2.3 Extent of chemical change

Learning objectives

2.3.1Dynamic equilibrium• Forward rate = backward rate (closed system)• Describe physical and chemical systems at equilibrium 2.3.2Equilibrium law• K determined from stoichiometry• Deduce K expressions for homogeneous reactions 2.3.3Equilibrium constant (K)• Magnitude indicates extent of reaction• Temperature dependent• Kreverse = 1/Kforward• Interpret K << 1, K < 1, K = 1, K > 1, and K >> 1 2.3.4Le Châtelier's principle• Predict effects of changes in concentration, temperature, pressure• Include effects on K and equilibrium composition 2.3.5(HL)—Reaction quotient (Q)• Non-equilibrium concentrations• Compare Q to K to predict direction• Calculate Q and predict direction toward equilibrium 2.3.6(HL)—Equilibrium calculations• K with initial and equilibrium concentrations• Use small-K approximations; quadratic equations are not expected 2.3.7(HL)—K and ΔG• Both measure equilibrium position• Calculate ΔG⦵ from ΔG⦵ = −RT ln K

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.

At equilibrium both directions continue at equal non-zero rates while reactant and product concentrations remain constant and need not be equal.

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.

1 mark

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.

1 mark

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

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.

Worked reading: if K = 0.0665 at 100 C, K < 1, so reactants are favoured and the forward reaction has a small extent. This describes equilibrium composition, not reaction speed. At the same temperature, reversing the equation gives K = 1/0.0665.

1 mark

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.

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.

the no-catalyst and catalyst rows begin from matching reactant and product levels; the catalyst row differs through a wider connecting path, representing faster equalization; both rows finish at matching equal liquid levels; the catalyst is not shown changing equilibrium position or the final state.

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.

2 marks

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

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

delta G equals RT ln(Q/K); Q less than K gives negative forward delta G and net progress toward products; Q greater than K gives positive forward delta G and net progress toward reactants; Q equals K gives delta G zero and dynamic equilibrium, not stopped reactions.

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.

Worked $Q$ example: for $\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ , $Q=[\ce{NH3}]^2/([\ce{N2}][\ce{H2}]^3)$ . If every current concentration is $0.50\,\mathrm{mol\,dm^{-3}}$ , $Q=(0.50)^2/[(0.50)(0.50)^3]=4.0$ . At 475 K, $K=0.59$ , so $Q>K$ : the mixture contains too much product relative to equilibrium and the reverse direction is favoured until $Q=K$ .

3 marks

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.

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.

Worked equilibrium example: for $\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$ , $K=3.0$ , $[\ce{SO2}]_{eq}=0.12$ and $[\ce{SO3}]_{eq}=0.18\,\mathrm{mol\,dm^{-3}}$ . From $3.0=(0.18)^2/[x(0.12)^2]$ , $x=[\ce{O2}]_{eq}=0.75\,\mathrm{mol\,dm^{-3}}$ . Forming $0.18\,\mathrm{mol\,dm^{-3}}$ of $\ce{SO3}$ consumes $0.18$ of $\ce{SO2}$ and $0.090$ of $\ce{O2}$ , so their initial concentrations were $0.30$ and $0.84\,\mathrm{mol\,dm^{-3}}$ , respectively.

3 marks

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.

$ΔG° = −RT ln K$

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.

Worked Gibbs example: for $\ce{2NO(g) <=> N2O2(g)}$ at 298 K, local spectroscopic data give $K=1.39\times10^{-5}$ . $\Delta G^\circ=-(8.31\,\mathrm{J\,K^{-1}\,mol^{-1}})(298\,\mathrm{K})\ln(1.39\times10^{-5})=+2.77\times10^4\,\mathrm{J\,mol^{-1}}=+27.7\,\mathrm{kJ\,mol^{-1}}$ . The positive value agrees with $K<1$ : reactants are favoured under standard conditions at this temperature.

2 marks

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.

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.

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—IB ChemistryRead

2.3 Extent of chemical change

Learning objectives

Dynamic Equilibrium

The Equilibrium Law

Interpreting K

Le Châtelier's Principle

The Reaction Quotient Q

RICE-Table Equilibrium Calculations

K and Gibbs Energy

Extent of Chemical Change Summary

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Recognizing Dynamic Equilibrium

Writing Kc Expressions

Using K to Describe Extent

Predicting Equilibrium Shifts

Comparing Q with K

Solving for Equilibrium Composition

Calculating ΔG° from K