Chemical Equilibrium ICE Tables: Kc, Kp, and Worked Examples
How to set up an ICE table, write the equilibrium expression, relate Kc and Kp, use the 5% approximation, and predict reaction direction with Q — worked step by step.
Learning Objectives
- ✓Write equilibrium expressions for Kc and Kp, omitting pure solids and liquids.
- ✓Build and solve an ICE table, including when to use the 5% approximation.
- ✓Use the reaction quotient Q to predict the direction of reaction.
1. Direct Answer: What an ICE Table Does
An ICE table (Initial, Change, Equilibrium) organizes the concentrations or pressures of every species as a reaction moves to equilibrium, so you can solve for the unknown shift x. The workflow is fixed: write the balanced equation, write the equilibrium expression K = [products]/[reactants] with each concentration raised to its stoichiometric coefficient (and OMITTING pure solids and pure liquids), fill in the Initial row, express the Change row in terms of x using the coefficients, add to get the Equilibrium row, then substitute into K and solve for x. Kc uses molar concentrations; Kp uses partial pressures of gases. A large K (≫1) means products are favored; a small K (≪1) means reactants dominate and x will be tiny — which is exactly when the 5% approximation saves you from a quadratic.
Key Points
- •ICE = Initial, Change, Equilibrium; solve for the shift x.
- •K expression: products over reactants, each raised to its coefficient; omit pure solids and liquids.
- •Large K favors products; small K favors reactants (and lets you approximate).
2. Writing Kc and Kp Correctly
For the general reaction aA + bB ⇌ cC + dD, the equilibrium constant is K = ([C]^c [D]^d) / ([A]^a [B]^b). Two rules trip students up. First, the COEFFICIENTS become EXPONENTS — a coefficient of 2 means the concentration is squared, not multiplied by 2. Second, PURE SOLIDS and PURE LIQUIDS are left out entirely because their activity is 1 (a solid’s concentration does not change), so for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = P(CO₂) alone. Kc uses molar concentrations; Kp uses gas partial pressures. They relate by Kp = Kc(RT)^Δn, where Δn is the moles of gaseous products minus moles of gaseous reactants. When Δn = 0, Kp = Kc. Getting the expression right is half the problem — a wrong expression guarantees a wrong x.
Key Points
- •Coefficients become exponents in the K expression.
- •Omit pure solids and pure liquids (their activity is 1).
- •Kp = Kc(RT)^Δn, where Δn = moles gaseous products − reactants; if Δn = 0, Kp = Kc.
3. Worked Example: A Dissociation with Kc
Consider N₂O₄(g) ⇌ 2 NO₂(g) with Kc = 4.0 × 10⁻³ at a given temperature, starting with 0.20 M N₂O₄ and no NO₂. ICE table: Initial [N₂O₄] = 0.20, [NO₂] = 0; Change −x for N₂O₄ and +2x for NO₂ (coefficient 2); Equilibrium [N₂O₄] = 0.20 − x, [NO₂] = 2x. Expression: Kc = [NO₂]²/[N₂O₄] = (2x)²/(0.20 − x) = 4.0 × 10⁻³. Because Kc is small, assume x ≪ 0.20 so 0.20 − x ≈ 0.20: 4x²/0.20 = 4.0 × 10⁻³, so 4x² = 8.0 × 10⁻⁴, x² = 2.0 × 10⁻⁴, x = 0.0141. Check the approximation: x/0.20 = 0.0141/0.20 = 7.1%, which exceeds 5%, so the approximation is borderline-invalid and you should solve the full quadratic 4x² + (4.0×10⁻³)x − 8.0×10⁻⁴ = 0 for a more accurate x ≈ 0.0136. Equilibrium: [NO₂] = 2x ≈ 0.027 M, [N₂O₄] ≈ 0.186 M.
Key Points
- •Change row uses coefficients: −x for N₂O₄, +2x for NO₂.
- •Try the 5% approximation first; here x/initial = 7.1% so the quadratic is needed.
- •Always validate the approximation before trusting the shortcut answer.
4. The 5% Approximation: When It Is Valid
When K is small, the shift x is tiny relative to the initial concentration, so (initial − x) ≈ initial, and you avoid the quadratic. The rule of thumb: the approximation is valid if x is less than 5% of the initial concentration from which it was subtracted. Compute x using the approximation, then check x/(initial) × 100%. If it is at or below 5%, accept the answer. If it exceeds 5% — as in the example above at 7.1% — the approximation introduces too much error and you must solve the quadratic with the quadratic formula. A useful shortcut to anticipate validity: if the initial concentration divided by K is greater than about 400, the approximation will almost certainly hold. This single check separates a quick answer from a wrong one.
Key Points
- •Approximation valid when x is ≤5% of the initial concentration.
- •Always compute the percentage and check before accepting.
- •If initial/K > ~400, the approximation will generally be valid.
5. Q vs K and Le Chatelier
The reaction QUOTIENT Q has the same form as K but uses CURRENT (not equilibrium) concentrations, and comparing them predicts direction: if Q < K the reaction proceeds FORWARD (toward products) to reach equilibrium; if Q > K it proceeds REVERSE (toward reactants); if Q = K the system is already at equilibrium. Le Chatelier’s principle then tells you how a disturbance shifts an existing equilibrium: adding a reactant or removing a product shifts forward; increasing pressure (decreasing volume) shifts toward the side with FEWER moles of gas; and raising temperature shifts in the endothermic direction (treat heat as a reactant for endothermic, a product for exothermic). Note that a catalyst speeds both directions equally and does NOT shift the equilibrium position — a frequently tested point.
Key Points
- •Q < K → forward; Q > K → reverse; Q = K → at equilibrium.
- •Pressure increase shifts toward fewer moles of gas; temperature shifts in the endothermic direction.
- •A catalyst does not change the equilibrium position, only the rate to reach it.
6. Solving Equilibrium Problems in ChemistryIQ
Snap a photo of an equilibrium problem and ChemistryIQ writes the correct Kc or Kp expression (omitting solids and liquids), builds the ICE table with the proper coefficient-based change row, tests the 5% approximation and solves the quadratic when needed, and uses Q to predict direction. It also walks through Le Chatelier shifts and the Kp = Kc(RT)^Δn conversion. This content is for educational purposes only.
Key Points
- •Builds the K expression and ICE table automatically with correct exponents.
- •Tests the 5% approximation and falls back to the quadratic when required.
- •Applies Q-vs-K direction and Le Chatelier reasoning.
High-Yield Facts
- ★ICE table: Initial, Change (in terms of x via coefficients), Equilibrium; substitute into K.
- ★Coefficients become exponents; omit pure solids and liquids from K.
- ★Kp = Kc(RT)^Δn; if Δn = 0 then Kp = Kc.
- ★5% approximation valid when x ≤ 5% of initial; otherwise solve the quadratic.
- ★Q < K → forward, Q > K → reverse; catalysts do not shift equilibrium position.
Practice Questions
1. For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), write the Kc expression.
2. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), what is Kp?
3. A reaction has Q = 0.5 and K = 2.0. Which direction does it proceed?
FAQs
Common questions about this topic
Because the activity of a pure solid or pure liquid is defined as 1 — its effective concentration does not change as the reaction proceeds, since adding more solid does not change its intrinsic density or molarity. Including a constant in K would be redundant, so by convention pure solids and pure liquids are omitted. Only gases (in Kp or Kc) and aqueous species (in Kc) appear in the expression. This is why a heterogeneous equilibrium like CaCO₃ decomposition has Kp = P(CO₂) alone.
By Kp = Kc(RT)^Δn, where R is the gas constant, T is temperature in Kelvin, and Δn is the change in moles of GAS (moles of gaseous products minus moles of gaseous reactants). When Δn = 0 — equal moles of gas on both sides — the (RT) term is 1 and Kp = Kc. When more gas moles are produced (Δn > 0), Kp exceeds Kc. Use only the gaseous species when counting Δn.
When the equilibrium constant is small enough that the shift x is negligible compared to the initial concentration. Compute x assuming (initial − x) ≈ initial, then check whether x is 5% or less of that initial concentration. If yes, accept the simplified answer; if x exceeds 5%, the approximation introduces too much error and you must solve the full quadratic. A quick pre-check: if initial concentration divided by K is greater than about 400, the approximation will usually hold.
No. A catalyst lowers the activation energy for both the forward and reverse reactions equally, so it speeds up how fast equilibrium is reached but does not change the equilibrium position or the value of K. Only temperature changes K. This is a commonly tested distinction: a catalyst affects rate (kinetics), not the equilibrium constant (thermodynamics).
Snap a photo and ChemistryIQ writes the correct Kc or Kp expression (omitting solids and liquids), builds the ICE table with coefficient-based change terms, tests the 5% approximation and solves the quadratic when needed, predicts direction with Q, and applies Le Chatelier and the Kp = Kc(RT)^Δn conversion. This content is for educational purposes only.