Chemical Equilibrium Constants

01 / UNDERSTAND & APPLY

1. What is balanced at equilibrium?

Imagine two connected rooms with people moving in both directions. The populations can stay steady while movement continues, provided the two flows have equal rates. Chemical equilibrium is similarly dynamic: forward and reverse transformations continue, but their rates match. It does not require equal amounts of reactants and products. A mixture may contain overwhelmingly more of one side and still have balanced opposing rates.

In a closed reacting system held at appropriate fixed conditions, equilibrium composition is stable in time on the macroscopic scale. Constancy alone is not a sufficient experimental proof: a reaction can be so slow that no change is detectable even though it is far from equilibrium. Kinetics tells you how quickly a system can approach its equilibrium; thermodynamics tells you which state is favored. A large equilibrium constant cannot promise a rapid reaction.

For an elementary reversible step A ⇌ B, rate equality k_f[A] = k_r[B] gives [B]/[A] = k_f/k_r at equilibrium. This is a useful microscopic illustration. Do not infer rate-law powers directly from the coefficients of every overall balanced reaction; multistep mechanisms need not have such rate laws. Equilibrium expressions, however, use the net stoichiometric coefficients regardless of the detailed mechanism.

Always attach the conditions and equation to the constant. Changing temperature usually changes the constant. Reversing or multiplying the written equation changes its numerical form even though you describe the same chemical system. Initial composition affects the amounts ultimately present, but it does not create a new equilibrium constant at the same temperature.

02 / UNDERSTAND & APPLY

2. Build the expression from activities and phases

For aA + bB ⇌ cC + dD, the equilibrium constant is a product of equilibrium activities raised to their stoichiometric powers: K = a(C)^c a(D)^d/[a(A)^a a(B)^b]. Activities are dimensionless measures of effective composition relative to standard states. For introductory ideal-gas work they reduce to partial-pressure ratios p/p°, and for sufficiently dilute solutions they are approximated using concentration ratios [X]/c°. These approximations should be stated rather than silently extended to every real mixture.

Pure solids and pure liquids have unit activity in the usual standard-state approximation and disappear from the written expression. They do not disappear from the chemical equation or the atom balance. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), K depends on CO₂ activity, provided the required solid phases coexist. Adding more CaCO₃ to that already equilibrated two-solid system does not change its equilibrium CO₂ pressure. Removing a phase completely can change what constraints remain available.

The concentration expression commonly called K_c writes molar concentrations directly; its pressure counterpart K_p writes gas partial pressures. Many school problems assign apparent units according to the powers in those expressions. Thermodynamic K is dimensionless. Keep the convention of the problem consistent, and never compare a number calculated from pressures in one unit with a constant based on another unannounced convention.

Liquid water is omitted when it is the nearly pure solvent in a dilute aqueous model, but gaseous water is included in a gas reaction. Water mixed substantially with other liquids may not have unit activity. A species’ formula alone therefore cannot tell you whether to omit it: inspect the physical state and the model.

03 / UNDERSTAND & APPLY

3. Transform equations, transform constants

Reversing a reaction interchanges numerator and denominator, so the reverse constant is 1/K. Multiplying every coefficient by a number r raises the constant to the power r. Adding two reactions multiplies their constants after intermediate species cancel. These operations follow from the algebra of activity products; they are not unrelated rules to memorize.

A large K means the equilibrium expression favors products for the reaction as written. It does not mean every reactant vanishes, and it cannot be compared naively with K for a differently scaled equation. If A ⇌ B has K = 100, then 2A ⇌ 2B has K = 10,000 without any change in the physical ratio [B]/[A]. The numerical magnitude depends on the equation normalization.

The cancellation rule is especially useful in thermodynamic cycles. It is also a diagnostic: after combining equations, check that the intended net reaction really remains. If a species appears on both sides with unequal coefficients, cancel only the common amount. A wrong direction for one subsidiary reaction can turn a multiplication into the required division.

04 / UNDERSTAND & APPLY

4. Relate Kp to Kc without losing Δn

For each ideal gas, p_i = [i]RT. Substitute that relation into every pressure factor in K_p. Each product factor contributes a power of RT in the numerator and each reactant factor contributes one in the denominator. The resulting net exponent is Δn_gas, the sum of gaseous product coefficients minus the sum of gaseous reactant coefficients. Solids, liquids and dissolved ions do not enter this gas-count difference.

In the familiar unnormalized textbook convention, K_p = K_c(RT)^Δn_gas. Choose R to match the pressure unit and volume unit used in the expressions. Temperature must be kelvin. If normalized dimensionless K_p and K_c are being used, the standard-state factors must also appear: the factor is (RTc°/p°)^Δn_gas. This distinction explains why apparently different unit conventions can produce different numerical textbook constants.

When Δn_gas = 0, the RT factor cancels and the conventional K_p and K_c are numerically equal with compatible conventions. Equality here says nothing about whether equilibrium contains equal gas amounts. For H₂ + I₂ ⇌ 2HI, there are equal total gaseous coefficients on each side, but HI can still greatly predominate.

05 / UNDERSTAND & APPLY

5. Q describes now; K describes equilibrium

The reaction quotient Q has the same algebraic form as K, but it uses the mixture’s current activities. You can calculate Q before equilibrium is reached or immediately after a perturbation. If Q is smaller than K, increasing product activities and decreasing reactant activities brings Q upward: the favored net direction is forward. If Q exceeds K, the reverse direction is favored. Equality identifies equilibrium with respect to that reaction.

This direction test is more dependable than vague language such as “the reaction opposes the change.” For instance, changing volume alters several gas partial pressures at once. Calculate how their powers combine rather than assuming the side with more molecules is always favored. Also distinguish the instantaneous composition change caused by the manipulation from the subsequent change due to reaction.

The thermodynamic statement is Δ_rG = Δ_rG° + RT ln Q = RT ln(Q/K), with dimensionless Q and K. Thus Q < K gives negative forward reaction Gibbs energy. This is a direction or driving-force statement at the specified state; it is not an activation-energy calculation. A catalyst can make that favored relaxation faster without changing its equilibrium endpoint.

06 / UNDERSTAND & APPLY

6. Turn conservation into an ICE calculation

An initial–change–equilibrium table is a bookkeeping device. It expresses all concentrations using one extent variable while respecting the balanced coefficients. For A + B ⇌ 2C starting with A₀, B₀ and C₀, a forward concentration change x gives A₀ − x, B₀ − x and C₀ + 2x. The factor two belongs in both the amount change and the squared equilibrium term if C has coefficient two.

Use Q to choose a sensible initial sign, then substitute the equilibrium expressions into K. Solve the resulting equation and reject roots that make any amount negative. A root is not valid merely because it satisfies an equation created after squaring; physical bounds must still hold. For a forward change in this example, 0 ≤ x ≤ min(A₀,B₀). For a reverse change, the available C imposes its own bound.

If K is extremely small, a small-change approximation can simplify a polynomial, but it must be tested after solving. Compare the neglected change with each quantity it was subtracted from, not just with one conveniently large concentration. A common 5% rule is a practical tolerance, not a fundamental chemical law. When the approximation fails, return to the unsimplified equation.

The ICE table normally uses concentrations at a fixed volume. If reaction changes a gas mixture’s total amount while pressure rather than volume is fixed, volume changes too; a mole-extent table followed by mole fractions and partial pressures is safer. Never combine a changing-volume problem with fixed-volume concentration changes without justification.

07 / UNDERSTAND & APPLY

7. Predict shifts by recalculating the quotient

At constant temperature, K is unchanged by ordinary concentration or pressure manipulations within the assumed model. Adding a reactant while leaving other activities initially unchanged lowers Q when that reactant appears in the denominator, so net forward reaction follows. Removing a product also lowers Q. These statements require the manipulation actually to change the relevant activity: adding more of a pure solid already present does not do so.

Compressing an ideal gaseous equilibrium at constant temperature multiplies every reacting partial pressure by the same factor f greater than one. Consequently Q_p becomes f^Δn_gas times its old value. For a reaction with fewer gas moles on the product side, Δn_gas is negative, Q drops and the forward direction is favored. Expansion gives the opposite effect. With Δn_gas = 0, uniform compression leaves Q unchanged in the ideal model.

Adding inert gas at fixed volume and temperature increases total pressure but does not change the reacting gases’ partial pressures p_i = n_iRT/V; no shift follows. At fixed total pressure and temperature, adding inert gas expands the mixture and lowers the reacting partial pressures, favoring the side with more gaseous moles if Δn_gas is nonzero. “Pressure increased” is therefore incomplete information unless the method is specified.

For dilute-solution equilibria, dilution similarly changes concentration factors, but solvent activity and coupled acid–base or precipitation reactions may need attention. A useful response should identify the actual changed variables and the relevant quotient, rather than apply a gas-mole slogan to aqueous or heterogeneous systems.

08 / UNDERSTAND & APPLY

8. Temperature changes K; catalysts do not

Heating favors the endothermic direction. For an exothermic forward reaction, increasing temperature generally decreases its equilibrium constant; for an endothermic forward reaction, it increases it. Treating heat as a qualitative bookkeeping term can help predict the direction, but heat is not a chemical species to insert into an activity expression.

The van ’t Hoff relation d(ln K)/dT = Δ_rH°/(RT²) makes the sign connection precise. Over a moderate temperature interval where Δ_rH° is approximately constant, ln(K₂/K₁) = −Δ_rH°/R(1/T₂ − 1/T₁). Use kelvin and consistent energy units. If heat capacity changes matter over a wide interval, a constant-enthalpy approximation can fail.

A catalyst changes the pathway and accelerates approach to equilibrium in both directions. It does not change the thermodynamic constant or the equilibrium composition at a specified temperature and other constraints. A high-pressure, lower-temperature condition may favor ammonia thermodynamically while slowing its uncatalyzed production; practical conditions reflect rate as well as equilibrium. A yield discussion that ignores this distinction confuses two different chemical questions.

Watch the wording “increases concentration” versus “shifts toward.” After compression, a reactant concentration can rise instantly from reduced volume even if the subsequent reaction consumes some of it. Similarly, heating may alter gas volume and composition simultaneously. State what is held fixed, which immediate change occurs, and which later reaction direction is predicted.

09 / UNDERSTAND & APPLY

9. Original practice: constants, direction and composition

These are original practice problems. Use ideal gases or dilute ideal solutions where stated, and treat the numerical constants as using consistent classroom concentration or pressure conventions. A correct solution should show both an algebraic result and a physical check. No example here is represented as a past-examination question.

Before calculating, write the equation and phase labels. Then ask whether you need an equilibrium constant, a current quotient or an extent. Most wrong answers in this chapter arise from mixing these roles rather than from difficult algebra.