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Inquiry Question 2: What factors affect equilibrium and how?

Investigate the effects of temperature, concentration, volume and/or pressure on a system at equilibrium and explain how Le Chatelier's principle can be used to predict such effects

A focused answer to the HSC Chemistry Module 5 dot point on Le Chatelier's principle. How concentration, pressure, volume and temperature shift equilibrium position, the role of catalysts, the Haber process worked example, and the past HSC questions markers reward.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-22

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

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What this dot point is asking
The answer
Examples in context
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In plain English

Picture an equilibrium as a see-saw that has settled level, with reactants on one side and products on the other, both sides still moving but no longer changing height. Le Chatelier's principle just says: if you push down on one side, the see-saw does not stay pushed down and it does not fully bounce back either, it settles at a NEW level position that partially cancels your push. Add more reactant (push down on the reactant side) and the see-saw tips a little toward products to use some of it up. Squeeze the gas into a smaller box (raise the pressure) and the see-saw tips toward whichever side has fewer gas molecules, because that side takes up less room. Add heat and the see-saw tips toward whichever direction "eats" that heat, the endothermic direction, because heat behaves like an extra ingredient added to that side. A catalyst is different: it just makes the see-saw wobble to its resting position faster, it does not change where that resting position ends up.

What this dot point is asking

NESA wants you to apply Le Chatelier's principle to predict how an equilibrium responds to changes in concentration, pressure, volume, and temperature, and to explain industrial applications (especially the Haber process). This builds directly on dynamic equilibrium and is one of the highest-frequency Module 5 questions, appearing every HSC paper in either short answer or extended response form.

The answer

Le Chatelier's principle

If a system at equilibrium is disturbed by a change in conditions, the system shifts in the direction that partially opposes the disturbance.

The principle predicts the direction of the shift but not its magnitude. The magnitude depends on Kc and the size of the disturbance.

Concentration changes

Adding a reactant (or removing a product) shifts the equilibrium to the right (toward products), because the system responds to "use up" the extra reactant.

Removing a reactant (or adding a product) shifts the equilibrium to the left (toward reactants).

Kc does not change. Concentration disturbances shift the position, not the constant.

Pressure changes (gas reactions)

Increasing pressure (by decreasing volume) shifts the equilibrium toward the side with fewer moles of gas.

Decreasing pressure (by increasing volume) shifts the equilibrium toward the side with more moles of gas.

If both sides have equal moles of gas (for example, $H_2 + I_2 \rightleftharpoons 2HI$ ), pressure changes have no effect on the position.

Adding an inert gas at constant volume does not shift the equilibrium because partial pressures of reactants and products are unchanged.

Temperature changes

This is the only disturbance that changes Kc.

Increasing temperature shifts the equilibrium in the endothermic direction (the system absorbs the added heat).

Decreasing temperature shifts the equilibrium in the exothermic direction.

For an exothermic forward reaction ( $\Delta H < 0$ ), the reverse is endothermic. Heating shifts left, cooling shifts right.

Catalysts

A catalyst increases the rate of both forward and reverse reactions equally. It does not shift the equilibrium position and does not change Kc. It only reduces the time taken to reach equilibrium.

An owned illustrative concentration-versus-temperature response curve shows how a system re-equilibrates after each type of disturbance:

Worked example

The Haber process is the canonical HSC application.

N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}, \quad \Delta H = -92 \text{ kJ/mol}

Industrial conditions are chosen to maximise the rate of ammonia production while keeping yield economically viable.

Pressure: 200 atm: Forward reaction goes from 4 moles of gas to 2. High pressure shifts equilibrium right (toward ammonia), increasing yield. Pressure is limited by reactor cost.
Temperature: 400°C: Forward reaction is exothermic. Le Chatelier favours low temperature for yield. But low temperature reduces the rate. 400°C is the kinetics-thermodynamics compromise.
Iron catalyst: Speeds up the approach to equilibrium without changing position.
Continuous removal of ammonia: $NH_3$ is liquefied and removed. Le Chatelier predicts the system shifts further right to replace the removed product.

Result: a sustained, economically viable rate of production at modest yield (around 15-20% per pass), with unreacted gases recycled.

Examples in context

Example 1. BHP Newcastle blast furnace carbon monoxide equilibrium. Iron oxide reduction in the blast furnaces formerly operated by BHP at Newcastle relied on $Fe_2O_{3(s)} + 3CO_{(g)} \rightleftharpoons 2Fe_{(l)} + 3CO_{2(g)}$ , $\Delta H < 0$ . Plant operators kept CO partial pressures high by burning coke against a hot air blast, pushing the equilibrium to the right and harvesting iron at the hearth. Excess heat from the bottom of the furnace would have shifted equilibrium left, lowering yield, so coolant water jackets stabilised the temperature. The HSC Le Chatelier framework explains why historical operators tuned blast-air ratios, coke addition rates and slag composition simultaneously rather than treating any one variable in isolation.

Example 2. Sulfuric acid contact process at Port Kembla. The contact-process step $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ , $\Delta H = -198 \text{ kJ mol}^{-1}$ , runs at the Incitec Pivot sulfuric acid plant near Port Kembla. Plant chemists are caught between two competing pressures: low temperatures shift equilibrium right (more $SO_3$ ) but slow the reaction unbearably. They compromise at 450 degrees C with a vanadium oxide catalyst that speeds the kinetics without changing $K_{eq}$ . They also feed slight excess oxygen to push the equilibrium further right under Le Chatelier, and remove $SO_3$ by absorption into 98 percent sulfuric acid, again driving net forward reaction.

Try this

Q1. State Le Chatelier's principle and explain why adding a catalyst does not change the position of equilibrium. [3 marks]

Cue. Statement of the principle (system partially opposes imposed change); catalyst speeds forward and reverse rates equally, so the equilibrium position is unchanged.

Q2. For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ , $\Delta H = +57 \text{ kJ mol}^{-1}$ , predict and justify the shift in equilibrium position when (a) the temperature is increased, (b) the volume is halved. [2+2 marks]

Cue. (a) Endothermic forward, so heating shifts right (more $NO_2$ ). (b) Halving volume doubles pressure; equilibrium shifts to the side with fewer moles of gas (left, towards $N_2O_4$ ).

Q3. Consider $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ , $\Delta H = -198 \text{ kJ mol}^{-1}$ . (a) State the effect on yield of $SO_3$ of increasing temperature. (b) State the effect of increasing pressure. (c) Explain why industry operates at a moderate 450 degrees C rather than the lowest possible temperature. [1+1+3 marks]

Cue. (a) Lower yield. (b) Higher yield. (c) Lower temperature would maximise yield but the rate becomes too slow to be economic; the chosen temperature is a kinetic-thermodynamic compromise.

Exam-style practice questions

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

2022 HSC5 marksThe industrial production of ammonia by the Haber process is represented by the equation N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = -92 kJ/mol. Apply Le Chatelier's principle to explain the choice of industrial conditions (200 atm, 400°C, iron catalyst) used in this process.

Show worked answer →

A 5 mark answer needs to apply Le Chatelier to each condition and resolve the temperature compromise.

Pressure (200 atm): The forward reaction converts 4 moles of gas (1 + 3) to 2 moles of gas. Increasing pressure shifts the equilibrium toward the side with fewer moles of gas, that is, toward ammonia. High pressure increases yield. The pressure is limited to ~200 atm because higher pressures require expensive, thick-walled reactors.
Temperature (400°C): The forward reaction is exothermic ( $\Delta H < 0$ ). Le Chatelier predicts that decreasing temperature shifts the equilibrium right (toward ammonia), increasing yield. However, lower temperature slows the rate of reaction. 400°C is a compromise. High enough for an acceptable rate, low enough to maintain a workable equilibrium position.
Iron catalyst: A catalyst does not shift the equilibrium position. It increases the rate of both forward and reverse reactions equally, allowing equilibrium to be reached faster at the chosen temperature.
Removing ammonia: Liquid ammonia is condensed and removed continuously, decreasing product concentration and shifting the equilibrium further right (Le Chatelier's response to product removal).

Markers reward (1) correct prediction of direction for each factor, (2) explicit reference to "fewer moles of gas" and "endothermic/exothermic", (3) recognising the temperature compromise (yield vs rate), (4) noting that a catalyst affects rate not position.

2019 HSC3 marksA sealed flask contains the equilibrium 2NO₂(g) ⇌ N₂O₄(g), ΔH = -57 kJ/mol. The flask is heated. Predict and explain the colour change observed.

Show worked answer →

$NO_2$ is brown; $N_2O_4$ is colourless. The forward reaction is exothermic ( $\Delta H = -57$ kJ/mol), so the reverse reaction is endothermic.

Le Chatelier's principle states that when temperature is increased, the equilibrium shifts in the endothermic direction to absorb the added heat. Here, the endothermic direction is the reverse (right to left), forming more $NO_2$ .

Therefore the equilibrium shifts left, $[NO_2]$ increases, and the gas mixture becomes a darker brown.

Markers reward (1) identifying that the reverse reaction is endothermic, (2) stating the equilibrium shifts in the endothermic direction, (3) connecting the shift to a darker brown colour.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksFor the equilibrium

2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}

\Delta H = -198\ \text{kJ mol}^{-1}

, state and justify the direction of shift when (a) the volume of the container is halved, (b) the temperature is decreased, (c) a catalyst is added.

Show worked solution →

A 3-mark identify-and-justify needs the correct direction AND the correct Le Chatelier reasoning for each part.

(a) Volume halved (pressure increased): The left side has 3 moles of gas (2 + 1); the right side has 2 moles of gas. Increasing pressure shifts the equilibrium toward the side with fewer moles of gas, so it shifts right, toward $SO_3$ .
(b) Temperature decreased: The forward reaction is exothermic ( $\Delta H < 0$ ). Decreasing temperature shifts the equilibrium in the exothermic direction, so it shifts right, toward $SO_3$ , increasing yield.
(c) Catalyst added: A catalyst increases the rate of the forward and reverse reactions equally. It does not shift the equilibrium position; there is no change, only a faster approach to the same equilibrium.

Marking criteria: 1 mark per part for the correct direction (or "no change") explicitly linked to a Le Chatelier justification; a bare direction with no reasoning earns no mark for that part.

foundation4 marksThe equilibrium

N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}

has

\Delta H = +57\ \text{kJ mol}^{-1}

. A sealed, rigid flask at equilibrium is heated. (a) State whether the mixture becomes darker or paler brown, with a full explanation. (b) Explain what happens to the value of

K_c

Show worked solution →

(a) Colour change. $NO_2$ is brown; $N_2O_4$ is colourless. The forward reaction ( $N_2O_4 \rightarrow 2NO_2$ ) is endothermic ( $\Delta H > 0$ ). Le Chatelier's principle states that increasing temperature shifts the equilibrium in the endothermic direction, which here is the forward direction. The equilibrium shifts right, $[NO_2]$ increases, and the mixture becomes a darker brown.

(b) Effect on $K_c$ . Temperature is the only disturbance that changes the equilibrium constant. Because the forward (endothermic) reaction is favoured by heating, more products form relative to reactants at the new equilibrium, so $K_c$ increases.

Marking criteria: (a) 1 mark for identifying the forward reaction as endothermic, 1 mark for stating the shift is in the endothermic (forward) direction, 1 mark for correctly concluding "darker brown". (b) 1 mark for stating $K_c$ increases (not "no change"), since concentration/pressure disturbances leave $K_c$ unchanged but temperature disturbances do not.

core5 marksAt a certain temperature, a 2.00 L rigid vessel at equilibrium for

N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}

contains 0.0400 mol of

N_2O_4

and 0.0600 mol of

NO_2

. Calculate

K_c

at this temperature, giving your answer to 2 significant figures with correct units.

Show worked solution →

Step 1: write the equilibrium expression.

K_c = \frac{[NO_2]^2}{[N_2O_4]}

Step 2: convert moles to equilibrium concentrations (divide by volume).

[N_2O_4] = \frac{0.0400\ \text{mol}}{2.00\ \text{L}} = 0.0200\ \text{mol L}^{-1}

[NO_2] = \frac{0.0600\ \text{mol}}{2.00\ \text{L}} = 0.0300\ \text{mol L}^{-1}

Step 3: substitute into the expression.

K_c = \frac{(0.0300\ \text{mol L}^{-1})^2}{0.0200\ \text{mol L}^{-1}} = \frac{9.00 \times 10^{-4}\ \text{mol}^2\text{L}^{-2}}{0.0200\ \text{mol L}^{-1}}

Step 4: evaluate.

K_c = 0.0450\ \text{mol L}^{-1}

Step 5: round to 2 significant figures (matching the least precise given value).

K_c = 4.5 \times 10^{-2}\ \text{mol L}^{-1}

Marking criteria: 1 mark for the correct equilibrium expression, 1 mark for converting both moles to concentrations using the volume, 1 mark for correct substitution, 1 mark for the correct numerical evaluation, 1 mark for the correct units and significant figures. Note that $K_c$ carries units here because the moles of gas differ across the equation (2 mol products vs 1 mol reactant).

core4 marksThe rate-vs-time graph below is an owned illustrative trace for the reaction

2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}

in a sealed vessel. At

t_1

, additional

O_2

is injected without changing the volume. (a) Describe the shape of the forward-rate and reverse-rate curves after

t_1

. (b) Explain, using Le Chatelier's principle, why the two rates become equal again at a new equilibrium.

Show worked solution →

(a) Description. Before $t_1$ , the forward rate and reverse rate are equal and constant (dynamic equilibrium). At $t_1$ , injecting extra $O_2$ instantly raises $[O_2]$ , so the forward rate jumps up immediately (more reactant collisions), while the reverse rate is momentarily unchanged. Over time, the forward rate falls and the reverse rate rises as reactants are consumed and product accumulates, until the two curves meet again at a new, higher common rate at $t_2$ .

(b) Explanation. Le Chatelier's principle predicts that adding a reactant ( $O_2$ ) shifts the equilibrium position to the right, toward $SO_3$ , to partially consume the added $O_2$ . As the forward reaction proceeds faster than the reverse immediately after $t_1$ , $[SO_2]$ and $[O_2]$ fall while $[SO_3]$ rises; this changes the forward and reverse rates until they again become equal, defining the new equilibrium position (with a higher $[SO_3]$ than before $t_1$ , and $K_c$ unchanged because temperature was not altered).

Marking criteria: (a) 1 mark for the immediate jump in forward rate at $t_1$ , 1 mark for describing convergence to a new equal rate. (b) 1 mark for correctly citing "adding reactant shifts right", 1 mark for linking this to the changing rates equalising at a new position, with $K_c$ explicitly unchanged.

core5 marksIn the Haber process

N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}

, an industrial reactor is fed 5.00 mol of

N_2

with excess

H_2

. At the operating equilibrium, 68.0% of the

N_2

is converted to

NH_3

. Calculate the mass of

NH_3

produced, to 3 significant figures. (

M(NH_3) = 17.03\ \text{g mol}^{-1}

Show worked solution →

Step 1: moles of $N_2$ reacted.

n(N_2\ \text{reacted}) = 5.00\ \text{mol} \times 0.680 = 3.40\ \text{mol}

Step 2: mole ratio from the balanced equation. 1 mol $N_2$ produces 2 mol $NH_3$ .

n(NH_3) = 2 \times 3.40\ \text{mol} = 6.80\ \text{mol}

Step 3: mass of $NH_3$ .

m(NH_3) = n \times M = 6.80\ \text{mol} \times 17.03\ \text{g mol}^{-1} = 115.804\ \text{g}

Step 4: round to 3 significant figures (matching 5.00 mol and 68.0%, both 3 s.f.).

m(NH_3) = 116\ \text{g}

Marking criteria: 1 mark for correct moles of $N_2$ reacted, 1 mark for the correct 1:2 mole ratio, 1 mark for correct moles of $NH_3$ , 1 mark for the mass calculation, 1 mark for the correct answer to 3 significant figures with units. This yield reflects the Le Chatelier compromise (200 atm, 400 degrees C) discussed above; a real single pass gives closer to 15 to 20% conversion, with unreacted gas recycled to approach the 68.0% figure used here cumulatively.

exam6 marksEvaluate the choice of 450 degrees C and a vanadium(V) oxide catalyst (rather than a much lower temperature) for the industrial contact process

2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}

\Delta H = -198\ \text{kJ mol}^{-1}

, with reference to Le Chatelier's principle.

Show worked solution →

This is a 6-mark EVALUATE: markers reward a judgement supported by Le Chatelier reasoning on both sides of the trade-off, not just a list of facts.

Band 6 plan.

State the thermodynamic case for low temperature: the forward reaction is exothermic, so Le Chatelier predicts that a lower temperature shifts the equilibrium right, increasing the equilibrium yield of $SO_3$ .
State the kinetic case against very low temperature: reaction rate falls sharply as temperature drops, so a very high-yield equilibrium reached only after an uneconomically long time is not commercially viable.
Introduce the catalyst: $V_2O_5$ increases the rate of both forward and reverse reactions equally, so equilibrium is reached faster at whatever temperature is chosen; it does not change the position or $K_c$ , so it cannot substitute for a temperature choice, only accelerate reaching whichever position temperature has set.
Judgement: 450 degrees C is chosen because it gives an acceptable reaction rate (helped further by the catalyst) while still retaining a good enough $SO_3$ yield (around 95 to 98 percent with the additional strategy of excess $O_2$ and $SO_3$ removal by absorption); it is therefore a deliberate compromise between the Le Chatelier-favoured low-temperature yield and the practical need for a viable rate, and is more effective than either extreme.

Model paragraph (excerpt). By Le Chatelier's principle, the exothermic contact-process equilibrium would give its highest yield of $SO_3$ at the lowest possible temperature, since cooling shifts the position toward the exothermic (forward) direction. However, operating at such a low temperature would make the rate of reaction commercially unworkable, since reaction rate depends strongly on temperature independent of the equilibrium position. The $V_2O_5$ catalyst increases the rate at which equilibrium is approached but, critically, does not alter the equilibrium position or $K_c$ , so it cannot be used instead of raising temperature; it can only make whichever temperature is chosen more time-efficient. The operating choice of 450 degrees C is therefore best understood as a deliberate compromise: it sacrifices some equilibrium yield (compared with a colder reactor) in exchange for a commercially viable rate, with the catalyst then recovering as much rate as possible without further sacrificing yield.

Marker's note: top-band answers (1) correctly separate the effect of temperature (changes position and $K_c$ ) from the effect of the catalyst (changes rate only), (2) explicitly name the trade-off as yield versus rate, (3) reach a clear, justified judgement rather than only describing both sides, and (4) reference at least one additional real strategy (excess $O_2$ or product removal) as further evidence of yield-maximising design around the temperature compromise.