A chemist shines light through a deep red iron thiocyanate solution, reads an absorbance value, and — using Beer's Law and an ICE table — calculates Keq to four significant figures without ever measuring individual concentrations directly.
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The equilibrium $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$ is exothermic in the forward direction (ΔH < 0). A student 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.
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.
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 × ΔH matrix — Keq change and shift direction for all four combinations
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:
Think First data — iron thiocyanate (exothermic forward, ΔH < 0):
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.
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: $\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq)$
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: $(8.50 \times 10^{-4})^2 = 7.225 \times 10^{-7}$
$$K_{eq} = \frac{1.50 \times 10^{-4}}{7.225 \times 10^{-7}} = 207.6 \approx 208 \checkmark$$
The equation ΔG° = −RT ln Keq is the bridge between thermodynamics and equilibrium — it connects the spontaneity of a reaction (from Module 4) to the position of that reaction's equilibrium.
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:
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.
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.
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.
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.
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.
| 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).
$$K_{eq} = \frac{4.80 \times 10^{-4}}{(1.52 \times 10^{-3})(1.52 \times 10^{-3})} = \frac{4.80 \times 10^{-4}}{2.3104 \times 10^{-6}} = 207.7 \approx \mathbf{208}$$
$(1.52 \times 10^{-3})^2 = 2.3104 \times 10^{-6}$ ✓; $4.80 \times 10^{-4} / 2.3104 \times 10^{-6} = 207.7$ ✓
For an exothermic reaction (ΔH < 0), increasing temperature decreases Keq because heat acts as a product. 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. A colourimetric experiment measures [FeSCN²⁺] at equilibrium, and from this you can calculate Keq using the ICE table and the initial concentrations.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
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?
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?
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?
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