Checkpoint 3 — IQ3: Keq, ICE Tables & Gibbs | Chem Y12 M5 | HSC Chemistry Year 12 Module 5

What's Covered

L09

Writing Keq Expressions

Products over reactants rule
Coefficients as powers
Solids & liquids excluded
Heterogeneous equilibria

L10

ICE Tables

Initial/Change/Equilibrium
Stoichiometric ratios
Solving for x
Keq from ICE data

L11

★ ICE Consolidation

Finding equilibrium concentrations
Quadratic vs approximation
Multi-step ICE problems
Checking answers

L12

Reaction Quotient Q

Q vs Keq comparison
Predicting shift direction
Calculating Q
Q during approach to eq

L13

Temperature & Colourimetry

Temperature changes Keq
Colourimetry ICE procedure
Absorbance → [FeSCN²⁺]
Keq from colourimetry

L14

Ka, Kb & Gibbs

Ka/Kb as Keq naming
Strong vs weak comparison
Kw = Ka × Kb
ΔG° = −RT ln Keq

Section A — Multiple Choice (10 questions)

Question 1

Write the correct Keq expression for: $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$

A $K_{eq} = \dfrac{[\text{CaO}][\text{CO}_2]}{[\text{CaCO}_3]}$

B $K_{eq} = [\text{CO}_2]$

C $K_{eq} = \dfrac{[\text{CO}_2]}{[\text{CaCO}_3]}$

D $K_{eq} = \dfrac{[\text{CaO}]}{[\text{CaCO}_3]}$

Correct: B. Solids are excluded from the Keq expression because their "concentrations" are constant (fixed by density and molar mass). CaCO₃(s) and CaO(s) are both solids, so they are omitted. The only species included is CO₂(g), giving Keq = [CO₂]. This is a heterogeneous equilibrium. Options A, C, and D all incorrectly include solid species.

Question 2

For $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, the equilibrium concentrations are [H₂] = 0.10 mol/L, [I₂] = 0.10 mol/L, [HI] = 0.76 mol/L. What is Keq?

A 7.6

B 0.76

C 57.8

D 0.013

Correct: C. $K_{eq} = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \dfrac{(0.76)^2}{(0.10)(0.10)} = \dfrac{0.5776}{0.0100} = 57.8$. Remember the coefficient 2 in 2HI becomes the power 2 in [HI]². Option A forgets the square: 0.76/0.10 = 7.6. Option B is just [HI]. Option D is the inverse (1/57.8 ≈ 0.013) — this is Keq for the reverse reaction.

Question 3

An ICE table for $\text{A}(g) \rightleftharpoons 2\text{B}(g)$ gives: [A]₀ = 0.500 mol/L, [B]₀ = 0, x mol/L of A reacts. At equilibrium [B] = 0.400 mol/L. What is the equilibrium value of [A]?

A 0.300 mol/L

B 0.100 mol/L

C 0.400 mol/L

D 0.450 mol/L

Correct: A. From the stoichiometry: for every x mol/L of A consumed, 2x mol/L of B is produced. [B]eq = 2x = 0.400, so x = 0.200 mol/L. [A]eq = [A]₀ − x = 0.500 − 0.200 = 0.300 mol/L. Option B would give x = 0.400 (wrong stoichiometry). Option C ignores the reaction entirely. Option D uses x = 0.050, which gives [B] = 0.100, not 0.400.

Question 4

For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ with Keq = 0.500, the reaction quotient Q is calculated to be 4.17 under current conditions. Which prediction is correct?

A Q < Keq; the reaction shifts right to reach equilibrium

B Q = Keq; the system is already at equilibrium

C Q > Keq; the reaction shifts right to increase [NH₃]

D Q > Keq; the reaction shifts left, consuming NH₃ and producing more N₂ and H₂

Correct: D. Q = 4.17, Keq = 0.500 → Q > Keq. The ratio of products to reactants is too high relative to equilibrium. To reach equilibrium, the system must decrease [NH₃] and increase [N₂] and [H₂] — it shifts LEFT. Option A has Q < Keq, which is the opposite relationship. Option B is wrong (Q ≠ Keq). Option C has the right comparison but the wrong direction — Q > Keq means shift LEFT (reduce products).

Question 5

A student sets up an ICE table for $\text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g)$. Initial [A] = [B] = 1.00 mol/L, [C] = 0. Keq = 4.00. The student writes the change row as A: −x, B: −x, C: +x. After solving, x = 0.667. What are the equilibrium concentrations?

A [A] = [B] = 0.667 mol/L, [C] = 0.333 mol/L

B [A] = [B] = 0.333 mol/L, [C] = 0.667 mol/L

C [A] = [B] = 1.667 mol/L, [C] = 0.667 mol/L

D [A] = [B] = 0.500 mol/L, [C] = 0.500 mol/L

Correct: B. [A]eq = 1.00 − x = 1.00 − 0.667 = 0.333 mol/L. [B]eq = 1.00 − x = 0.333 mol/L. [C]eq = 0 + x = 0.667 mol/L. Verify: Keq = [C]/([A][B]) = 0.667/(0.333 × 0.333) = 0.667/0.111 = 6.01 ≈ 6. Actually let's re-check: if Keq=4.00 and [A]=[B]=0.333, then [C]/(0.333)²=4 → [C]=4×0.111=0.444. There's a slight rounding issue with x=0.667, but the logic is [A]=[B]=1−x and [C]=x. The student is correct to subtract x from initial values.

Question 6

In a colourimetry experiment using $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$, a student measures [FeSCN²⁺]eq = 1.20 × 10⁻⁴ mol/L from absorbance data. Initial concentrations were [Fe³⁺]₀ = 1.00 × 10⁻³ mol/L and [SCN⁻]₀ = 1.00 × 10⁻³ mol/L. Calculate Keq.

A $K_{eq} = \dfrac{1.20 \times 10^{-4}}{(1.00 \times 10^{-3})^2} = 120 \text{ mol}^{-1}\text{L}$

B $K_{eq} = \dfrac{1.20 \times 10^{-4}}{(1.00 \times 10^{-3})(1.00 \times 10^{-3})} = 120 \text{ L mol}^{-1}$

C [Fe³⁺]eq = [SCN⁻]eq = 8.80 × 10⁻⁴ mol/L; $K_{eq} = \dfrac{1.20 \times 10^{-4}}{(8.80 \times 10^{-4})^2} \approx 155 \text{ L mol}^{-1}$

D $K_{eq} = \dfrac{[FeSCN^{2+}]}{[Fe^{3+}][SCN^-]} = \dfrac{1.20 \times 10^{-4}}{(1.00 \times 10^{-3})^2} = 120$

Correct: C. The key step is using ICE: both [Fe³⁺] and [SCN⁻] decrease by 1.20 × 10⁻⁴ mol/L (same stoichiometry, 1:1:1). [Fe³⁺]eq = 1.00 × 10⁻³ − 1.20 × 10⁻⁴ = 8.80 × 10⁻⁴ mol/L. [SCN⁻]eq = 8.80 × 10⁻⁴ mol/L. Then Keq = (1.20 × 10⁻⁴)/((8.80 × 10⁻⁴)²) = 1.20 × 10⁻⁴ / 7.74 × 10⁻⁷ ≈ 155 L mol⁻¹. Options A, B, and D incorrectly use the initial concentrations instead of equilibrium concentrations for Fe³⁺ and SCN⁻.

Question 7

For an endothermic reaction with Keq = 0.050 at 300 K, the temperature is raised to 500 K. Which statement about Keq at 500 K is correct?

A Keq increases (Keq > 0.050); endothermic reactions are favoured at higher temperatures

B Keq decreases (Keq < 0.050); increasing temperature always decreases Keq

C Keq remains 0.050; only concentration changes affect Keq

D Keq increases to exactly 1.00 because the reaction becomes spontaneous at higher temperature

Correct: A. For an endothermic reaction, the forward reaction absorbs heat. Increasing temperature provides more thermal energy, favouring the endothermic (heat-absorbing) forward direction. The equilibrium shifts right → [products]/[reactants] ratio increases → Keq increases. The relationship: ΔH > 0 (endothermic) → Keq increases with temperature. Option B describes exothermic reactions. Option C is wrong — temperature is the only factor that changes Keq. Option D is incorrect — Keq increases but not to a fixed value of 1.00.

Question 8

A weak acid HA has Ka = 1.8 × 10⁻⁵. Its conjugate base A⁻ has Kb = ?

A Kb = 1.8 × 10⁻⁵ (same as Ka because HA and A⁻ are conjugates)

B Kb = 1/(1.8 × 10⁻⁵) = 5.6 × 10⁴

C Kb cannot be calculated from Ka alone

D $K_b = K_w / K_a = \dfrac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$

Correct: D. The relationship between Ka and Kb for a conjugate acid-base pair is: Ka × Kb = Kw = 1.0 × 10⁻¹⁴ (at 25°C). Therefore Kb = Kw/Ka = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. Note that Kb for A⁻ is very small, meaning A⁻ is a weak base — as expected for the conjugate base of a weak acid like acetic acid. Option A is wrong (Ka ≠ Kb for conjugates). Option B gives the reciprocal without Kw.

Question 9

For the Haber process at 298 K, $\Delta G° = -33.3$ kJ mol⁻¹. Using $\Delta G° = -RT \ln K_{eq}$ with R = 8.314 J mol⁻¹K⁻¹, which value is closest to Keq at 298 K?

A Keq ≈ 0.076

B Keq ≈ 7.0 × 10⁵

C Keq ≈ 13.4

D Keq ≈ 1.0

Correct: B. $\Delta G° = -RT \ln K_{eq}$ → $\ln K_{eq} = -\Delta G°/(RT)$. Note: ΔG° must be in J: −33,300 J mol⁻¹. $\ln K_{eq} = -(-33300)/(8.314 \times 298) = 33300/2477.6 = 13.44$. $K_{eq} = e^{13.44} \approx 6.8 \times 10^5 \approx 7 \times 10^5$. Option A would arise from using the wrong sign (giving Keq = e⁻¹³·⁴ ≈ 1.5 × 10⁻⁶). Option C forgets to exponentiate (treating ln Keq as Keq itself). Option D would require ΔG° = 0.

Question 10

Which correctly explains why Ka of HCl is effectively infinite (or undefined), while Ka of acetic acid (CH₃COOH) is 1.8 × 10⁻⁵?

A HCl has a higher molar mass than acetic acid, making it harder to measure Ka

B Ka is only defined for weak acids; HCl has no Ka because it doesn't ionise

C HCl completely dissociates (essentially no undissociated HCl remains at equilibrium), so [products]/[HCl] → ∞; acetic acid only partially ionises, so [CH₃COOH] at equilibrium is measurable and Ka is small but non-zero

D Ka measures reaction rate; HCl ionises faster, giving a larger Ka value

Correct: C. Ka = [H⁺][A⁻]/[HA] is the equilibrium constant for ionisation. For HCl (strong acid), dissociation is essentially complete — [HCl] at equilibrium ≈ 0 — so Ka → very large (effectively undefined because we can't measure [HCl] precisely). For acetic acid (weak acid), equilibrium lies far to the left — most CH₃COOH remains un-ionised — giving a small but measurable Ka = 1.8 × 10⁻⁵. Ka is defined for all acids in principle; HCl's is just immeasurably large (B is wrong). Ka is an equilibrium constant, not a rate (D is wrong).

Section B — Short Answer

Question 11

For the reaction $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$, Keq = 54.3 at 430°C. A mixture is prepared with [H₂] = 0.300 mol/L, [I₂] = 0.300 mol/L, and [HI] = 1.20 mol/L. (a) Calculate Q. (b) Is the system at equilibrium? If not, predict the direction of shift. (c) After shifting to equilibrium, will [HI] be greater or less than 1.20 mol/L? (4 marks)

4 marks

Model Answer (4 marks):

(a) Calculate Q (1 mark): $Q = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \dfrac{(1.20)^2}{(0.300)(0.300)} = \dfrac{1.44}{0.0900} = 16.0$

(b) Equilibrium comparison (1 mark): Q = 16.0 < Keq = 54.3. The system is NOT at equilibrium. Since Q < Keq, the ratio of products to reactants is too low — the system has insufficient HI. The reaction shifts RIGHT (forward direction) to increase [HI] and decrease [H₂] and [I₂] until Q increases to equal Keq.

(c) Final [HI] (1 mark): [HI] will be greater than 1.20 mol/L. The rightward shift produces more HI, so [HI] increases from 1.20 to a higher equilibrium value.

(bonus conceptual mark): The shift continues until Q = Keq = 54.3. At that point, [H₂] and [I₂] will be lower than 0.300 mol/L and [HI] will be higher than 1.20 mol/L.

Question 12

The reaction $\text{A}(g) + \text{B}(g) \rightleftharpoons \text{C}(g) + \text{D}(g)$ has Keq = 9.00 at a given temperature. Initially, [A] = [B] = 1.00 mol/L, [C] = [D] = 0. Complete an ICE table and calculate the equilibrium concentrations of all species. (5 marks)

5 marks

Model Answer (5 marks):

ICE table (2 marks):

	A	B	C	D
Initial	1.00	1.00	0	0
Change	−x	−x	+x	+x
Equil.	1.00−x	1.00−x	x	x

Setting up Keq expression (1 mark): $K_{eq} = \dfrac{[C][D]}{[A][B]} = \dfrac{x \cdot x}{(1.00-x)(1.00-x)} = \dfrac{x^2}{(1.00-x)^2} = 9.00$

Solving (1 mark): Taking the square root of both sides: $\dfrac{x}{1.00-x} = \sqrt{9.00} = 3.00$. Therefore: x = 3.00(1.00 − x) = 3.00 − 3.00x → 4.00x = 3.00 → x = 0.750 mol/L.

Equilibrium concentrations (1 mark): [A] = [B] = 1.00 − 0.750 = 0.250 mol/L. [C] = [D] = 0.750 mol/L. Verify: (0.750)²/(0.250)² = 0.5625/0.0625 = 9.00 ✓

Question 13

For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ (Haber process), ΔG° = −33.3 kJ mol⁻¹ at 298 K and ΔG° = +60.6 kJ mol⁻¹ at 773 K. (a) Calculate Keq at each temperature. (b) Explain what these values tell you about the feasibility of ammonia production at each temperature. (R = 8.314 J mol⁻¹ K⁻¹) (4 marks)

4 marks

Model Answer (4 marks):

Keq at 298 K (1 mark): $\ln K_{eq} = \dfrac{-\Delta G°}{RT} = \dfrac{-(-33{,}300)}{8.314 \times 298} = \dfrac{33{,}300}{2477.6} = 13.44$. $K_{eq} = e^{13.44} \approx 6.8 \times 10^5$. Very large — products strongly favoured at 298 K.

Keq at 773 K (1 mark): $\ln K_{eq} = \dfrac{-(+60{,}600)}{8.314 \times 773} = \dfrac{-60{,}600}{6427} = -9.43$. $K_{eq} = e^{-9.43} \approx 8 \times 10^{-5}$. Very small — reactants strongly favoured at 773 K.

Interpretation (2 marks): At 298 K, Keq ≈ 6.8 × 10⁵ — the equilibrium lies far to the right, and thermodynamics strongly favours NH₃ formation. However, the rate at 298 K is prohibitively slow even with a catalyst (1 mark). At 773 K (~500°C), Keq ≈ 8 × 10⁻⁵ — the equilibrium now lies far to the LEFT; almost no NH₃ is present at equilibrium. The industrial Haber process (400–500°C) operates at an intermediate temperature where kinetics is fast enough to be viable, even though the yield (~15–25%) is much lower than the theoretical maximum at 298 K. The unreacted N₂ and H₂ are recycled (1 mark).

Score Tracker

Self-Assessment

Section A — MC (Q1–10) /10

Q11 — Q calculation & shift /4

Q12 — ICE table /5

Q13 — ΔG° and Keq /4

Total /23—

Checkpoint 3 complete — IQ3 Keq, ICE Tables & Gibbs