Concentration, pressure and catalysts can all move an equilibrium, but not one of them changes Keq. Temperature is the only thing that does, and colour is how you measure the result.
Today's hook, A flask of iron thiocyanate is deep red at room temperature and noticeably paler when warmed. Nothing was added and nothing was removed, so the equilibrium constant itself must have changed, and a colourimeter can put a number on it.
1
You’re here
Get oriented
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Beyond the syllabus. Determining Keq at different temperatures, and the qualitative effect of temperature on equilibrium, is core — and colourimetry is a well-supported method. The van 't Hoff equation and the ΔG°–Keq relationship are extension: the exam will not require them.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
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
Deep dive, Extension 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
Deep dive, ExtensionGibbs–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
The Gibbs–Keq relationship and the van ’t Hoff equation are Extension for this course. They explain why only temperature changes Keq, which is worth understanding, but the NSW Chemistry Stage 6 syllabus does not require you to calculate with them. The qualitative temperature rules above are the assessable content.
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.
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 Keq increase or decrease?
50%
Actual answer
Darker (more brown NO&sub2;); Keq decreases.
Increasing temperature shifts the exothermic reaction in the reverse (endothermic) direction → more NO&sub2; produced → darker brown colour. Because the reverse reaction is favoured, the ratio of products to reactants falls → Keq decreases. Temperature is the ONLY factor that changes Keq.
How close was your prediction?
Excellent, the colour change + Keq direction link is a classic HSC question.
For exothermic reactions: higher T → reverse shift → more reactant → Keq decreases. This is always true.
2
Why only temperature changes Keq
01
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.
Start from what Keq actually is. For a given reaction, Keq is fixed at a given temperature. It is a property of the reaction and the conditions, not of the particular mixture you happen to have in the flask. That single sentence is what does the work here, and it is worth reading twice.
Ask what each disturbance actually changes. Adding a reactant, compressing a gas mixture or adding a catalyst all change the composition of the mixture, or how quickly it gets where it is going. None of them changes what counts as equilibrium for that reaction. So the system moves back to satisfy the same Keq, which is exactly why we describe those changes as shifting the position of equilibrium while Keq itself stays put.
Temperature is different in kind. Keq is set by the thermodynamics of the reaction through $\Delta G^\circ = -RT \ln K_{eq}$, and $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Combining the two gives $\ln K_{eq} = -\dfrac{\Delta H^\circ}{RT} + \dfrac{\Delta S^\circ}{R}$. Here ΔH° and ΔS° are properties of the reaction itself, so the only quantity on the right that a chemist can vary is T. Change the temperature and you change the value of Keq that the system is heading towards; change anything else and you have only moved the mixture relative to an unchanged target.
Careful with the reasoning
It is not enough to say "the equation contains T, so temperature changes Keq". A concentration term does not appear in that equation at all, and that absence is the real argument: Keq is fixed by ΔH°, ΔS° and T, so no amount of adding, removing or compressing can move it, while a temperature change must.
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
Use the predictor. For 2NO2(g) ⇌ N2O4(g), ΔH < 0 (brown to colourless), the mixture is heated. What happens to the colour, and to Keq?
3
Reading Keq values across temperatures
02
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?
4
Measuring Keq by colourimetry
03
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.
1
Prepare calibration standards, known [FeSCN²⁺] measured with colorimeter → plot A vs [FeSCN²⁺] → straight line through origin
2
Prepare equilibrium mixture from known initial [Fe³⁺] and [SCN⁻]
3
Measure absorbance of equilibrium mixture → use calibration curve to find [FeSCN²⁺]eq
4
Set up ICE table with known initial [Fe³⁺] and [SCN⁻] and measured [FeSCN²⁺]eq → calculate [Fe³⁺]eq and [SCN⁻]eq
5
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.
Safety (school-risk controls)
If this investigation is run as a class practical, it must be teacher-supervised with appropriate controls. Use dilute solutions only (about 0.002 mol/L Fe(NO₃)₃ and KSCN); wear safety glasses and gloves; iron(III) and thiocyanate solutions are irritants. Never add acid to thiocyanate solutions, acidified thiocyanate can release toxic hydrogen cyanide gas. Collect all solutions as inorganic/heavy-metal waste; do not pour down the sink. Where reagents or a colorimeter are unavailable, the same calculation can be completed from supplied secondary absorbance data.
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?
5
Calculating Keq from colourimetry data
04
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.
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.
5b
Investigation, Determining Keq by Colourimetry
Syllabus: conduct an investigation to determine the equilibrium constant of a chemical equilibrium system
We just saw how absorbance data is processed into a Keq value. That raises a question: where does that data come from, and how do you know the number you get is worth quoting? This card answers it → the full run, from calibration to an uncertainty you can defend.
Inquiry question. What is the value of Keq for Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) at room temperature, and how reliable is that value?
Safety
Teacher supervision and a written risk assessment required. Safety glasses and gloves throughout. Thiocyanate must never contact acid, which can release toxic hydrogen cyanide. Iron(III) nitrate is an irritant and stains. Use a water bath, never a naked flame, near these solutions. All iron/thiocyanate waste goes to the labelled residues container.
Equipment. Colourimeter or spectrophotometer set near 447 nm (blue filter), matched cuvettes, 0.200 mol L⁻¹ Fe(NO₃)₃ (for standards, in large excess), 0.00200 mol L⁻¹ KSCN, 0.00200 mol L⁻¹ Fe(NO₃)₃ (for the equilibrium mixtures), volumetric pipettes or graduated syringes, 5 volumetric flasks or measuring cylinders, distilled water, lint-free tissue.
1
Blank the instrument with distilled water at 447 nm. Handle cuvettes by the ribbed faces only, a fingerprint on the optical face is a real absorbance error.
2
Prepare the calibration standards. Mix a known small volume of 0.00200 mol L⁻¹ KSCN with a large excess of 0.200 mol L⁻¹ Fe(NO₃)₃. The excess Fe³⁺ drives the reaction essentially to completion, so [FeSCN²⁺] can be taken as the SCN⁻ concentration after mixing. Make five standards spanning the range you expect.
3
Measure the absorbance of each standard. Plot absorbance against [FeSCN²⁺] and draw the line of best fit through the origin. Record the gradient, this is your calibration.
4
Prepare the equilibrium mixtures. Combine measured volumes of 0.00200 mol L⁻¹ Fe(NO₃)₃ and 0.00200 mol L⁻¹ KSCN in five different ratios, made up to the same total volume with distilled water. Neither reactant is in excess here, which is the whole point, these mixtures sit at equilibrium.
5
Measure the absorbance of each equilibrium mixture and record the temperature of the room. Keq is temperature dependent, so a value quoted without a temperature is incomplete.
6
Use the calibration line to convert each absorbance into [FeSCN²⁺]eq.
7
For each mixture, build an ICE table. Initial concentrations are the concentrations after mixing, so apply the dilution factor first. Subtract [FeSCN²⁺]eq from each initial concentration to obtain [Fe³⁺]eq and [SCN⁻]eq, then calculate Keq.
8
Average the five Keq values. Quote the mean, the range, and the temperature.
Part 1, calibration standards (large excess Fe³⁺, so [FeSCN²⁺] ≈ [SCN⁻] after mixing)
Standard
V(KSCN) / mL
V(Fe³⁺ excess) / mL
[FeSCN²⁺] / mol L⁻¹
Absorbance
1
2
3
4
5
Part 2, equilibrium mixtures. Temperature: ______ °C
Mixture
[Fe³⁺]₀ after mixing
[SCN⁻]₀ after mixing
Absorbance
[FeSCN²⁺]eq
[Fe³⁺]eq
[SCN⁻]eq
Keq
1
2
3
4
5
Analysis and evaluation.
Your five mixtures had different starting concentrations. Should they give the same Keq? Say what it means about your data if they do, and what it means if they do not.
Calculate the percentage spread of your Keq values about the mean. Which single mixture contributes most to the spread, and is there a reason in the raw absorbance for that?
Identify the largest source of uncertainty. Consider volume measurement, the calibration fit, cuvette handling, and the assumption that the standards went to completion. Justify your choice with a number where you can.
The calibration assumes the excess Fe³⁺ converts all the SCN⁻. If that assumption is slightly wrong, does your Keq come out too high or too low? Reason it through.
Propose one specific change to the method that would most reduce the uncertainty you identified, and say what you expect it to improve.
Conclusion. Quote Keq as a mean with its range and the temperature at which it was measured. State how confident you are in the value, on the evidence of the spread across your five mixtures, not on how close it lands to a textbook number.
The dilution step is where most marks are lost: the initial concentrations in the ICE table are not the bottle concentrations. Mixing 5.00 mL of 0.00200 mol L⁻¹ Fe³⁺ into a 10.00 mL total gives [Fe³⁺]₀ = 0.00100 mol L⁻¹. Apply the dilution before the ICE table, every time.
Common Error: Using the calibration standards as if they were equilibrium mixtures. They are deliberately not at a useful equilibrium, the large excess of Fe³⁺ is there to force the reaction to completion so that [FeSCN²⁺] is known independently. Calculating Keq from a standard gives a meaningless number.
Method: blank at 447 nm → calibrate with excess Fe³⁺ standards where [FeSCN²⁺] ≈ [SCN⁻] → plot A against [FeSCN²⁺] → measure five equilibrium mixtures → read [FeSCN²⁺]eq off the line → dilution-corrected ICE table for each → Keq per mixture → mean, range and temperature.
Pause, write the highlighted sequence into your book.
6
Extension: Gibbs free energy and Keq
05
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.
Extension, not assessable here
This card is enrichment. The two dot points this lesson must deliver are the qualitative effect of temperature on Keq and determining Keq experimentally, both covered in Cards 01 to 04. Gibbs free energy itself is assumed knowledge from Module 4 and is worth connecting up, but the quantitative relationship ΔG° = −RT ln Keq sits beyond the Module 5 dot points. Read this card for the connection, and do not expect to be examined on the calculation in this module.
A reaction with ΔG° = −200 kJ/mol strongly favours products at equilibrium. A reaction with ΔG° = −1 kJ/mol barely favours them. A reaction with ΔG° = 0 has Keq = 1, meaning the equilibrium expression evaluates to 1 (which is not the same as saying the concentrations are equal, unless the stoichiometry happens to make it so). All three follow from the standard relationship ΔG° = −RT ln Keq, which connects 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:
$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(g)$, 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?
7
Worked examples and activities
Worked Examples
Worked Example 1 Analyse
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.
1
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.
2
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.
3
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 Analyse/Evaluate
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.
1
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).
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, Position or Constant?
Which changes: the position, or Keq itself?
Card 01 argued that only temperature changes Keq, because Keq is fixed by ΔH°, ΔS° and T. Use that argument on each case below. For each one, state (a) whether the position of equilibrium moves, (b) whether the value of Keq changes, and (c) the one-sentence reason. Then check yourself against the model answers before moving on.
More Fe³⁺(aq) is added to the equilibrium Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) at a fixed temperature, and the solution darkens.
The same solution is warmed from 25°C to 45°C and fades. The forward reaction is exothermic.
A student reports that warming the solution "used up the FeSCN²⁺, so there is less of it, so Keq must have gone down". Identify what is right and what is wrong in that sentence.
Check your answers
1. Position moves right (the darker colour shows more FeSCN²⁺). Keq is unchanged: adding a species changes the composition, not the value the ratio must return to at that temperature.
2. Position moves left and Keq does change, decreasing. The forward reaction is exothermic, so raising T lowers Keq, and the fading colour is the visible consequence.
3. Right: less FeSCN²⁺ is present, and Keq did decrease. Wrong: the reasoning runs backwards. The fall in Keq is not caused by FeSCN²⁺ being used up; raising the temperature lowered Keq, and the system responded by shifting left, which is why less FeSCN²⁺ remains.
🧮 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.
✓
Practice questions
01
Multiple Choice
+5 XP
A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct
02
Short Answer
(4 marks) 1. 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.
(3 marks) 2. 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.
Show all answers
Short Answer Model Answers
SA1 (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.
SA2 (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.
✓
Retrieve and reflect
Check what actually stuck
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?
