2.3 Extent of chemical change

Dynamic equilibrium occurs in a closed system when the forward and reverse processes continue at equal rates. Macroscopic amounts remain constant, but reactants and products need not be equal in amount.

The same rate-balance idea applies to physical equilibria such as vaporization and condensation as well as to reversible chemical reactions.

In a sealed liquid–vapour system at equilibrium, molecules continue evaporating and condensing at equal rates, so pressure and amounts are constant on average. Equal rates do not mean equal concentrations, and opening the system can prevent equilibrium by allowing matter to escape.

Representative question

[Maximum number: 1]

Ammonia is manufactured by the Haber process.

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad \Delta H_{\mathrm{r}}^{\ominus}=-92.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

Outline what is meant by dynamic equilibrium.

$Kc = [C]^r[D]^s / ([A]^p[B]^q) for pA + qB ⇌ rC + sD$

For a homogeneous reaction, place product concentrations over reactant concentrations and use each balanced-equation coefficient as the exponent.

Build the expression only after balancing the equation, and use equilibrium rather than initial concentrations. The numerical value of K changes with temperature; changing starting amounts can move the equilibrium composition without changing K.

Representative question

[Maximum number: 1]

Deduce the $K_{\mathrm{c}}$ expression for the reaction in part (d)(i).

$\left[\mathrm{NH}_{3}\right]^{2} /\left(\left[\mathrm{N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}\right)$

K range	Equilibrium tendency
K << 1	reactants strongly favoured
K < 1	reactants favoured
K = 1	comparable amounts
K > 1	products favoured
K >> 1	products strongly favoured

$Kreverse = 1 / Kforward$

K describes a ratio, not reaction speed: a very large K can still belong to a slow reaction. Reversing the equation gives 1/K, while multiplying every coefficient by a factor raises K to that factor. Interpret 'favoured' as equilibrium composition, not complete conversion.

Representative question

[Maximum number: 1]

At $100^{\circ} \mathrm{C} \mathrm{K}_{\mathrm{c}}$ for this reaction is 0.0665 . Outline what this indicates about the extent of this reaction.

reaction hardly proceeds

OR

reverse reaction/formation of $\mathrm{NO}_{2}$ is favoured

OR

«concentration of» reactants greater than «concentration of» products «at equilibrium»

Marking guidance:

Accept equilibrium lies to the left.

An equilibrium shifts to partially counteract an imposed change. Pressure favours the side with fewer gaseous molecules; temperature favours the endothermic direction; concentration changes alter composition.

At fixed temperature, concentration and pressure changes do not change K. Temperature changes K. A catalyst changes rates in both directions and does not change equilibrium position.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), compression favours the two-mole gas side, but K is unchanged if temperature is fixed. Heating favours the endothermic direction and changes K; a catalyst reaches the same equilibrium faster by accelerating both directions.

Disturbance	Immediate evidence	K at fixed/new T	Direction check
concentration or pressure change	Q changes before composition readjusts	unchanged if T is fixed	compare the new Q with K
raise temperature	heat favours the endothermic direction	K increases if the forward reaction is endothermic; decreases if it is exothermic	use the stated forward ΔH
catalyst	both forward and reverse rates increase	unchanged	equilibrium composition is unchanged; it is reached sooner

For gas pressure, count gaseous coefficients only. Use Q/K or opposing-rate evidence to justify the shift rather than the phrase “counteracts the change” alone.

Representative question

[Maximum number: 2]

Explain why an increase in pressure shifts the position of equilibrium towards the products and how this affects the value of the equilibrium constant, $K_{\mathrm{c}}$ .

shifts to the side with fewer moles «of gas» OR shifts to right as there is a reduction in volume $\boldsymbol{}$

«value of » $K_{c}$ unchanged

Marking guidance:

Accept " $K_{c}$ only affected by changes in temperature".

$Q = product-over-reactant concentration expression using current concentrations$

Q uses concentrations at any time, not necessarily equilibrium. Q < K means the forward direction is needed; Q > K means the reverse direction is needed; Q = K means equilibrium.

Calculate Q with the same expression as K but using the current concentrations. If Q = 0.20 and K = 5.0, too little product is present relative to equilibrium, so the forward direction lowers the mismatch. Recalculate after composition changes; Q is a snapshot, not a new constant.

Representative question

[Maximum number: 3]

0.200 mol sulfur dioxide, 0.300 mol oxygen and 0.500 mol sulfur trioxide were mixed in a $1.00 \mathrm{dm}^{3}$ flask at 1000 K .

Predict the direction of the reaction showing your working.

$\text { «reaction quotient } / Q=» \frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]} / \frac{0.500^{2}}{0.200^{2} \times 0.300} / 20.8 \vee$

reaction quotient/Q/20.8/answer < $K_{\mathrm{c}}$ /280 OR mixture needs more product for the number to equal $K_{\mathrm{c}} \vee$

reaction proceeds to the right/products

Marking guidance:

Do not award M3 without valid reasoning.

Write the balanced reaction, record initial concentrations, express changes as coefficient multiples of x, and substitute the equilibrium row into Kc. Use a small-K approximation when justified; quadratic equations are not expected here.

Equilibrium reactions do not use the limiting-reactant idea: both directions remain possible, so calculate the equilibrium composition instead.

After using an approximation such as C₀ − x ≈ C₀, validate it by checking that x/C₀ is small, commonly below 5% for the stated course method. If the check fails, the approximation is not justified; revise the setup rather than treating equilibrium as a limiting-reactant completion.

Representative question

[Maximum number: 3]

The equilibrium constant, $K_{\mathrm{c}}$ , for the reaction

\mathrm{CO}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})

was found to be 10.0 at $420^{\circ} \mathrm{C}$ .
1.00 mol of CO(g) and 1.00 mol of $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ are mixed in a $1.00 \mathrm{dm}^{3}$ container at $420^{\circ} \mathrm{C}$ . Calculate the equilibrium concentration of each component in the mixture, showing your working.
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$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})$
$(1.00-x)(1.00-x) \quad x \quad x$
$K_{c}=10.0=\frac{x^{2}}{(1.00-x)^{2}} / \sqrt{10.0}=\frac{x}{(1.00-x)}$ ;
x=0.760 /(1.00-x)=0.240;
$[\mathrm{CO}]=0.240\left(\mathrm{moldm}^{-3}\right)$ and $\left[\mathrm{H}_{2} \mathrm{O}\right]=0.240\left(\mathrm{moldm}^{-3}\right)$ and $\left[\mathrm{H}_{2}\right]=0.760\left(\mathrm{moldm}^{-3}\right)$
and $\left[\mathrm{CO}_{2}\right]=0.760\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)$ ;

$ΔG° = −RT ln K$

K and ΔG both describe equilibrium position. K > 1 corresponds to product-favoured equilibrium and ΔG° < 0; K = 1 corresponds to ΔG° = 0 and approximately equal equilibrium tendencies.

Use a dimensionless K, T in kelvin and R = 8.31 J mol⁻¹ K⁻¹. The sign link applies to ΔG°: the actual ΔG away from standard conditions also depends on the reaction quotient.

Representative question

[Maximum number: 2]

Determine the equilibrium constant, K , for this reaction at $25^{\circ} \mathrm{C}$ , referring to section 1 of the data booklet.

If you did not obtain an answer in (c)(iii), use $\Delta G=-43.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ , but this is not the correct answer.

$\Delta G=-RT\ln K$

$-41.8\ \mathrm{kJ\ mol^{-1}} =-\frac{8.31}{1000}(298)\ln K$

$\ln K=16.9$

$K=e^{16.9}=2.19\times10^7$

Award [2] for correct final answer.

Retrieve the route: define dynamic equilibrium, write K, interpret its magnitude, predict Le Châtelier shifts, compare Q with K, solve a RICE table, and connect K with ΔG.

Check closed-system and equal-rate language, exponents and direction, whether a change affects K, current versus equilibrium concentrations, stoichiometric x changes, and kelvin/unit consistency in ΔG calculations.

ConceptIB Chemistry HL

2.3 Extent of chemical change

Dynamic Equilibrium

Recognizing Dynamic Equilibrium

Assessment in practice

The Equilibrium Law

Writing Kc Expressions

Assessment in practice

Interpreting K

Using K to Describe Extent

Assessment in practice

Le Châtelier's Principle

Predicting Equilibrium Shifts

Assessment in practice

The Reaction Quotient Q

Comparing Q with K

Assessment in practice

RICE-Table Equilibrium Calculations

Solving for Equilibrium Composition

Assessment in practice

K and Gibbs Energy

Calculating ΔG° from K

Assessment in practice

Extent of Chemical Change Summary

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