{2NO2}$ , $\Delta n = 1$ , so at 25 °C $K_p = 0.148$ (0.1484 more precisely) and $K_c = 0.1484/24.79 = 5.99\times10^{-3}$ .

What the size of K tells you

K spans hundreds of orders of magnitude. At 25 °C, $\ce{2H2} + \ce{O2} \mathrel{\ce{->}} \ce{2H2O(l)}$ has K ≈ $10^{83}$ : the equilibrium lies utterly on the side of water. $\ce{N2} + \ce{O2} \mathrel{\ce{<=>}} \ce{2NO}$ has K ≈ $2\times10^{-31}$ : air holds essentially no equilibrium NO. With 0.148, $\ce{N2O4} \mathrel{\ce{<=>}} \ce{2NO2}$ keeps sizeable amounts of both. As a rule of thumb, $K > 10^3$ means “essentially complete” and $K < 10^{-3}$ “hardly goes”.

But K says only where the reaction ends, not how fast it gets there. A mixture of hydrogen and oxygen can sit quietly for years — that is a question of kinetics; see reaction rates.

The reaction quotient Q: which way will it go?

Q = \frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}\qquad Q<K:\ \longrightarrow\qquad Q>K:\ \longleftarrow\qquad Q=K:\ \rightleftharpoons

Q has the same form as K but uses the concentrations at any moment. A mixture always reacts in the direction that brings Q towards K.

$Q$: reaction quotient: the “products/reactants” ratio right now
$K$: equilibrium constant: the value Q takes at equilibrium

Worked example

Will the brown deepen or fade?

At 25 °C a flask contains $p_{\ce{N2O4}} = 0.80$ bar and $p_{\ce{NO2}} = 0.10$ bar. Given $K_p = 0.148$ , which way does the reaction go?

Write the K expression · compare Q with K

Reaction quotient Q and equilibrium constant K

$\ce{CaCO3(s)} \mathrel{\ce{<=>}} \ce{CaO(s)} + \ce{CO2(g)}$

K_p = p_{\ce{CO2}}

Q = (0.3) = 0.3

Temperature · 800 °C (1073 K) Amount of solid (any units) · CaCO₃ 1.0 Amount of solid (any units) · CaO 1.0 p(CO₂) · 0.3 bar

$\ce{CaCO3(s)}$ $\ce{CaO(s)}$ Pure solid or liquid: activity 1, left out of K Change the amount of solid as you like: Q does not move — which is why solids are left out of K.

Reaction quotient Q 0.3

Equilibrium constant K (800 °C (1073 K)) 0.229

Q > K: net reaction runs in reverse

K near 1: sizeable amounts on both sides

Converting between Kc and Kp

K_p = K_c\,(RT)^{\Delta n},\quad \Delta n = 1,\quad RT = 89.23\ \mathrm{L\,bar\,mol^{-1}}

K_c = 0.00257 · K_p = 0.229

Limestone: drag the amount of solid and Q does not budge; only the CO₂ pressure and the temperature matter. Switch to other systems to compare Kc and Kp.

Concept check

The equilibrium-constant expression for $\ce{CaCO3(s)} \mathrel{\ce{<=>}} \ce{CaO(s)} + \ce{CO2(g)}$ is:

K_p = p_{\ce{CO2}}

K = \dfrac{[\ce{CaO}]\,p_{\ce{CO2}}}{[\ce{CaCO3}]}

K_p = 1/p_{\ce{CO2}}

K = [\ce{CaO}]/[\ce{CaCO3}]

Try the calculation

At 25 °C, $K_p = 0.148$ for $\ce{N2O4(g)} \mathrel{\ce{<=>}} \ce{2NO2(g)}$ . What is $K_p$ for $\ce{2NO2(g)} \mathrel{\ce{<=>}} \ce{N2O4(g)}$ ?

Try the calculation

At 500 °C, $K_p = 1.56\times10^{-5}$ for $\ce{N2} + \ce{3H2} \mathrel{\ce{<=>}} \ce{2NH3}$ (pressures in bar). Find $K_c$ at the same temperature (concentrations in mol/L).

Key takeaways

The equilibrium constant is products over reactants, each raised to its coefficient; it changes only with temperature.
Pure solids and liquids stay out of K; for heterogeneous equilibria K is often just one gas pressure.
$K_p = K_c(RT)^{\Delta n}$ ; reversing inverts K and multiplying the coefficients by n raises K to the n-th power.
Q < K runs forward, Q > K runs in reverse, Q = K is at equilibrium. K tells you the destination, not the speed.

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Data and references

P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 6A (the equilibrium constant).
D. D. Wagman et al., “The NBS tables of chemical thermodynamic properties”, J. Phys. Chem. Ref. Data 11, Suppl. 2 (1982) (N₂O₄, NO₂, CaCO₃, CaO, CO₂, H₂O).
W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed. (2016) (vapour pressure of water; ΔfG° of NO).
NIST Chemistry WebBook, Shomate parameters for N₂, H₂ and NH₃ (Kp of ammonia at 500 °C).