Equilibrium Constants: Writing K, Reading Q and Converting Kc to Kp

Dynamic equilibrium: forward and reverse at equal rates

In a reversible reaction, products can react to re-form reactants. As the forward reaction uses up reactants it slows, and as products build up the reverse reaction speeds up. Equilibrium is reached when the two rates are equal. From then on, the concentrations stay constant even though both reactions continue. That is why chemists call it a dynamic equilibrium and write it with a double arrow, ⇌.

The same equilibrium composition can be reached from either side. A sealed flask that starts with pure NO₂ and one of the same volume that starts with the equivalent amount of pure N₂O₄ end at the same mixture when held at the same temperature. Equilibrium is only reached in a closed system; if a product escapes, as CO₂ does from an open container, the reaction never settles.

Building the K expression from a balanced equation

For a general reaction aA + bB ⇌ cC + dD, the law of mass action gives the equilibrium constant in terms of molar concentrations: Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ). Products go on top, reactants on the bottom, and each concentration is raised to the power of its coefficient. Unlike a rate law, this expression comes straight from the balanced equation.

Pure solids and pure liquids are left out, because their effective concentration does not change as the reaction proceeds. Water is also left out when it is the solvent in a dilute aqueous solution. For example, the decomposition CaCO₃(s) ⇌ CaO(s) + CO₂(g) has Kc = [CO₂]. At a given temperature, the equilibrium pressure of CO₂ above the solids is fixed, whatever the amounts of the two solids.

Strictly, an equilibrium constant is built from activities, which have no units, so K is reported without units. Using concentrations in mol/L or pressures in atm or bar is the usual approximation in general chemistry.

Reading the magnitude of K

A large K means that at equilibrium the products dominate; a small K means the reactants dominate. As a rough guide, K greater than about 10³ means the reaction is close to complete, K smaller than about 10⁻³ means little product forms, and values in between give a mixture with significant amounts of both.

OpenStax section 13.4 gives Kc = 4.1 × 10⁻⁴ for N₂(g) + O₂(g) ⇌ 2 NO(g) at 2000 °C. Even at that temperature, air contains only a little NO at equilibrium. At room temperature the constant is far smaller still, which is why nitrogen and oxygen coexist in the atmosphere.

K says nothing about speed. A reaction with an enormous K can still be too slow to observe. Kinetics, not the equilibrium constant, decides how quickly equilibrium is reached.

Worked example: Kc and Kp for nitrogen dioxide and dinitrogen tetroxide

For reactions involving gases, the equilibrium constant can also be written with partial pressures, giving Kp. The two are related by Kp = Kc(RT)^Δn, which means Kc multiplied by RT raised to the power Δn, where Δn is the moles of gaseous products minus the moles of gaseous reactants, R = 0.08206 L atm mol⁻¹ K⁻¹ when pressures are in atm, and T is in kelvins. When Δn = 0, Kp equals Kc.

OpenStax section 13.2 describes 0.10 mol of NO₂ placed in a 1.0 L flask at 25 °C, which comes to equilibrium with [NO₂] = 0.016 mol/L and [N₂O₄] = 0.042 mol/L. As a check, 0.016 + 2(0.042) = 0.100 mol/L, so all the nitrogen is accounted for. For N₂O₄(g) ⇌ 2 NO₂(g), Kc = [NO₂]²/[N₂O₄] = (0.016)² / 0.042 = 6.1 × 10⁻³, to two significant figures.

Here Δn = 2 − 1 = 1, and RT = 0.08206 × 298.15 = 24.47. So Kp = (6.1 × 10⁻³)(24.47) = 0.15, with pressures in atm. The value is larger than Kc because Δn is positive and RT is greater than 1.

Changing the equation changes K

Because K is built from the balanced equation, any change to the equation changes K in a predictable way. These rules are the equilibrium counterpart of the rules for manipulating ΔH in Hess's law, with multiplication in place of addition.

Worked example. Start from N₂(g) + O₂(g) ⇌ 2 NO(g), Kc = 4.1 × 10⁻⁴ at 2000 °C. Reversing it gives 2 NO(g) ⇌ N₂(g) + O₂(g) with Kc = 1/(4.1 × 10⁻⁴) = 2.4 × 10³. Halving the forward equation gives ½ N₂(g) + ½ O₂(g) ⇌ NO(g) with Kc = √(4.1 × 10⁻⁴) = 2.0 × 10⁻². Doing both, NO(g) ⇌ ½ N₂(g) + ½ O₂(g) has Kc = 1/√(4.1 × 10⁻⁴) = 49.

• Reverse the equation: the new K is 1/K.
• Multiply every coefficient by n: the new K is Kⁿ. Halving the equation takes the square root.
• Add two equations: multiply their K values. Both K values must be for the same temperature.

Comparing Q with K to predict direction

The reaction quotient, Q, has exactly the same form as K but uses the concentrations present at any moment, not only at equilibrium. Comparing Q with K tells you which way the reaction must go. If Q < K, there is too little product, and the reaction proceeds forward. If Q > K, there is too much product, and the reaction proceeds in reverse. If Q = K, the system is at equilibrium.

Worked example. OpenStax section 13.2 gives Kc = 0.640 at 800 °C for CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g). A mixture contains [CO] = 0.0500, [H₂O] = 0.0400, [CO₂] = 0.0300 and [H₂] = 0.0600 mol/L. Q = (0.0300)(0.0600) / [(0.0500)(0.0400)] = 0.00180/0.00200 = 0.900. Q is larger than K, so the mixture shifts toward CO and H₂O.

A second mixture has the same CO and H₂O but [CO₂] = 0.0200 and [H₂] = 0.0250 mol/L. Now Q = (0.0200)(0.0250)/(0.00200) = 0.250, which is less than 0.640, so this mixture reacts forward and makes more CO₂ and H₂.

Worked example: K from a measured equilibrium mixture

Measuring one equilibrium amount is often enough to find K, because the balanced equation fixes how every other amount changed. The data here are illustrative. At one fixed temperature, 0.500 mol of PCl₅ is sealed in an empty 2.00 L flask and allowed to reach PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At equilibrium the flask holds 0.180 mol of Cl₂.

Work in concentrations. The starting [PCl₅] is 0.500 mol / 2.00 L = 0.250 mol/L, and the equilibrium [Cl₂] is 0.180 mol / 2.00 L = 0.0900 mol/L. Each mole of Cl₂ formed also forms one mole of PCl₃ and uses one mole of PCl₅, so [PCl₃] = 0.0900 mol/L and [PCl₅] = 0.250 − 0.0900 = 0.160 mol/L. Then Kc = (0.0900)(0.0900)/0.160 = 0.0506. At this temperature, 36.0 percent of the PCl₅ decomposed. The ICE-table guide extends this bookkeeping to problems where you must solve for the equilibrium amounts.

Equilibrium-expression mistakes

A short checklist catches most errors in writing and using K.

• Putting reactants on top. K is always products over reactants for the equation as written.
• Leaving out the coefficients as exponents, or using them as multipliers instead.
• Including a pure solid, a pure liquid or solvent water in the expression.
• Using initial concentrations in K. Only equilibrium values belong in K; any other set gives Q.
• Forgetting that K depends on temperature. A K value is meaningless without the temperature it applies to.
• Converting Kc to Kp with Δn counted over all species instead of gases only, or with T in Celsius.