The law of mass action

Result

The law of mass action

T-044Home CU-203Threads equilibrium

Statement

The equilibrium constant expression from forward and reverse rates.

Why it matters

Rate laws describe how fast a reaction proceeds in one direction; the law of mass action shows what happens once a reaction runs in both directions simultaneously and settles into a dynamic balance — the ratio of forward and reverse rate constants collapses into a single number, the equilibrium constant \(K\), expressed purely in terms of concentrations. Almost everything else in this unit builds directly on that expression: le-chatelier interprets how a system already at this balance responds to disturbance, reaction-quotient generalises the identical algebraic form to non-equilibrium compositions to predict direction, kp-kc-relation translates it into partial pressures, and solubility-product is simply this same expression applied to a saturated ionic solid.

It also explains, rather than merely asserts, why equilibrium expressions take the specific form of products over reactants raised to stoichiometric coefficients — a form that otherwise looks like an arbitrary convention.

Hypotheses

The forward and reverse reactions are each treated as elementary (single-step) processes when writing their rate laws directly from stoichiometry.Only for a genuinely elementary step do a rate law's exponents equal the reaction's stoichiometric coefficients; for a multistep overall reaction the rate law generally does not match the overall stoichiometry, even though (Discussion) the equilibrium expression itself still takes the simple product/reactant form. The system has reached true dynamic equilibrium: forward and reverse rates are equal and concentrations are constant in time.This is the condition actually being imposed algebraically in the Proof — setting the forward and reverse rate expressions equal to each other is only valid once the system has stopped changing net composition, even though individual forward and reverse reaction events continue occurring. Concentrations stand in adequately for thermodynamic activities.Strictly, \(K\) should be built from dimensionless activities referenced to a standard state, not raw concentrations; the concentration form used throughout this unit is an excellent approximation in dilute solution and at low gas pressure, but deviates measurably at high ionic strength or high pressure.

Proof

a\text{A}+b\text{B} \rightleftharpoons c\text{C}+d\text{D}, \qquad \text{rate}_f = k_f[\text{A}]^a[\text{B}]^b

For an elementary forward step, the rate law's exponents equal the stoichiometric coefficients directly (Hypotheses); this is the standard starting point, not a general fact about all rate laws. A

\text{rate}_r = k_r[\text{C}]^c[\text{D}]^d

The reverse elementary step is treated identically, with its own rate constant \(k_r\) and its own stoichiometry-matched exponents. A

\text{At equilibrium: } \text{rate}_f=\text{rate}_r \ \Rightarrow\ k_f[\text{A}]^a[\text{B}]^b = k_r[\text{C}]^c[\text{D}]^d

Dynamic equilibrium (Hypotheses) means forward and reverse rates are equal, not zero — both reactions continue at equal, opposing rates, so net concentrations no longer change. A

K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} = \frac{k_f}{k_r}

Rearranging Step 3 isolates the equilibrium constant as exactly the ratio of the two rate constants; since \(k_f\) and \(k_r\) each depend only on temperature (Arrhenius behaviour), \(K_c\) itself depends only on temperature, never on the specific starting concentrations used to reach equilibrium. A

Result

K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} = \frac{k_f}{k_r}

Reading. A single, temperature-fixed number, the ratio of forward to reverse rate constants, sets the exact concentration ratio a system settles into once forward and reverse rates balance.

Scope. Directly derivable in this simple stoichiometric form for elementary reactions; for a multistep overall reaction the same product/reactant expression still holds at true equilibrium (Discussion), even though the reaction's own rate law generally does not mirror its overall stoichiometry.

Corollaries & converses

reaction-quotient uses the identical algebraic expression evaluated at any concentrations, equilibrium or not; comparing it against \(K_c\) is the rigorous basis for predicting a reaction's net direction.
kp-kc-relation is the same result rewritten in partial pressures for gas-phase reactions, related back to \(K_c\) via the ideal gas law.
solubility-product is this expression applied to a sparingly soluble ionic solid, with the solid's own (constant, activity-one) contribution folded into \(K_{sp}\) rather than written explicitly.
Converse: measuring \(K_c\) experimentally gives the ratio \(k_f/k_r\) without needing to measure either rate constant separately.

Fails without

Drop the elementary-step assumption (Hypotheses) and apply stoichiometric exponents directly to a multistep overall reaction's rate law: the resulting rate law is generally wrong, since a non-elementary reaction's true rate law is set by its slowest step's molecularity, not by the overall balanced equation — even though (Discussion) the equilibrium expression itself still takes the simple product/reactant form regardless.
Compare \(K_c\) values measured at two different, unstated temperatures as if they were the same fixed constant: since \(K_c=k_f/k_r\) and both rate constants are themselves temperature-dependent (Hypotheses), such a comparison silently conflates two genuinely different numbers, producing spurious or contradictory conclusions about the reaction's equilibrium position.

Common errors

Including a pure solid's or pure liquid's concentration explicitly in the expression, rather than treating its (constant) activity as \(1\) and omitting it.
Assuming rate-law exponents equal stoichiometric coefficients for every reaction, rather than only for elementary steps (Hypotheses).
Believing that changing a concentration changes the value of \(K_c\) itself; only temperature changes \(K_c\) (le-chatelier addresses this explicitly).
Writing reactants over products, or omitting the stoichiometric exponents, when constructing the expression.

Discussion

Cato Guldberg and Peter Waage proposed the law of mass action in 1864, from empirical rate observations, well before the elementary-step kinetic derivation given above was available — another instance, alongside Hess's law, of a chemical regularity discovered empirically before its full theoretical justification existed.

For a genuinely multistep overall reaction, the equilibrium expression still takes the simple product/reactant form derived here, even though the reaction's measured rate law does not match its overall stoichiometry. This is because the overall \(K\) equals the product of the individual elementary steps' own mass-action ratios, and every intermediate species' concentration cancels out of that product, leaving only reactants and products of the net, overall equation.

Common misconception: that a catalyst changes the value of \(K_c\). A catalyst speeds up the forward and reverse elementary steps by an identical factor (it lowers the same activation-energy barrier for both directions), so \(k_f/k_r\), and hence \(K_c\), is completely unaffected; only the time taken to reach equilibrium changes.

Worked examples

\text{A} \rightleftharpoons \text{B}, \qquad k_f = 0.50\,\text{s}^{-1},\quad k_r=0.10\,\text{s}^{-1}

A simple unimolecular isomerisation, treated as elementary in both directions. By Step 4, \(K_c=k_f/k_r=0.50/0.10=5.0\). A

\text{Total concentration } [\text{A}]+[\text{B}]=0.60\,\text{M}, \qquad K_c=\frac{[\text{B}]}{[\text{A}]}=5.0

Combine the mass-balance constraint with the equilibrium expression: \([\text{B}]=5.0[\text{A}]\), so \([\text{A}]+5.0[\text{A}]=0.60\), giving \([\text{A}]=0.10\,\text{M}\) and \([\text{B}]=0.50\,\text{M}\). A

[\text{A}]_{eq}=0.10\,\text{M}, \quad [\text{B}]_{eq}=0.50\,\text{M}

Reading. Knowing only the two rate constants and total material fixes the exact equilibrium composition, without ever needing to solve a differential rate equation directly.

Scope. The identical two-step method (find \(K_c\) from rate constants, then combine with a mass balance) applies to any reaction whose forward and reverse steps are elementary.

Problems

For an elementary reaction \(\text{A}+\text{B}\rightleftharpoons \text{C}\) with \(k_f=4.0\times10^{-2}\,\text{M}^{-1}\text{s}^{-1}\) and \(k_r=8.0\times10^{-3}\,\text{s}^{-1}\), find \(K_c\).
Solution
\(K_c=k_f/k_r=(4.0\times10^{-2})/(8.0\times10^{-3})=5.0\,\text{M}^{-1}\).
Explain why the equilibrium expression for \(\text{CaCO}_3(s)\rightleftharpoons \text{CaO}(s)+\text{CO}_2(g)\) is written simply as \(K_c=[\text{CO}_2]\), with no solid terms appearing.
Solution
Pure solids have constant, activity-one contributions regardless of how much solid is present (their "concentration," being a fixed density, does not vary), so by convention they are omitted from the mass-action expression entirely; only the gas-phase species, whose concentration genuinely can vary, appears (Common errors, first bullet).
A student measures \(K_c\) for a reaction at two different temperatures and finds different values. Explain whether this contradicts Step 4's claim that \(K_c\) is fixed.
Solution
No contradiction: Step 4 shows \(K_c=k_f/k_r\), and both rate constants are themselves temperature-dependent (Arrhenius behaviour), so \(K_c\) is fixed only at a given, specified temperature. Measuring different \(K_c\) values at different temperatures is expected and is, in fact, the basis of the van't Hoff relation used elsewhere to extract reaction enthalpy from temperature-dependent equilibrium data.