CHEM1011 · Fundamentals Of Chemistry 1a
Chemical Equilibrium
This chapter covers chemical equilibrium: writing the equilibrium constant K and the reaction quotient Q, predicting which way a reaction will shift, solving for unknown concentrations with an ICE table, and applying Le Châtelier's principle. It is one of the most reliable sources of marks in CHEM1011 because almost every question reduces to three moves — write K, compare Q to K, or set up an ICE table — and the datasheet gives you constants but never the method.
What this chapter covers
- 011. The equilibrium-constant expression: K = [products]/[reactants], each raised to its coefficient
- 022. The phase rule: pure solids and pure liquids are omitted (activity = 1)
- 033. Reaction quotient Q and the direction test (Q < K forward, Q > K reverse, Q = K at equilibrium)
- 044. Magnitude of K: large K favours products, small K favours reactants
- 055. K manipulation: reverse to 1/K, scale to Kⁿ, add reactions to multiply K's
- 066. The ICE method: Initial-Change-Equilibrium in terms of one unknown x
- 077. Solving: the quadratic formula, small-K and excess approximations, and the compulsory 5% check
- 088. Le Châtelier's principle: shifts under concentration, pressure/volume, and temperature
ICE table for H₂ + I₂ ⇌ 2HI (Kc = 25)
- +1Build the ICE table. Initial: [H₂] = [I₂] = 0.10, [HI] = 0. Change: reactants −x each, product +2x. Equilibrium: [H₂] = [I₂] = 0.10 − x, [HI] = 2x.
- +1Substitute the equilibrium row into K: K = [HI]²/([H₂][I₂]) = (2x)²/(0.10 − x)² = 25.
- +1Both sides are perfect squares, so take the square root: 2x/(0.10 − x) = √25 = 5. (Do NOT write = 25.)
- +1Rearrange and solve: 2x = 5(0.10 − x) = 0.50 − 5x → 7x = 0.50 → x = 0.0714 (the positive, physical root). Back-substitute: [HI] = 2x = 0.143 M; [H₂] = [I₂] = 0.0286 M.
Key terms
- Dynamic equilibrium
- The state of a reversible reaction where the forward and reverse rates are equal, so concentrations stay constant. The concentrations are fixed, not necessarily equal.
- Equilibrium constant (K)
- For aA + bB ⇌ cC + dD, K = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ using equilibrium concentrations. A pure number that depends only on temperature; a large K favours products, a small K favours reactants.
- Reaction quotient (Q)
- The same algebraic expression as K but evaluated with the current (not necessarily equilibrium) concentrations. Comparing Q to K tells you which way the reaction must shift: Q < K → forward, Q > K → reverse, Q = K → at equilibrium.
- ICE table
- A bookkeeping table of Initial, Change, and Equilibrium concentrations. Reactants change by −x and products by +x, each multiplied by its coefficient, expressing every equilibrium concentration in terms of one unknown x.
- Le Châtelier's principle
- A system at equilibrium responds to an imposed change in concentration, pressure/volume, or temperature by shifting in the direction that partly opposes that change.
- Small-K approximation
- When K is very small (≲10⁻³), only a little reactant reacts, so x ≪ c₀ and (c₀ − x) ≈ c₀, simplifying the algebra. Must be validated with the 5% check: x/c₀ × 100% < 5%; otherwise solve exactly.
Chemical Equilibrium FAQ
Why do pure solids and pure liquids get left out of K and Q?
Their activity is taken as 1, so they make no contribution to the ratio. For CaCO₃(s) ⇌ CaO(s) + CO₂(g) the expression is simply K = [CO₂] — both solids drop out. Water acting as a pure solvent is omitted too; dissolved [aq] species and gases [g] always stay in.
How do I tell which way a reaction will shift?
Compute Q with the current concentrations using the same expression as K, then compare. If Q < K there is too little product, so it shifts forward; if Q > K there is too much product, so it shifts reverse; if Q = K it is already at equilibrium. The system always moves to drive Q toward K — a large Q means going backward, not forward.
When can I drop the x and use the small-K approximation?
When K is very small (about 10⁻³ or less) or a reactant is in large excess, its change is negligible, so you replace (c₀ − x) with c₀ and solve the simpler equation. Then you MUST run the 5% check — if x/c₀ × 100% < 5% the shortcut is valid; if it is 5% or more, go back and solve the full quadratic. Markers award the check itself.
Does Le Châtelier change the value of K?
Only temperature changes K. Adding or removing a species or changing pressure shifts the position of equilibrium but K is unchanged — the system simply moves until Q = K again. A catalyst reaches equilibrium faster without shifting it, and an inert gas added at constant volume changes nothing.
What's on the exam for chemical equilibrium?
Almost every item is one of three: (1) write K, correctly omitting pure solids and liquids; (2) compare Q to K to state the direction of shift; (3) set up an ICE table and solve for an unknown concentration or for K. Le Châtelier predictions (concentration, pressure by counting gas moles, temperature by the sign of ΔH) and K-manipulation problems are the other staples.
Exam move
Drill the three core moves until they are automatic, because the exam recycles them: write K (products over reactants, each to its coefficient, pure solids/liquids omitted); compute Q and call the direction by driving Q toward K; and build an ICE table, substitute into K, then solve. For solving, recognise the shortcut first — a perfect-square K means square-root both sides, and a tiny K means drop x and run the 5% check (state the check, it earns a mark). For Le Châtelier, count gas moles on each side for pressure questions and put "heat" on the product side if exothermic / reactant side if endothermic for temperature questions, then treat the change like adding or removing a reagent. Remember the trap: only temperature changes K. The datasheet hands you constants, not the method, so the method is what you memorise.
