Chemistry: Atoms First 2e · Fundamental Equilibrium Concepts
Equilibrium Constants
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The equilibrium constant K quantifies where a reversible reaction settles. For a reaction aA + bB ⇌ cC + dD, the law of mass action Rule relating equilibrium concentrations to K Full entry → gives, for molar concentrations,
Kc = [C]c[D]d[A]a[B]b
and the analogous expression Kp for gas-phase reactions uses partial pressures instead of concentrations. Products go on top, reactants on the bottom, and every concentration is raised to its stoichiometric coefficient. Pure solids and pure liquids do not appear in the expression. The magnitude of K tells you whether products or reactants dominate at equilibrium, and comparing the reaction quotient Q — the same expression evaluated at the current, possibly non-equilibrium concentrations — with K predicts the direction of change. Because K depends only on the reaction as written and on temperature, it is a powerfully predictive number.
Why this matters
The equilibrium constant turns "which way does this go?" into arithmetic. Engineers use K to compute the maximum possible yield of ammonia, sulfuric acid, and pharmaceuticals. Chemists rank acid and base strength with Ka and Kb values (Chapter 14) and predict precipitation with the solubility product Ksp (Chapter 15). And on exams, virtually every equilibrium calculation — and most of acid-base chemistry — begins with writing the correct K expression.
The college version
Core Concepts
The law of mass action
For aA + bB ⇌ cC + dD, the equilibrium expression is written with product concentrations in the numerator and reactant concentrations in the denominator, each raised to the stoichiometric coefficient from the balanced equation. A coefficient of 2 means the concentration is squared; a coefficient of 3 means it is cubed. It follows from setting the forward rate equal to the reverse rate at equilibrium: Kc = kf/kr.
What the magnitude of K means
A very large K (e.g., 105) means products dominate — the reaction "goes far." A very small K (e.g., 10-5) means reactants dominate. A K near 1 means comparable amounts of both sides. Crucially, K says nothing about speed: a huge-K reaction can still be impossibly slow.
The reaction quotient Q and the direction of change
Q uses the same expression as K, but with current concentrations at any moment:
Qc = [C]c[D]d[A]a[B]b
- Q < K: the system is "too reactant-heavy"; the forward reaction runs until Q reaches K.
- Q > K: the system is "too product-heavy"; the reverse reaction runs.
- Q = K: the system is at equilibrium; no net change.
Kp versus Kc
For gas-phase reactions, Kp uses partial pressures and Kc uses molarities. They are related by
Kp = Kc(RT)Δn
where Δn = (moles of gaseous products) - (moles of gaseous reactants) and R = 0.08206 L atm mol⁻¹ K⁻¹ when pressures are in atm. If Δn = 0, the two constants are equal.
Heterogeneous equilibria
Pure solids and pure liquids have constant concentrations (fixed density and molar mass), so they are built into K and omitted from the expression. For the thermal decomposition of calcium carbonate,
CaCO3(s) ⇌ CaO(s) + CO2(g),
the equilibrium expression is simply Kc = [CO2]. Only gases and aqueous solutes appear.
Manipulating reactions and their constants
If a reaction is reversed, the new constant is 1/K. If the coefficients are multiplied by n, the constant is raised to the n-th power (Kn). If two reactions are added, their constants multiply. These rules let you combine known equilibria into new ones without new experiments.
How It Works / Step-by-Step Process
- Write the balanced equation with physical states; include only gases and aqueous solutes (omit pure solids and pure liquids).
- Raise each concentration (or partial pressure) to its stoichiometric coefficient.
- To predict direction: compute Q from current conditions and compare with K.
- To convert between Kc and Kp: use Kp = Kc(RT)Δn with Δn from the balanced equation.
Common Confusions
| Common Confusion | Correct Understanding |
|---|---|
| "K and Q are interchangeable." | K is the equilibrium value; Q is the current value. Only when Q = K is the system at equilibrium. |
| "K changes when concentrations change." | Concentrations change along the path to equilibrium; K changes only with temperature. |
| "Include solids and pure liquids in the expression." | They are omitted; their concentrations are constant and absorbed into K. |
| "The coefficients are multiplied, not raised to powers." | A coefficient c means [C] raised to the power c. |
| "A huge K means the reaction is fast." | It means product-favored; the rate is governed by kinetics (activation energy), not by K. |
| "K depends on the starting amounts." | At a given temperature and reaction as written, K is the same regardless of initial amounts. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
The equilibrium constant is like a final score ratio: 100-to-1 means products win; 1-to-100 means reactants win. The reaction quotient is the score right now — if it is 10-to-1 but the final ratio should be 100-to-1, the game keeps going until the score reaches the right ratio.
Worked examples
For N2(g) + 3H2(g) ⇌ 2NH3(g):
Kc = [NH3]2[N2][H2]3
The coefficient 3 on H2 cubes the hydrogen concentration. Now convert between forms: Here Δn = 2 - (1 + 3) = -2, so Kp = Kc(RT)-2, or rearranged, Kc = Kp(RT)2. At 25 °C, Kp ≈ 6.0 × 105 atm⁻². With RT = (0.08206 L atm mol-1K-1)(298.15 K) = 24.5 L atm mol-1:
Kc = (6.0 × 105)(24.5)2 = (6.0 × 105)(598) = 3.6 × 108
Dimensional analysis: atm⁻² × (L atm mol⁻¹)² = L² mol⁻² = M⁻². The huge value of K confirms that at room temperature ammonia strongly dominates at equilibrium — which is why the industrial challenge is speed, not thermodynamics.
At 445 °C, Kc = 50.2 for H2(g) + I2(g) ⇌ 2HI(g). Two mixtures are analyzed:
Mixture A: [H2] = 0.100 M, [I2] = 0.100 M, [HI] = 0.300 M.
Q = [HI]2[H2][I2] = (0.300)2(0.100)(0.100) = 0.09000.0100 = 9.0
Q = 9.0 < K = 50.2, so the forward reaction runs: HI forms until Q climbs to 50.2.
Mixture B: [H2] = 0.050 M, [I2] = 0.050 M, [HI] = 0.400 M.
Q = (0.400)2(0.050)(0.050) = 0.1600.00250 = 64
Q = 64 > K = 50.2, so the reverse reaction runs: HI decomposes. Same reaction, same temperature — only the current composition differs, and Q versus K gives the direction.
For the Haber reaction, Kc = 3.6 × 108 at 25 °C. The reverse reaction 2NH3(g) ⇌ N2(g) + 3H2(g) has
K' = 1Kc = 13.6 × 108 = 2.8 × 10-9
Ammonia barely decomposes at room temperature. Doubling all coefficients (2N2 + 6H2 ⇌ 4NH3) squares the constant: K'' = (3.6 × 108)2 = 1.3 × 1017. A K value is meaningless without the exact balanced equation it belongs to.
Key takeaways
- Kc = [C]c[D]d/[A]a[B]b — products over reactants, coefficients as exponents.
- Large K → product-favored; small K → reactant-favored; K near 1 → comparable amounts.
- K depends only on temperature and on how the reaction is written — never on initial amounts, catalysts, or how equilibrium is approached.
- Q < K → forward; Q > K → reverse; Q = K → equilibrium.
- Kp = Kc(RT)Δn, with Δn = (moles gaseous products) − (moles gaseous reactants).
- Pure solids and pure liquids are omitted from K and Q expressions.
- Reverse reaction: K' = 1/K. Coefficients multiplied by n: K' = Kn. Added reactions: multiply K values.
- Big K ≠ fast reaction. K is thermodynamics; rate is kinetics.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Write the Kc expression for N2(g) + 3H2(g) ⇌ 2NH3(g) and explain the exponent on [H2].
Show answer
Kc = [NH3]2/([N2][H2]3); the exponent 3 is the stoichiometric coefficient of H₂ in the balanced equation.
For the same reaction, what is Δn, and does Kp equal Kc?
Show answer
Δn = 2 - 4 = -2, so Kp = Kc(RT)-2 — they are not equal.
A mixture has Q > K. Which direction does the reaction shift, and why?
Show answer
Reverse. The numerator is too large relative to K; the reverse reaction consumes products until Q = K.
Why are CaCO3(s) and CaO(s) absent from the expression for CaCO3(s) ⇌ CaO(s) + CO2(g)?
Show answer
Pure solids have constant concentration (fixed density and molar mass), so they are built into K and omitted.
If K = 4.0 × 10-3 for a reaction, what is K for the reverse reaction?
Show answer
K' = 1/K = 1/(4.0 × 10-3) = 2.5 × 102.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- equilibrium constant (K)
- The fixed value of the mass-action ratio at a given temperature
- reaction quotient (Q)
- The mass-action ratio at the current, possibly non-equilibrium conditions
- Kc / Kp
- Equilibrium constant using molarities / partial pressures
- law of mass action
- Rule relating equilibrium concentrations to K
- product-favored / reactant-favored
- Equilibrium lying toward products / reactants
Sources & references
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