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)

Extension questions are marked. Questions carrying a “Deep dive, Extension” tag go beyond the core route for this module. Attempt them for enrichment, and score them separately, they should not count towards your core mastery for this checkpoint.

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. Solving 4.00 = x/(1.00 − x)² gives x = 0.610. What are the equilibrium concentrations?

A [A] = [B] = 0.610 mol/L, [C] = 0.390 mol/L

B [A] = [B] = 0.390 mol/L, [C] = 0.610 mol/L

C [A] = [B] = 1.610 mol/L, [C] = 0.610 mol/L

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

Correct: B. [A]eq = 1.00 − x = 1.00 − 0.610 = 0.390 mol/L. [B]eq = 1.00 − x = 0.390 mol/L. [C]eq = 0 + x = 0.610 mol/L. Verify: Keq = [C]/([A][B]) = 0.610/(0.390 × 0.390) = 0.610/0.152 = 4.01 ≈ 4.00 ✓ (rounds to the stated Keq). Option A swaps the reactant and product values; Option C adds x to the reactants instead of subtracting; Option D ignores the stoichiometry.

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 $K_{eq} = \dfrac{1.20 \times 10^{-4}}{(1.00 \times 10^{-3})^2} = 120$ (initial concentrations used unchanged)

B $K_{eq} = \dfrac{1.20 \times 10^{-4}}{(8.80 \times 10^{-4})(1.00 \times 10^{-3})} \approx 136$ (change subtracted from Fe³⁺ only)

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

D $K_{eq} = \dfrac{[Fe^{3+}][SCN^-]}{[FeSCN^{2+}]} = \dfrac{(8.80 \times 10^{-4})^2}{1.20 \times 10^{-4}} \approx 6.5 \times 10^{-3}$ (expression inverted)

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. A uses the initial concentrations unchanged, skipping ICE entirely. B subtracts the change from Fe³⁺ but leaves SCN⁻ at its initial value. D inverts the expression, putting reactants over products. Keq is quoted as a dimensionless number throughout this course, see Lesson 10.

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 9Deep dive, Extension

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 the Ka of HCl in dilute aqueous solution is not quoted as an ordinary finite value, 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 is essentially completely ionised in dilute aqueous solution, so almost no undissociated HCl remains and the ratio [products]/[HCl] is too large to determine meaningfully in water; 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), ionisation in dilute aqueous solution is essentially complete, so [HCl] at equilibrium is vanishingly small and Ka is very large, too large to determine reliably in water. Treat HCl as completely ionised rather than assigning it an exact or infinite aqueous Ka: water levels all very strong acids to H₃O⁺, so it cannot distinguish between them. 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