The problems are on the chemical equilibrium worksheet. Try them on your own first, then compare with the answers below.
Quick answer: At chemical equilibrium the forward and reverse reaction rates are equal, so concentrations stay constant. These worked answers correct the 20 problems of the AP Chemistry worksheet on chemical equilibrium, from writing K expressions to cobalt chloride equilibrium: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the expression, the substitution and the result in bold. In ICE problems the check step (substituting back into K) is included; do the same in your work. Small differences in the last digit come from rounding intermediate values.
Level: AP Chemistry · Curriculum: AP Chem Unit 7
Problem 1 Writing K expressions
1. Kc = [NH₃]²/([N₂][H₂]³).2. Kc = [CO₂] (solids omitted).
3. Kc = [SO₃]²/([SO₂]²[O₂]).
Problem 2 What the size of K says
1. 3.0 × 10¹² goes almost completely to products.2. 1.0 has comparable amounts.
3. 2.0 × 10⁻²⁰ hardly reacts.
Problem 3 Which way does it go?
1. Q = (0.10)²/((0.10)² × 0.10) = 0.010/0.0010 = 10.2. Q < K (10 < 25), so the forward reaction runs; more SO₃ forms.
Problem 4 Le Chatelier directions
1. Right. 2. Right. 3. Right (4 mol of gas to 2 mol). 4. Left (the reverse is endothermic). 5. No shift.Problem 5 Reading an equilibrium graph
1. About 30 s (concentrations stop changing).2. [A] = 0.40 M and [B] = 0.60 M.
3. K = [B]/[A] = 0.60/0.40 = 1.5.
4. No; both reactions continue at equal rates.
Problem 6 Changing the equation
1. 1/5.0 = 0.20.2. 5.0² = 25.
3. 5.0½ = 2.2.
Problem 7 K from measurements
1. Kc = [HI]²/([H₂][I₂]).2. Kc = (0.40)²/(0.050 × 0.050) = 0.160/0.00250 = 64.
Problem 8 Kp from Kc
1. Δn = 2 − 1 = +1.2. Kp = 0.20 × 0.08206 × 350 = 5.7.
3. 1/5.74 = 0.17.
Problem 9 A perfect-square ICE table
1. H₂: 0.50, −x, 0.50 − x; I₂: same; HI: 0, +2x, 2x.2. 2x/(0.50 − x) = 8, so 2x = 4.0 − 8x and x = 0.40.
3. [H₂] = [I₂] = 0.10 M; [HI] = 0.80 M. Check: 0.64/0.010 = 64 ✓.
Problem 10 Finding K from an ICE table
1. CO changes by −0.30: [CO] = 0.70 M. H₂ changes by −0.60: [H₂] = 1.40 M.2. Kc = 0.30/(0.70 × 1.40²) = 0.30/1.372 = 0.22 M⁻².
Problem 11 Starting from a mixture
1. Q = 0.30²/0.10 = 0.90 > 0.20, so the reverse reaction runs.2. NO₂ falls by 2y, N₂O₄ rises by y: (0.30 − 2y)²/(0.10 + y) = 0.20, so 4y² − 1.4y + 0.07 = 0, y = 0.0604. [NO₂] = 0.18 M, [N₂O₄] = 0.16 M. Check: 0.179²/0.160 = 0.20 ✓.
Problem 12 Temperature and colour
1. The tube turns darker brown.2. The forward reaction is exothermic, so heating makes K decrease.
3. After heating, K falls below the old Q (Q > K), so the reverse reaction runs and more NO₂ forms.
Problem 13 Compression and Q
1. Kc = 0.40²/(0.20² × 0.10) = 0.160/0.0040 = 40.2. New concentrations 0.40, 0.20, 0.80 M: Q = 0.80²/(0.40² × 0.20) = 0.64/0.032 = 20.
3. Q < K, so the shift is to the right, towards 2 mol of gas instead of 3.
Problem 14 A heterogeneous equilibrium
1. Kp = P(CO)²/P(CO₂) = 0.80²/0.20 = 3.2.2. No effect; the amount of a pure solid does not appear in K.
3. Q = 0.64/0.50 = 1.28 < 3.2, so the shift is to the right.
Problem 15 Adding equilibria
1. N₂ + O₂ + 2 NO + O₂ ⇌ 2 NO + 2 NO₂, and cancelling 2 NO gives N₂(g) + 2 O₂(g) ⇌ 2 NO₂(g).2. K = K1K2 = 4.0 × 10⁻³¹ × 6.0 × 10¹² = 2.4 × 10⁻¹⁸.
Problem 16 A quadratic ICE table
1. x²/(0.200 − x) = 0.0500, so x² + 0.0500x − 0.0100 = 0.2. x = [−0.0500 + √(0.00250 + 0.0400)]/2 = 0.0781.
3. [PCl₃] = [Cl₂] = 0.078 M; [PCl₅] = 0.122 M; dissociation = 0.0781/0.200 = 39 %.
Problem 17 The small-x approximation
1. (2x)²/((0.80 − x)(0.20 − x)) = 1.0 × 10⁻⁴.2. With x small: 4x² = 1.0 × 10⁻⁴ × 0.16 = 1.6 × 10⁻⁵, x = 2.0 × 10⁻³, so [NO] = 2x = 4.0 × 10⁻³ M.
3. x/0.20 = 1.0 % < 5 %: valid.
Problem 18 A stress and a new equilibrium
1. [CO] = 1.333 M; Q = 0.667²/(1.333 × 0.333) = 1.0.2. Q < K (1.0 < 4.0), so the shift is forward.
3. (0.667 + y)²/((1.333 − y)(0.333 − y)) = 4.0 gives 3y² − 8y + 1.333 = 0, y = 0.179. [CO₂] = [H₂] = 0.845 M, [CO] = 1.155 M, [H₂O] = 0.155 M.
Problem 19 Deducing ΔH from K
1. K falls as T rises, so the forward reaction is exothermic: ΔH < 0.2. The yield decreases.
3. Higher pressure favours the side with fewer gas moles: the yield increases.
4. A catalyst does not change K but lets the reaction reach equilibrium quickly at a moderate temperature.
Problem 20 Cobalt chloride equilibrium
1. Kc = [CoCl₄²⁻]/([Co(H₂O)₆²⁺][Cl⁻]⁴) (water omitted).2. HCl: blue (more Cl⁻, shift right). Water: pink (dilution shifts left). Heating: blue (endothermic direction, K rises).
3. Q = 0.0020/(0.010 × 0.50⁴) = 0.0020/(6.25 × 10⁻⁴) = 3.2 < 5.0, so the shift is forward (more blue).
