Write the K_p expression for N_2(g) + 3H_2(g) 2NH_3(g) and state its units. [2 marks]

2.0 mol of A and 2.0 mol of B in a 1 dm cubed flask reach equilibrium A + B C with 1.5 mol of C formed. Calculate K_c. [3 marks]

SingaporeChemistrySyllabus dot point

How is the position of a dynamic equilibrium described quantitatively, and how does it respond to changing conditions?

Explain dynamic equilibrium and write expressions for Kc and Kp, calculate equilibrium constants and equilibrium amounts, and apply Le Chatelier's principle to predict the effect of concentration, pressure, temperature and catalysts on the position of equilibrium

A focused answer to the H2 Chemistry learning outcome on equilibrium. Dynamic equilibrium, writing and calculating Kc and Kp, the ICE-table method, the effect of changing conditions through Le Chatelier's principle, and why a catalyst does not shift the position.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to describe dynamic equilibrium, write and calculate $K_c$ and $K_p$ , find equilibrium amounts using an ICE table, and predict how concentration, pressure, temperature, and catalysts shift the position of equilibrium through Le Chatelier's principle. Calculating $K_c$ from an ICE table and explaining industrial conditions (Haber, Contact) are guaranteed exam content.

The answer

Dynamic equilibrium

A reversible reaction reaches dynamic equilibrium when the forward and backward reactions proceed at equal rates in a closed system, so the concentrations of all species stay constant (though the reactions continue). Both reactions are still occurring; nothing has stopped.

Writing $K_c$ and $K_p$

For a general equilibrium $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ :

K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}, \qquad K_p = \frac{p_\text{C}^c\, p_\text{D}^d}{p_\text{A}^a\, p_\text{B}^b}

$K_c$ uses equilibrium concentrations; $K_p$ uses partial pressures (for gases). The partial pressure of a gas is its mole fraction times the total pressure. Pure solids and liquids are omitted from the expression.

Calculating with an ICE table

The standard method:

Initial: write starting moles of each species.
Change: use the stoichiometry to express the change with a single unknown.
Equilibrium: add the change to the initial moles.
Convert to concentrations (divide by volume) or partial pressures, then substitute.

The meaning of $K$

A large $K$ ( $\gg 1$ ) means products are favoured at equilibrium; a small $K$ ( $\ll 1$ ) means reactants are favoured. The value of $K$ depends only on temperature. Changing concentration or pressure shifts the position of equilibrium but does not change $K$ .

Le Chatelier's principle

If a system at equilibrium is disturbed, it shifts to partially oppose the change:

Increase concentration of a reactant: shifts forward (toward products).
Increase total pressure: shifts toward the side with fewer gas moles.
Increase temperature: shifts in the endothermic direction (and changes $K$ ).
Catalyst: no shift; it speeds both directions equally and only reaches equilibrium faster.

Only a temperature change alters the value of $K$ . For an exothermic forward reaction, raising temperature decreases $K$ and the yield.

Applying to industry

Haber process ( $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ , exothermic): high pressure favours $\text{NH}_3$ (fewer gas moles), but a moderate temperature (around $450$ degrees Celsius) is a compromise between yield (favoured by low $T$ ) and rate (favoured by high $T$ ).
Contact process ( $2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$ , exothermic): similar compromise, with a $\text{V}_2\text{O}_5$ catalyst.

Calculating Kc from a percentage dissociation

For $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ , $0.100$ mol of $\text{N}_2\text{O}_4$ is placed in a $1.00$ dm cubed flask. At equilibrium, $40\%$ has dissociated. Calculate $K_c$ .

Step 1: Find the equilibrium amounts using an ICE table

Because the volume is exactly $1.00\ \text{dm}^3$ , the number of moles equals the concentration in mol dm $^{-3}$ directly. Start by working out how much $\text{N}_2\text{O}_4$ reacts: $40\%$ of $0.100$ mol is $0.040$ mol. The stoichiometry of the equation ( $1$ mol $\text{N}_2\text{O}_4$ gives $2$ mol $\text{NO}_2$ ) fixes how much product forms.

                N2O4      NO2
Initial         0.100     0
Change          -0.040    +0.080
Equilibrium     0.060     0.080

Step 2: Write the $K_c$ expression and substitute

The equilibrium constant expression puts products over reactants with stoichiometric exponents. The coefficient $2$ on $\text{NO}_2$ becomes a square in the expression.

K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \frac{(0.080)^2}{0.060} = \frac{0.0064}{0.060} = 0.107\ \text{mol dm}^{-3}

Final answer: $K_c = 0.107\ \text{mol dm}^{-3}$ .

Try it: Equilibrium constant calculator - build an ICE table and get $K_c$ or $K_p$ with the correct units.

Examples in context

Example 1. Optimising the Haber process. Engineers choose about $200$ atmospheres and $450$ degrees Celsius with an iron catalyst. High pressure pushes the equilibrium toward ammonia (fewer gas moles) and the moderate temperature balances a reasonable yield against an acceptable rate; the catalyst gets the system to equilibrium quickly. Unreacted gases are recycled to raise the overall conversion. SEAB expects this compromise reasoning, not just a single "best" condition.

Example 2. Predicting the colour change in the cobalt equilibrium. The pink-to-blue cobalt chloride equilibrium is endothermic in the blue-forming direction. Heating shifts it toward blue (absorbing heat) and adding concentrated acid (more chloride) also shifts it toward blue. This visible demonstration is a favourite for testing Le Chatelier reasoning on both temperature and concentration in one question.

Try this

Q1. Write the $K_p$ expression for $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ and state its units. [2 marks]

Cue. $K_p = p_{\text{NH}_3}^2 / (p_{\text{N}_2}\, p_{\text{H}_2}^3)$ ; units Pa $^{-2}$ (or atm $^{-2}$ ).

Q2. State the effect on the value of $K_c$ of (a) adding a catalyst and (b) raising the temperature of an endothermic reaction. [2 marks]

Cue. (a) No change. (b) $K_c$ increases (shift toward products for an endothermic forward reaction).

Q3. $2.0$ mol of A and $2.0$ mol of B in a $1$ dm cubed flask reach equilibrium $\text{A} + \text{B} \rightleftharpoons \text{C}$ with $1.5$ mol of C formed. Calculate $K_c$ . [3 marks]

Cue. Equilibrium: A = B = $0.5$ , C = $1.5$ ; $K_c = 1.5/(0.5 \times 0.5) = 6.0$ mol $^{-1}$ dm $^3$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Specimen (9729)5 marksFor the equilibrium

\text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g})

, 1.00 mol of

\text{H}_2

and 1.00 mol of

\text{I}_2

were mixed in a

2.00\ \text{dm}^3

vessel and allowed to reach equilibrium at a fixed temperature. At equilibrium 1.60 mol of

\text{HI}

was present. Calculate

K_c

Show worked answer →

Use an ICE table (initial, change, equilibrium) in moles, then convert to concentrations.

Initial: H2 = 1.00, I2 = 1.00, HI = 0.
HI formed = 1.60 mol, so by the 1:2 stoichiometry, H2 used = I2 used = 0.80 mol.
Equilibrium moles: H2 = 0.20, I2 = 0.20, HI = 1.60.

Concentrations (divide by 2.00 dm cubed): [H2] = 0.10, [I2] = 0.10, [HI] = 0.80 mol per dm cubed.

$K_c = \dfrac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \dfrac{(0.80)^2}{(0.10)(0.10)} = \dfrac{0.64}{0.01} = 64$ .

Since equal moles of gas on each side, Kc is dimensionless here.

Markers reward the ICE table, the correct stoichiometric changes, conversion to concentration, and the value of Kc with a comment on units.

2023 (style)4 marksFor the exothermic equilibrium 2SO2(g) + O2(g) <=> 2SO3(g), state and explain the effect on the equilibrium yield of SO3 of (a) increasing the pressure and (b) increasing the temperature.

Show worked answer →

Apply Le Chatelier: the system shifts to oppose the change.

(a) Increasing pressure: there are 3 moles of gas on the left and 2 on the right. The system shifts to the side with fewer gas moles (right) to reduce pressure, so the yield of SO3 increases.

(b) Increasing temperature: the forward reaction is exothermic. The system shifts in the endothermic (backward) direction to absorb the added heat, so the yield of SO3 decreases (and Kc/Kp falls).

Markers reward the gas-mole comparison for pressure, the correct shift, the exothermic reasoning for temperature, and the effect on yield (and that K changes only with temperature).