Temperature & Keq, Colourimetry | HSC Chemistry Year 12 Module 5

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Think First — Before You Read

In 1884, Jacobus van 't Hoff at the University of Amsterdam published the van 't Hoff equation, showing that d(ln Keq)/dT = ΔH°/(RT²) — meaning Keq changes with temperature in a direction determined by the sign of ΔH. A student reproduces his result for Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) (exothermic, ΔH < 0) and measures Keq at three temperatures: 25°C → Keq = 895; 35°C → Keq = 580; 45°C → Keq = 360.

Before reading on: (1) Do these values make sense given the sign of ΔH? (2) If the student measured at 15°C, would Keq be greater or less than 895? (3) What trend do you notice in how Keq changes per 10°C? Write your predictions.

Know

Temperature is the only factor that changes the value of Keq
Endothermic reactions have Keq increase with temperature; exothermic reactions have Keq decrease
The Beer-Lambert law and calibration curve method for colourimetry

Understand

Why temperature affects Keq through collision theory and energy distribution
How colourimetry experimentally determines equilibrium concentrations
The qualitative connection between Gibbs free energy and Keq

Can Do

Predict and justify the direction of Keq change with temperature
Calculate Keq from colourimetry data using a calibration curve
Use the van't Hoff equation qualitatively to compare Keq values at different temperatures

Module 5 — Key Formulas: Lesson 13

Temperature effect on Keq: only temperature changes Keq

Exothermic forward (ΔH < 0): increase T → Keq decreases; decrease T → Keq increases

Endothermic forward (ΔH > 0): increase T → Keq increases; decrease T → Keq decreases

Beer-Lambert law: A = εlc — A ∝ c for fixed ε and path length l

Calibration curve: absorbance → [FeSCN²⁺]_eq → ICE table → Keq

Gibbs–Keq: ΔG° = −RT ln Keq (qualitative + introductory quantitative in L14)

Large Keq → large negative ΔG° → forward reaction strongly spontaneous

Small Keq → large positive ΔG° → reverse reaction spontaneous

Keq = 1 → ΔG° = 0 → neither direction favoured under standard conditions

Key Terms — scan these before reading

Temperature effect on Keq

Unlike LCP shifts, changing temperature changes the value of Keq itself.

Exothermic reaction and Keq

Increasing temperature decreases Keq (less product favoured at higher T) for exothermic reactions.

Endothermic reaction and Keq

Increasing temperature increases Keq (more product favoured at higher T) for endothermic reactions.

van 't Hoff relationship

Links change in Keq with temperature change; higher T gives larger Keq for endothermic reactions.

Colourimetry

A technique using absorbance of light to determine concentration of a coloured species in solution (Beer–Lambert law).

Absorbance

A = εlc; proportional to concentration; used to track changes in coloured equilibrium systems like NO₂/N₂O₄.

Misconceptions to Fix

✗ Wrong: A large Keq means the reaction happens quickly.

✓ Right: Keq indicates the equilibrium position — a large Keq means products are favoured at equilibrium. It says nothing about reaction rate. Rate depends on activation energy, temperature, and catalysts, not thermodynamic favourability. A reaction can have a huge Keq but be extremely slow.

Content

Temperature Is the Only Factor That Changes Keq

Every lesson in IQ2 and IQ3 has repeated that only temperature changes Keq — this card explains exactly why, at the thermodynamic level, and why no other factor can do the same.

Keq is related to the standard Gibbs free energy change: $\Delta G^\circ = -RT \ln K_{eq}$, which rearranges to $K_{eq} = e^{-\Delta G^\circ / RT}$.

Because $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, and this expression contains the term $T\Delta S^\circ$, ΔG° itself depends on temperature. As temperature changes, ΔG° changes — and therefore Keq changes.

No other factor — concentration, pressure, or catalyst — changes the thermodynamic energy landscape of the reaction (the stability of reactants vs products). They only change the system's current state (how far it currently is from equilibrium), not the thermodynamic destination.

Temperature increases

Keq decreases → shifts left (exo fwd)

Keq increases → shifts right (endo fwd)

Temperature decreases

Keq increases → shifts right (exo fwd)

Keq decreases → shifts left (endo fwd)

HSC Extended Response Wording

"Keq is determined by the thermodynamic stability of products relative to reactants (ΔG° = −RT ln Keq). Only temperature changes the thermodynamic energy landscape of the reaction — concentration, pressure, and catalyst changes affect the system's position relative to equilibrium but do not change the stability of reactants or products."

Common Error

"Increasing pressure makes more products form, so Keq must increase." Wrong — Keq is the ratio at equilibrium. Yes, more products form when pressure is increased (for reactions with fewer gas moles on the product side), but the denominator also changes. The ratio at the new equilibrium is the same Keq — satisfied with different absolute concentrations. The ratio is invariant at constant temperature.

Temperature × ΔH matrix — Keq change and shift direction for all four combinations

Exam Tip

When explaining equilibrium shifts, always state the direction (left or right) and justify using Le Chatelier's Principle language — simply stating the direction alone will not earn full marks.

Only temperature changes Keq (via ΔG° = −RT ln Keq). For exothermic forward reactions: increase T → Keq decreases (shift left); decrease T → Keq increases (shift right). For endothermic forward reactions: increase T → Keq increases (shift right); decrease T → Keq decreases (shift left). Concentration, pressure, and catalyst never change Keq.

Copy the 2×2 temperature × ΔH → Keq matrix into your notes before the check below.

The Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is exothermic. Which correctly describes the effect of increasing temperature on Keq?

Interactive — Temperature-Keq Colour Predictor

Temperature and Keq: Calculating and Interpreting Values

We just saw that only temperature changes Keq — the 2×2 matrix shows the direction of Keq change for exothermic and endothermic forward reactions under heating or cooling. That raises a question: when a data table gives Keq values at multiple temperatures, how do you read that data to determine ΔH sign and make predictions? This card answers it → by applying the temperature-Keq pattern to iron thiocyanate data and identifying the three required HSC response points.

Temperature-dependent Keq data is one of the most frequently tested quantitative aspects of IQ3 — the calculation itself is simple, but interpreting what the changing Keq values tell you about ΔH requires careful reasoning.

From a table of Keq at different temperatures, you can determine:

Keq decreases as T increases → forward reaction is exothermic
Keq increases as T increases → forward reaction is endothermic

Think First data — iron thiocyanate (exothermic forward, ΔH < 0):

Keq

895 (at 25°C)

580 (at 35°C)

360 (at 45°C)

> 895 (prediction at 15°C)

Interpretation

Higher Keq → more FeSCN²⁺ at equilibrium

↓ decreasing as T increases → exothermic forward ✓

↓ confirmed exothermic trend

Lower T → higher Keq for exothermic forward

Must Know (3 Required Points)

In HSC questions providing a Keq-vs-temperature table, always explicitly state: (1) the direction of Keq change with temperature; (2) what this implies about the exo/endothermic nature of the forward reaction; (3) predictions for Keq outside the given temperature range. All three points are commonly assessed.

Beyond the Syllabus

The Van't Hoff equation formalises the temperature–Keq relationship: $\ln(K_2/K_1) = -\Delta H^\circ/R \times (1/T_2 - 1/T_1)$. This is beyond HSC scope but the qualitative principle — larger |ΔH°| means Keq is more sensitive to temperature — is useful context for understanding why some equilibria shift dramatically with temperature while others barely change.

Reading Keq-vs-temperature tables: if Keq decreases as T increases → forward reaction exothermic; if Keq increases as T increases → forward reaction endothermic. Three required HSC points: (1) state the direction of Keq change with T; (2) infer the sign of ΔH; (3) predict Keq outside the given temperature range using the same trend.

Write the three-point HSC response structure for temperature-Keq data tables into your notes.

A reaction has Keq = 0.25 at 300°C, Keq = 1.10 at 400°C, and Keq = 4.80 at 500°C. What can you conclude?

Colourimetry: Measuring Keq Experimentally

We just saw how to read Keq-vs-temperature data tables to determine ΔH sign and make predictions. That raises a question: how do chemists actually measure Keq in the laboratory — what experimental technique converts a colour observation into a usable concentration for calculation? This card answers it → by explaining Beer-Lambert law and the five-step colourimetry procedure for the iron thiocyanate equilibrium.

Colourimetry converts the invisible (an equilibrium constant) into the visible (the intensity of a colour) — and by connecting absorbance to concentration, it turns a qualitative observation into a quantitative measurement.

The iron(III) thiocyanate equilibrium is ideal because FeSCN²⁺ is intensely red while Fe³⁺ and SCN⁻ are essentially colourless. The Beer-Lambert law states: for a fixed path length and molar absorptivity, absorbance (A) is proportional to the concentration of the absorbing species.

$$A = \varepsilon l c$$

where ε = molar absorptivity, l = path length (cm), c = concentration (mol/L).

Using a calibration curve (absorbance vs known [FeSCN²⁺] from standards), the equilibrium [FeSCN²⁺] in any sample can be read directly from the measured absorbance.

Prepare calibration standards — known [FeSCN²⁺] measured with colorimeter → plot A vs [FeSCN²⁺] → straight line through origin

Prepare equilibrium mixture from known initial [Fe³⁺] and [SCN⁻]

Measure absorbance of equilibrium mixture → use calibration curve to find [FeSCN²⁺]_eq

Set up ICE table with known initial [Fe³⁺] and [SCN⁻] and measured [FeSCN²⁺]_eq → calculate [Fe³⁺]_eq and [SCN⁻]_eq

Substitute all equilibrium concentrations into Keq expression → calculate Keq

Must Know (NESA Investigation)

The colourimetry Keq experiment is a NESA-specified investigation. Describe the method using the 5 steps above and explain Beer-Lambert: "Absorbance is proportional to [FeSCN²⁺] — measuring absorbance allows [FeSCN²⁺]_eq to be determined from the calibration curve, which is then used in the ICE table to find all equilibrium concentrations and calculate Keq."

Common Error

Describing colourimetry as "measuring the colour to find Keq" without explaining the quantitative connection. The key step is the calibration curve — it converts the absorbance reading (directly measurable) into a concentration (needed for the Keq calculation). Without the calibration step, the measurement gives no numerical Keq.

Beer-Lambert law: A = εlc (absorbance proportional to concentration for fixed path length and molar absorptivity). Five-step colourimetry Keq procedure: (1) prepare calibration standards → plot A vs [FeSCN²⁺]; (2) prepare equilibrium mixture; (3) measure absorbance → read [FeSCN²⁺]eq from calibration curve; (4) ICE table to find all equilibrium concentrations; (5) substitute into Keq expression.

Write Beer-Lambert and the five-step colourimetry procedure into your notes before the check below.

In a colourimetry experiment for Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺, why is a calibration curve necessary?

Calculating Keq from Colourimetry Data

We just saw the five-step colourimetry procedure where the calibration curve converts absorbance to [FeSCN²⁺]eq. That raises a question: once you have that equilibrium concentration from the instrument, how exactly do you build the ICE table and calculate Keq from actual experimental numbers? This card answers it → by working through a complete numerical example with initial concentrations 1.00 × 10⁻³ mol/L and measured [FeSCN²⁺]eq = 1.50 × 10⁻⁴ mol/L.

The colourimetry calculation is an ICE table problem where one equilibrium concentration (the coloured product) is given directly by the absorbance measurement — exactly the Type 2 ICE problem from L10.

Experimental setup: Initial [Fe³⁺] = 1.00 × 10⁻³ mol/L, initial [SCN⁻] = 1.00 × 10⁻³ mol/L. Absorbance measured → calibration curve gives: [FeSCN²⁺]_eq = 1.50 × 10⁻⁴ mol/L.

ICE Table: $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$

Initial (mol/L)

Fe³⁺: 1.00 × 10⁻³

SCN⁻: 1.00 × 10⁻³

FeSCN²⁺: 0

Change (mol/L)

Fe³⁺: −1.50 × 10⁻⁴

SCN⁻: −1.50 × 10⁻⁴

FeSCN²⁺: +1.50 × 10⁻⁴

Equilibrium (mol/L)

Fe³⁺: 8.50 × 10⁻⁴

SCN⁻: 8.50 × 10⁻⁴

FeSCN²⁺: 1.50 × 10⁻⁴

Keq calculation:

$$K_{eq} = \frac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^-]} = \frac{1.50 \times 10^{-4}}{(8.50 \times 10^{-4})(8.50 \times 10^{-4})}$$

Denominator: $(8.50 \times 10^{-4})^2 = 7.225 \times 10^{-7}$

$$K_{eq} = \frac{1.50 \times 10^{-4}}{7.225 \times 10^{-7}} = 207.6 \approx 208 \checkmark$$

Must Know (Stoichiometry)

In the Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺ ICE table, the stoichiometric ratio is 1:1:1 — one mole of Fe³⁺ reacts with one mole of SCN⁻ to produce one mole of FeSCN²⁺. The Change row entries for Fe³⁺ and SCN⁻ are both equal (in magnitude) to the change in FeSCN²⁺. Never assume all ICE tables have equal changes — always check stoichiometric coefficients first.

Common Error

Calculating equilibrium [Fe³⁺] correctly but forgetting to apply the same change to [SCN⁻]. In this symmetric case (equal initial concentrations, 1:1:1 stoichiometry), [Fe³⁺]_eq = [SCN⁻]_eq. In asymmetric cases, the changes are equal in magnitude but the equilibrium concentrations will differ.

Colourimetry ICE workflow: absorbance → calibration curve → [FeSCN²⁺]eq; fill Change row using stoichiometric ratios (1:1:1 for Fe³⁺+SCN⁻⇌FeSCN²⁺); equilibrium [Fe³⁺] = [SCN⁻] = initial − [FeSCN²⁺]eq; then Keq = [FeSCN²⁺]eq / ([Fe³⁺]eq[SCN⁻]eq). Always check stoichiometric coefficients before writing Change row.

Add the colourimetry ICE workflow steps to your notes before the check below.

In the colourimetry ICE table for Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺, if [FeSCN²⁺]eq = 2.00 × 10⁻⁴ mol/L and initial [Fe³⁺] = [SCN⁻] = 1.00 × 10⁻³ mol/L, what are [Fe³⁺]eq and [SCN⁻]eq?

Cross-lesson links: Van 't Hoff's 1884 temperature-Keq relationship introduced here quantifies the qualitative LCP temperature shift you learned in L05. The colourimetry ICE table in Card 4 is the standard NESA practical investigation connecting L10 (ICE tables) to experimental data. The ΔG° = −RT ln Keq equation in Card 5 links back to Module 4 Gibbs free energy and forward to Ka/Kb calculations in L14.

Gibbs Free Energy and Keq: The Qualitative Connection

We just saw how to calculate Keq from colourimetry data using the ICE table and equilibrium concentrations. That raises a question: what does the numerical value of Keq actually tell you about the thermodynamics — is there a deep connection between Keq and the free energy of the reaction? This card answers it → by introducing ΔG° = −RT ln Keq and explaining what different Keq magnitudes mean about product vs reactant stability.

A reaction with ΔG° = −200 kJ/mol strongly favours products at equilibrium. A reaction with ΔG° = −1 kJ/mol barely favours products. A reaction with ΔG° = 0 has exactly equal concentrations of reactants and products at equilibrium. These three observations — all explained by van 't Hoff's 1884 equation ΔG° = −RT ln Keq — connect the thermodynamics you studied in Module 4 to the equilibrium position you calculate in this module.

The relationship $\Delta G^\circ = -RT \ln K_{eq}$ encodes the following qualitative relationships:

Keq value	ln Keq	ΔG°	Interpretation
Large (>> 1)	Large positive	Large negative	Products strongly favoured; forward reaction highly spontaneous
= 1	= 0	= 0	Neither direction favoured under standard conditions
Small (<< 1)	Large negative	Large positive	Reactants strongly favoured; reverse reaction spontaneous

Examples:

$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(g)$, Keq ≈ 10⁴⁰ → ΔG° = very large negative → water formation massively spontaneous.
$\text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}(g)$, Keq = 10⁻³⁰ → ΔG° = very large positive → NO formation massively non-spontaneous at 25°C.

HSC Scope

You need the qualitative relationship only for most questions: large Keq → negative ΔG° → forward reaction spontaneous; small Keq → positive ΔG° → forward reaction non-spontaneous; Keq = 1 → ΔG° = 0. The full quantitative calculation ΔG° = −RT ln Keq is introduced in L14.

Insight — Resolves a Module 4 Confusion

In M4 you learned ΔG < 0 means a reaction is spontaneous — but many spontaneous reactions don't go to completion. Resolution: ΔG° < 0 means the forward direction is spontaneous under standard conditions, but equilibrium is reached before completion if Keq is not extremely large. The system reaches minimum free energy at the equilibrium composition, not at complete conversion. ΔG = 0 at equilibrium (not ΔG°).

ΔG° = −RT ln Keq links thermodynamics to equilibrium: Keq >> 1 → large negative ΔG° → products strongly favoured (forward reaction spontaneous); Keq = 1 → ΔG° = 0 → neither direction favoured; Keq << 1 → large positive ΔG° → reactants strongly favoured. Note: ΔG = 0 at equilibrium for any system; ΔG° = 0 only when Keq = 1.

Pause — record the three Keq-magnitude → ΔG° → spontaneity entries in your notes before the check below.

For the reaction 2NO₂(g) ⇌ 2NO(g) + O₂(g), ΔG° = +70 kJ/mol at 298 K. Which statement is consistent with this value?

Worked Examples

Worked Example 1 Band 5

The following Keq values are measured for $\text{A}(g) + \text{B}(g) \rightleftharpoons 2\text{C}(g)$: at 300°C, Keq = 0.25; at 400°C, Keq = 1.10; at 500°C, Keq = 4.80. (a) Is the forward reaction exothermic or endothermic? (b) At 200°C, would Keq be greater or less than 0.25? (c) A chemist claims Keq = 4.80 at 500°C means the equilibrium "strongly favours products." Evaluate this claim.

Part (a): Determine ΔH Direction

Keq increases as temperature increases: 0.25 → 1.10 → 4.80. Increasing temperature shifts the equilibrium in the endothermic direction. Since Keq increases (more products) as T increases, higher temperature favours the forward reaction → the forward reaction is endothermic (ΔH > 0). Consistent with LCP: adding heat shifts toward endothermic direction (right), producing more C.

Part (b): Predict Keq at 200°C

For an endothermic forward reaction, decreasing T decreases Keq. At 200°C (lower than 300°C): Keq < 0.25. Equilibrium would favour reactants even more than at 300°C.

Part (c): Evaluate the Claim

Keq = 4.80. Is 4.80 >> 1? No — 4.80 is moderately greater than 1. "Strongly favours products" implies Keq >> 1 (e.g. 10³ or higher). Keq = 4.80 is better described as "products are moderately favoured" — significant concentrations of both reactants and products would be present at equilibrium.

Answer: (a) Endothermic forward — Keq increases with T. (b) Keq < 0.25 at 200°C. (c) Claim overstated — Keq = 4.80 is moderately > 1; products somewhat favoured but significant reactants remain. ✓

Worked Example 2 Band 5–6

A student prepares an equilibrium mixture for $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$ with initial [Fe³⁺] = 2.00 × 10⁻³ mol/L and initial [SCN⁻] = 2.00 × 10⁻³ mol/L. After equilibrium, using the calibration curve: [FeSCN²⁺]_eq = 4.80 × 10⁻⁴ mol/L. (a) Set up and complete the ICE table. (b) Calculate Keq. (c) Verify by substitution.

Part (a): ICE Table

	Fe³⁺	SCN⁻	FeSCN²⁺
Initial (mol/L)	2.00 × 10⁻³	2.00 × 10⁻³	0
Change (mol/L)	−4.80 × 10⁻⁴	−4.80 × 10⁻⁴	+4.80 × 10⁻⁴
Equilibrium (mol/L)	1.52 × 10⁻³	1.52 × 10⁻³	4.80 × 10⁻⁴

Equilibrium [Fe³⁺] = 2.00 × 10⁻³ − 4.80 × 10⁻⁴ = 1.52 × 10⁻³ mol/L (same for [SCN⁻] by symmetry).

Part (b): Calculate Keq

$$K_{eq} = \frac{4.80 \times 10^{-4}}{(1.52 \times 10^{-3})(1.52 \times 10^{-3})} = \frac{4.80 \times 10^{-4}}{2.3104 \times 10^{-6}} = 207.7 \approx \mathbf{208}$$

Part (c): Verify

$(1.52 \times 10^{-3})^2 = 2.3104 \times 10^{-6}$ ✓; $4.80 \times 10^{-4} / 2.3104 \times 10^{-6} = 207.7$ ✓

Answer: (a) ICE table as above. (b) Keq = 208 at this temperature. (c) Verified ✓. Keq >> 1 → products significantly favoured — consistent with the deep red colour of FeSCN²⁺ visible even at low initial concentrations.

Activities

⚖️ Activity 1 — Predict + Explain

Applying Le Chatelier's Principle

For each disturbance below, predict the direction of equilibrium shift and explain your reasoning using collision theory.

An equilibrium system with more moles of gas on the right side is subjected to an increase in pressure. Predict the shift and explain.
For an endothermic reaction, the temperature is increased. Predict the shift and explain using energy considerations, and state whether Keq increases or decreases.
A catalyst is added to an equilibrium system. Explain what happens to the rate of attainment of equilibrium, the equilibrium position, and the value of Keq.

🧮 Activity 2 — Calculate + Analyse

Equilibrium Constant Calculations

Use the equilibrium expression and ICE table method to solve the problems below.

Question A

Write the equilibrium constant expression (Keq) for the reaction: CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g).

Question B

The reaction CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) has Keq = 3.90 × 10⁶ at 300°C and Keq = 1.85 × 10⁻² at 800°C. Is the forward reaction exothermic or endothermic? Justify your answer.

Question C

The reaction quotient Q is calculated to be 7.54. Given that Keq = 3.96, predict the direction the reaction will shift to reach equilibrium. Justify your answer.

Check Your Understanding

Multiple Choice

Q1. The equilibrium $\text{CO}(g) + 3\text{H}_2(g) \rightleftharpoons \text{CH}_4(g) + \text{H}_2\text{O}(g)$ has Keq = 3.90 × 10⁶ at 300°C and Keq = 1.85 × 10⁻² at 800°C. Which correctly interprets this data?

A The forward reaction is endothermic — Keq decreases as temperature increases

B The forward reaction is exothermic — Keq decreases as temperature increases, consistent with equilibrium shifting left at higher temperature

C Temperature has decreased Keq, so increasing pressure would restore the original Keq

D The forward reaction is endothermic because Keq is large at 300°C

Correct answer: B. Keq decreases from 3.90 × 10⁶ to 1.85 × 10⁻² as temperature increases — a dramatic decrease. Keq decreasing with increasing T → forward reaction is exothermic (LCP: adding heat shifts left; Keq decreases). Option A states the correct direction of Keq change but wrong conclusion about ΔH — endothermic would give increasing Keq with T. Option C incorrectly states pressure can restore Keq — only temperature changes Keq. Option D confuses the magnitude of Keq with ΔH.

Q2. In a colourimetry experiment for $\text{Fe}^{3+} + \text{SCN}^- \rightleftharpoons \text{FeSCN}^{2+}$, a calibration curve is used to determine [FeSCN²⁺]_eq from the measured absorbance. Why is the calibration curve necessary?

A To measure the temperature dependence of Keq at each wavelength

B To convert the absorbance reading (directly measurable) into a concentration (needed for the Keq calculation), using the Beer-Lambert relationship A ∝ c

C To correct for the colour of Fe³⁺ and SCN⁻ which interfere with the measurement

D To determine the wavelength at which FeSCN²⁺ absorbs light most strongly

Correct answer: B. The calibration curve establishes the relationship between absorbance and [FeSCN²⁺] using standards of known concentration. Beer-Lambert law: A = εlc → A ∝ c for fixed ε and l. The equilibrium [FeSCN²⁺] is read off this curve from the measured absorbance — converting the measurable (absorbance) into the needed quantity (concentration) for the Keq calculation.

Q3. For the equilibrium $2\text{NO}_2(g) \rightleftharpoons 2\text{NO}(g) + \text{O}_2(g)$, ΔG° = +70 kJ/mol at 298 K. Which statement is consistent with this value?

A Keq > 1; products are favoured at 298 K

B Keq < 1; reactants are favoured; ΔG° = −RT ln Keq gives a positive ΔG° when Keq < 1

C Keq = 1; ΔG° = 0 at equilibrium so Keq = 1

D Keq > 1; positive ΔG° means the reaction is exothermic and products are stable

Correct answer: B. $\Delta G^\circ = -RT\ln K_{eq}$. For ΔG° = +70 kJ/mol (positive): $\ln K_{eq} = -70000/(8.314 \times 298) = -28.25 \Rightarrow K_{eq} \approx 5 \times 10^{-13} \ll 1$. Very small Keq → reactants strongly favoured → consistent with ΔG° = +70 kJ/mol (positive ΔG° → non-spontaneous forward reaction). Option A is wrong — large positive ΔG° gives Keq < 1. Option C confuses ΔG (= 0 at equilibrium for any system) with ΔG° (only 0 when Keq = 1).

Short Answer Questions

SAQ 1 (4 marks). Describe the full colourimetry procedure for determining Keq for the Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺ equilibrium. Include the role of the calibration curve and explain how the Beer-Lambert law connects absorbance to concentration.

SAQ 2 (3 marks). A chemist measures Keq = 120 at 25°C and Keq = 35 at 50°C for a reaction. (a) Is the forward reaction exothermic or endothermic? (b) Predict Keq qualitatively at 75°C (greater, equal, or less than 35). (c) State why changing the pressure cannot restore Keq to 120.

Reveal Answers

SAQ 1: (1) Prepare calibration standards of known [FeSCN²⁺]; measure absorbance of each → plot A vs [FeSCN²⁺] → straight line through origin (Beer-Lambert: A = εlc; A ∝ c for fixed ε, l). (2) Prepare equilibrium mixture from known initial [Fe³⁺] and [SCN⁻]. (3) Measure absorbance of equilibrium mixture → read [FeSCN²⁺]eq from calibration curve (absorbance → concentration). (4) Set up ICE table with known initial concentrations and measured [FeSCN²⁺]eq → calculate [Fe³⁺]eq and [SCN⁻]eq. (5) Substitute all equilibrium concentrations into Keq = [FeSCN²⁺]/([Fe³⁺][SCN⁻]) → calculate Keq.

SAQ 2: (a) Keq decreases as temperature increases (120 → 35 as T increases from 25 to 50°C) → forward reaction is exothermic (ΔH < 0). (b) At 75°C (even higher temperature), Keq would be less than 35 — the exothermic forward reaction is further disfavoured at higher temperatures. (c) Keq is a thermodynamic constant determined only by temperature. Changing pressure changes the position of equilibrium (which concentrations are at equilibrium) but does not change the ratio of product to reactant concentrations at equilibrium at that temperature — Keq is unchanged.

Predict then reveal+8 XP

1 · Predict

2 · Reveal

3 · Compare

For the exothermic reaction 2NO&sub2;(g) ⇌ N&sub2;O&sub4;(g), brown NO&sub2; forms colourless N&sub2;O&sub4;. The temperature is increased. Predict: does the mixture get darker or lighter, and does K_eq increase or decrease?

Confidence 50%

Complete the Learn phase to unlock practice questions.

Extended Response

A student performing the colourimetry experiment for Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺ measures a higher absorbance at 35°C than at 25°C. (a) Using the relationship ΔG° = −RT ln Keq and the Beer-Lambert law, explain what this observation tells you about ΔH for the forward reaction. (b) Sketch (describe in words) what the calibration curve looks like and explain why using the wrong temperature calibration curve would give an incorrect Keq value. (c) Explain why it is essential to use the calibration curve rather than calculating concentration directly from the Beer-Lambert law in practice. (6 marks)

How did your thinking change?

For an exothermic reaction (ΔH < 0), increasing temperature decreases Keq — exactly as van 't Hoff predicted in 1884 with his equation d(ln Keq)/dT = ΔH°/(RT²). The data showed Keq dropping from 895 at 25°C to 580 at 35°C to 360 at 45°C, confirming the forward reaction is exothermic. At 15°C, Keq would be greater than 895. Were your three predictions correct?

I have completed this lesson

← Lesson 12 Lesson 14 →