Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
Learn
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
R1.4.1 Entropy, S HL
Entropy, S
A measure of the dispersal or distribution of matter and/or energy in a system: the more ways the energy of the system can be distributed among its particles, the higher its entropy. Under the same conditions a gas has a higher entropy than a liquid, which has a higher entropy than a solid. Molar entropies are expressed in J K⁻¹ mol⁻¹.
Standard entropy, S⦵
The entropy of one mole of a substance in its standard state, in J K⁻¹ mol⁻¹. S⦵ is an absolute value, not a value relative to the elements, so every pure element and compound at 298 K has a positive S⦵, including elements: for example S⦵ of H₂(g) is 130.7 J K⁻¹ mol⁻¹ and of O₂(g) is 205.2 J K⁻¹ mol⁻¹. This is unlike ΔH⦵f, which is zero for an element in its standard state by definition.
Standard entropy change, ΔS⦵
The entropy change of the system when the reaction occurs in the molar amounts shown in the equation, with all substances in their standard states: ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants), each S⦵ multiplied by its stoichiometric coefficient. Units: J K⁻¹ mol⁻¹.
Predicting the sign of ΔS
Because a gas has far higher entropy than a liquid or solid, the sign of ΔS for a change is predicted mainly from the change in the amount of gas: an increase in moles of gas gives a positive ΔS, a decrease gives a negative ΔS. Changes of state from solid to liquid to gas increase entropy. Dissolving an ionic solid usually increases entropy; dissolving a gas decreases it, because the gas particles become confined in the liquid.
Students often think The sign of ΔS follows the total number of particles (formula units) on each side of the equation, whatever their states. In fact No. A gas has far higher entropy than a liquid or a solid, so the sign of ΔS is predicted from the change in the amount (moles) of gas. Solids and liquids contribute little.
Students often think Dissolving always increases entropy, because the substance spreads out through the solvent. In fact No. Dissolving an ionic solid usually increases entropy, but dissolving a gas decreases it: the gas particles move from the gas phase into a liquid, where they are confined among solvent molecules.
R1.4.2 Gibbs energy change, ΔG HL
Gibbs energy change, ΔG
The quantity that relates the energy that can be obtained from a reaction to its enthalpy change, its entropy change and the absolute temperature: ΔG⦵ = ΔH⦵ − TΔS⦵. Units: kJ mol⁻¹. Because ΔH⦵ is in kJ mol⁻¹ and ΔS⦵ in J K⁻¹ mol⁻¹, ΔS⦵ must be divided by 1000 (or ΔH⦵ multiplied by 1000) before the two terms are combined.
Absolute temperature, T
Temperature on the kelvin scale, which starts at absolute zero. T/K = temperature/°C + 273. T in ΔG⦵ = ΔH⦵ − TΔS⦵, in T = ΔH⦵/ΔS⦵ and in RT ln K must be in kelvin.
Students often think Data can be substituted into ΔG⦵ = ΔH⦵ − TΔS⦵, T = ΔH⦵/ΔS⦵, ΔG⦵ = −RT ln K and ΔG = ΔG⦵ + RT ln Q as printed, because the equations do not show any conversion between J and kJ. In fact No. ΔH⦵ and ΔG⦵ are in kJ mol⁻¹, but ΔS⦵ and R are in J K⁻¹ mol⁻¹, so the units must be made to match (for example by dividing the J value by 1000) before the terms are combined.
Students often think The temperature can be substituted in whatever units the question gives it, including °C. In fact No. T is the absolute temperature and must be in kelvin: T/K = temperature/°C + 273.
R1.4.3 Spontaneous change HL
Spontaneous change
At constant pressure (and temperature), a change is spontaneous (thermodynamically feasible) if ΔG is negative. The sign of ΔG says nothing about the rate: a spontaneous change may be too slow to observe if its activation energy is high.
Temperature at which a reaction becomes spontaneous
When ΔH⦵ and ΔS⦵ have the same sign, the sign of ΔG⦵ changes at T = ΔH⦵/ΔS⦵ (the temperature at which ΔG⦵ = 0), assuming ΔH⦵ and ΔS⦵ do not vary with temperature. If both are positive the reaction is spontaneous above this temperature; if both are negative it is spontaneous below it. If ΔH⦵ < 0 and ΔS⦵ > 0 it is spontaneous at all temperatures; if ΔH⦵ > 0 and ΔS⦵ < 0 it is spontaneous at none.
Entropy change of the surroundings
Heat released by a reaction at constant pressure disperses into the surroundings and increases their entropy: ΔS(surroundings) = −ΔH/T. The total entropy change is ΔS(system) + ΔS(surroundings), and ΔG = −TΔS(total). ΔG therefore accounts for both the direct entropy change of the chemicals and the indirect entropy change of the surroundings caused by the transfer of heat.
Students often think A reaction that releases heat is spontaneous whatever its entropy change, so the sign of ΔH alone decides whether a reaction is spontaneous. In fact No. Spontaneity depends on the sign of ΔG = ΔH − TΔS. An exothermic reaction with a negative ΔS becomes non-spontaneous above the temperature at which TΔS outweighs ΔH.
Students often think A change can only be spontaneous if the entropy of the system increases, so any reaction with a negative ΔS is never spontaneous. In fact No. The total entropy (system plus surroundings) must increase. A change in which the system's entropy decreases is spontaneous if enough heat is released to increase the entropy of the surroundings by more.
R1.4.4 ΔG and the reaction quotient, Q HL
ΔG and the reaction quotient, Q
For a reaction mixture of any composition, ΔG = ΔG⦵ + RT ln Q, where Q has the form of the equilibrium expression but uses the actual concentrations (or pressures) present, R = 8.31 J K⁻¹ mol⁻¹ and T is in kelvin. ΔG is negative when Q < K, so the forward reaction proceeds; as the reaction approaches equilibrium ΔG becomes less negative.
ΔG at equilibrium and ΔG⦵ = −RT ln K
At equilibrium ΔG = 0 and Q = K, so ΔG⦵ = −RT ln K. A negative ΔG⦵ corresponds to K > 1 (products favoured), a positive ΔG⦵ to K < 1, and ΔG⦵ = 0 to K = 1. The logarithm is the natural logarithm (ln), not log₁₀.
ΔG compared with ΔG⦵
ΔG⦵ is a fixed value for a reaction at a given temperature, referring to all substances in their standard states; it is zero only if K = 1. ΔG refers to the actual mixture, changes as the composition changes, and becomes zero at equilibrium.
Students often think The 'log' key on the calculator can be used for ln K, since both are logarithms. In fact No. It is the natural logarithm, ln (base e). Using log₁₀ gives a wrong value unless a factor of 2.303 is included.
Students often think At equilibrium ΔG⦵ = 0, because there is no longer any driving force for the reaction. In fact No. ΔG (for the actual mixture) is zero at equilibrium. ΔG⦵ is a fixed value for the reaction at that temperature, equal to −RT ln K; it is zero only if K = 1.
Diagnostic a bearings check, not a test
8 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Consider the four changes W, X, Y and Z below at 298 K. Which change results in an increase in the entropy of the system?
W: 2H₂S(g) + O₂(g) → 2S(s) + 2H₂O(l) (exothermic)
X: NH₃(g) → NH₃(aq) (exothermic)
Y: 2C(s) + O₂(g) → 2CO(g) (exothermic)
Z: N₂(g) + 2O₂(g) → 2NO₂(g) (endothermic) HL
Answer and reasoning
Change W — A student who counts every formula unit sees 3 on the left and 4 on the right and predicts an increase. But the states matter: 3 mol of gas become a solid and a liquid, with no gas left, so the entropy of the system falls sharply.
Change X — A student who believes that dissolving always increases entropy picks this. That is usually true for an ionic solid, but here a gas dissolves: its molecules lose the freedom of the gas phase and are confined in the water, so the entropy of the system decreases.
Change Y — The amount of gas doubles: 1 mol O₂(g) becomes 2 mol CO(g), and solid carbon is converted into part of a gas. Gases have far higher entropy than solids, so ΔS is positive (from standard entropies, ΔS⦵ ≈ +179 J K⁻¹ mol⁻¹). There are fewer formula units on the right (2) than on the left (3), which shows why counting all species misleads: only moles of gas decide the sign.
Change Z — A student who treats entropy as heat content expects a reaction that absorbs heat to gain entropy. But ΔS of the system depends on the distribution of the chemicals: 3 mol of gas become 2 mol of gas, so ΔS is negative. The heat absorbed affects the entropy of the surroundings, not the sign of ΔS of the system.
2 For the Haber process reaction, N₂(g) + 3H₂(g) → 2NH₃(g), ΔH⦵ = −92.2 kJ mol⁻¹ and ΔS⦵ = −198.1 J K⁻¹ mol⁻¹. Assuming these values do not change with temperature, what is ΔG⦵ for the reaction at 450 °C? HL
Answer and reasoning
+51.0 kJ mol⁻¹ — T = 450 + 273 = 723 K and ΔS⦵ = −0.1981 kJ K⁻¹ mol⁻¹. ΔG⦵ = ΔH⦵ − TΔS⦵ = −92.2 − 723(−0.1981) = −92.2 + 143.2 = +51.0 kJ mol⁻¹. At this temperature the unfavourable entropy term outweighs the favourable enthalpy term.
−3.1 kJ mol⁻¹ — A student who substitutes the Celsius temperature gets −92.2 − 450(−0.1981) = −3.1 kJ mol⁻¹. T in ΔG⦵ = ΔH⦵ − TΔS⦵ is the absolute temperature, 723 K.
−235.4 kJ mol⁻¹ — A student who subtracts TΔS⦵ as a positive amount, ignoring that ΔS⦵ is negative, gets −92.2 − 143.2 = −235.4 kJ mol⁻¹. Because ΔS⦵ is negative, −TΔS⦵ = −723(−0.1981) = +143.2 kJ mol⁻¹.
+143134.1 kJ mol⁻¹ — A student who substitutes ΔS⦵ in J K⁻¹ mol⁻¹ alongside ΔH⦵ in kJ mol⁻¹ gets −92.2 − 723(−198.1) = +143134.1. ΔS⦵ must first be converted to −0.1981 kJ K⁻¹ mol⁻¹.
3 For a reaction, ΔH⦵ is negative and ΔS⦵ is negative. Assuming that neither value changes with temperature, which statement about the reaction is correct? HL
Answer and reasoning
Spontaneous only when the temperature is low enough — ΔG⦵ = ΔH⦵ − TΔS⦵. With ΔS⦵ negative, −TΔS⦵ is positive and grows with T. At low T the negative ΔH⦵ dominates and ΔG⦵ < 0; above T = ΔH⦵/ΔS⦵ the entropy term dominates and ΔG⦵ > 0.
Spontaneous only when the temperature is high enough — A student who believes heating always makes reactions more feasible picks this. Heating increases the rate, but with a negative ΔS⦵ raising T makes −TΔS⦵ larger and positive, so ΔG⦵ becomes less negative and eventually positive.
Spontaneous at every temperature, however high or low — A student who thinks an exothermic reaction is spontaneous whatever its entropy change picks this. With ΔS⦵ negative, the positive −TΔS⦵ term outweighs ΔH⦵ above T = ΔH⦵/ΔS⦵, so the reaction is not spontaneous at high temperature.
Not spontaneous at any temperature, however high or low — A student who thinks the entropy of the system must increase for a change to be spontaneous picks this. The heat released increases the entropy of the surroundings; at low temperature this outweighs the decrease in the system's entropy, and ΔG⦵ is negative.
4 A reaction mixture reaches equilibrium at constant temperature and pressure. Which statement about the equilibrium mixture is correct? HL
Answer and reasoning
ΔG for the reaction in this mixture has become zero. — As a reaction approaches equilibrium, ΔG becomes less negative and finally reaches zero: at equilibrium there is no net driving force in either direction, and Q = K.
ΔG⦵ for the reaction has become equal to zero. — A student who confuses ΔG with ΔG⦵ picks this. ΔG⦵ is a fixed value for the reaction at that temperature, equal to −RT ln K; it does not change as the reaction proceeds and is zero only if K = 1.
ΔG for the mixture has reached its most negative value. — A student who confuses ΔG with G picks this. G of the mixture is at a minimum at equilibrium, but ΔG, the driving force for further reaction, becomes less negative as equilibrium is approached and is zero at equilibrium.
The forward and reverse reactions have both come to a stop. — A student who pictures equilibrium as static picks this. Equilibrium is dynamic: both reactions continue at equal rates. ΔG = 0 means there is no net change, not that reaction has ceased.
5 Nitrogen and hydrogen react to form ammonia:
N₂(g) + 3H₂(g) → 2NH₃(g)
Standard entropies: S⦵(N₂(g)) = 191.6 J K⁻¹ mol⁻¹, S⦵(H₂(g)) = 130.7 J K⁻¹ mol⁻¹, S⦵(NH₃(g)) = 192.8 J K⁻¹ mol⁻¹.
What is the standard entropy change, ΔS⦵, for the reaction? HL
Answer and reasoning
+385.6 J K⁻¹ mol⁻¹ — A student who takes S⦵ of the elements N₂ and H₂ as zero, as for ΔH⦵f, uses only the product: 2(192.8) = +385.6. Standard entropies are absolute values; N₂(g) and H₂(g) have positive S⦵ values that must be included.
+198.1 J K⁻¹ mol⁻¹ — A student who subtracts products from reactants gets 583.7 − 385.6 = +198.1. ΔS⦵ is products minus reactants, and a reaction that reduces the moles of gas from 4 to 2 must have a negative ΔS⦵.
−129.5 J K⁻¹ mol⁻¹ — A student who adds each S⦵ value once, ignoring the coefficients, gets 192.8 − (191.6 + 130.7) = −129.5. S⦵ values are per mole: 3 mol of H₂ and 2 mol of NH₃ appear in the equation.
−198.1 J K⁻¹ mol⁻¹ — ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants) = 2(192.8) − [191.6 + 3(130.7)] = 385.6 − 583.7 = −198.1 J K⁻¹ mol⁻¹. The negative sign agrees with the prediction: 4 mol of gas become 2 mol of gas.
6 For the reaction 2NO(g) + O₂(g) → 2NO₂(g) at 150 °C, ΔG⦵ = −52.2 kJ mol⁻¹ and ΔS⦵ = −146.6 J K⁻¹ mol⁻¹. What is ΔH⦵ for the reaction? HL
Answer and reasoning
−74.2 kJ mol⁻¹ — A student who substitutes the Celsius temperature gets −52.2 + 150(−0.1466) = −74.2 kJ mol⁻¹. T must be the absolute temperature, 423 K.
+9.8 kJ mol⁻¹ — A student who moves the TΔS⦵ term across the equals sign without changing its sign writes ΔH⦵ = ΔG⦵ − TΔS⦵ = −52.2 + 62.0 = +9.8 kJ mol⁻¹. From ΔG⦵ = ΔH⦵ − TΔS⦵, ΔH⦵ = ΔG⦵ + TΔS⦵.
−62064.0 kJ mol⁻¹ — A student who uses ΔS⦵ in J K⁻¹ mol⁻¹ with ΔG⦵ in kJ mol⁻¹ gets −52.2 + 423(−146.6) = −62064.0. ΔS⦵ must be converted to −0.1466 kJ K⁻¹ mol⁻¹ first.
7 Silicon is manufactured by heating silica with carbon:
SiO₂(s) + 2C(s) → Si(s) + 2CO(g) ΔH⦵ = +690 kJ mol⁻¹
Standard entropies in J K⁻¹ mol⁻¹: SiO₂(s) 41.5, C(s) 5.7, Si(s) 18.8, CO(g) 197.7.
Assuming ΔH⦵ and ΔS⦵ do not change with temperature and ignoring any changes of state, above what temperature does the reaction become spontaneous? HL
Answer and reasoning
1950 K — A student who takes S⦵ of the elements C(s) and Si(s) as zero gets ΔS⦵ = 2(197.7) − 41.5 = +353.9 J K⁻¹ mol⁻¹ and T = 690 000/353.9 = 1950 K. Elements have positive standard entropies, which must be included.
4076 K — A student who adds each S⦵ value once, ignoring the coefficients, gets ΔS⦵ = 18.8 + 197.7 − 41.5 − 5.7 = +169.3 J K⁻¹ mol⁻¹ and T = 690 000/169.3 = 4076 K. The equation shows 2 mol of C and 2 mol of CO.
1910 K — ΔS⦵ = [18.8 + 2(197.7)] − [41.5 + 2(5.7)] = 414.2 − 52.9 = +361.3 J K⁻¹ mol⁻¹ = +0.3613 kJ K⁻¹ mol⁻¹. ΔG⦵ = 0 when T = ΔH⦵/ΔS⦵ = 690/0.3613 = 1910 K. Both ΔH⦵ and ΔS⦵ are positive, so ΔG⦵ is negative above this temperature. (In practice silicon melts at 1687 K, so the product forms as a liquid; the calculation ignores this change of state.)
1.91 K — A student who divides ΔH⦵ in kJ mol⁻¹ by ΔS⦵ in J K⁻¹ mol⁻¹ gets 690/361.3 = 1.91 K. The units must match: 690 000 J mol⁻¹ ÷ 361.3 J K⁻¹ mol⁻¹ = 1910 K.
Working ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants) = [18.8 + 2(197.7)] − [41.5 + 2(5.7)] = 414.2 − 52.9 = +361.3 J K⁻¹ mol⁻¹. At the temperature where the reaction becomes spontaneous, ΔG⦵ = 0, so T = ΔH⦵/ΔS⦵ = (690 000 J mol⁻¹)/(361.3 J K⁻¹ mol⁻¹) = 1910 K (to the nearest kelvin). Above 1910 K, TΔS⦵ > ΔH⦵ and ΔG⦵ < 0.
8 For a reaction at 25 °C, the equilibrium constant, K, is 5.00 × 10¹⁸. What is ΔG⦵ for the reaction at this temperature? (R = 8.31 J K⁻¹ mol⁻¹) HL
Answer and reasoning
−46.3 kJ mol⁻¹ — A student who uses log₁₀ in place of ln gets −2.476 × 18.70 = −46.3 kJ mol⁻¹. The equation uses the natural logarithm: ln(5.00 × 10¹⁸) = 43.06.
−106.6 kJ mol⁻¹ — T = 25 + 273 = 298 K. ΔG⦵ = −RT ln K = −(8.31 × 298 × ln(5.00 × 10¹⁸))/1000 = −2.476 × 43.06 = −106.6 kJ mol⁻¹. A very large K corresponds to a large negative ΔG⦵.
−8.9 kJ mol⁻¹ — A student who substitutes the Celsius temperature gets −(8.31 × 25 × 43.06)/1000 = −8.9 kJ mol⁻¹. T must be the absolute temperature, 298 K.
−106622.9 kJ mol⁻¹ — A student who calculates −RT ln K with R in J K⁻¹ mol⁻¹ gets −106622.9 J mol⁻¹ and writes it as kJ mol⁻¹. The answer must be divided by 1000: −106.6 kJ mol⁻¹.
Working T = 25 + 273 = 298 K. ln K = ln(5.00 × 10¹⁸) = 43.06. ΔG⦵ = −RT ln K = −(8.31 J K⁻¹ mol⁻¹)(298 K)(43.06) = −1.066 × 10⁵ J mol⁻¹ = −106.6 kJ mol⁻¹.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
4 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 For the change C(diamond) → C(graphite), ΔG⦵ = −2.9 kJ mol⁻¹ at 298 K. Yet diamonds show no detectable conversion into graphite at room temperature, even over centuries. Which statement accounts for both facts? HL
Answer and reasoning
The change is not spontaneous, as it does not happen quickly. — A student who thinks 'spontaneous' means 'happens quickly' picks this. The sign of ΔG decides spontaneity: ΔG⦵ is negative, so the change is spontaneous. Its rate is a separate, kinetic question.
The change is spontaneous, but at 298 K its rate is negligible. — A negative ΔG⦵ means the change is thermodynamically feasible. Whether it is observed depends on its rate: the activation energy for rearranging the covalent network of diamond is very high, so at 298 K the conversion is immeasurably slow. Spontaneous does not mean fast.
The change needs a catalyst at 298 K to make its ΔG⦵ negative enough. — A student who thinks a catalyst makes a reaction more favourable picks this. A catalyst lowers the activation energy and changes the rate only; it cannot change ΔG⦵, which is already negative.
Diamond is favoured at equilibrium, since ΔG⦵ is negative. — A student who drops the minus sign in ΔG⦵ = −RT ln K reads a negative ΔG⦵ as K < 1. In fact a negative ΔG⦵ gives K > 1, so graphite, the product, is favoured at equilibrium.
2 For the reaction 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH⦵ = −197.8 kJ mol⁻¹ and ΔS⦵ = −188.0 J K⁻¹ mol⁻¹. Which statement about the reaction at 298 K is correct? HL
Answer and reasoning
Not spontaneous: the entropy of the system falls in this reaction. — A student who thinks the system's entropy must increase for a change to be spontaneous picks this. ΔG takes into account the entropy change of the surroundings as well; here the heat released increases the entropy of the surroundings by much more than the system loses.
Spontaneous: an exothermic reaction is spontaneous whatever its ΔS⦵. — A student who thinks the sign of ΔH⦵ alone decides spontaneity picks this. The conclusion is right at 298 K but the reason is false: with ΔS⦵ negative, the reaction stops being spontaneous above T = ΔH⦵/ΔS⦵ ≈ 1050 K.
Not spontaneous: ΔG⦵ = −197.8 − 298(−188.0), which is positive. — A student who substitutes ΔS⦵ in J K⁻¹ mol⁻¹ with ΔH⦵ in kJ mol⁻¹ gets a large positive value. Converted consistently, ΔG⦵ = −197.8 − 298(−0.1880) = −141.8 kJ mol⁻¹, which is negative.
Spontaneous: the surroundings gain more entropy than the system loses. — The heat released raises the entropy of the surroundings by −ΔH⦵/T = 197 800/298 = +664 J K⁻¹ mol⁻¹, far more than the system loses (188.0 J K⁻¹ mol⁻¹). Equivalently, ΔG⦵ = −197.8 − 298(−0.1880) = −141.8 kJ mol⁻¹, which is negative.
3 For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) at 450 °C, ΔG⦵ = −61.9 kJ mol⁻¹. In a particular mixture of the three gases at this temperature, [SO₂] = 0.500 mol dm⁻³, [O₂] = 0.250 mol dm⁻³ and [SO₃] = 0.00250 mol dm⁻³. What is ΔG for the reaction in this mixture? (R = 8.31 J K⁻¹ mol⁻¹) HL
Answer and reasoning
−85.9 kJ mol⁻¹ — A student who calculates Q = 1.0 × 10⁻⁴ correctly but uses log₁₀ in place of ln gets −61.9 + 6.008 × (−4.00) = −85.9 kJ mol⁻¹. The equation uses the natural logarithm: ln(1.0 × 10⁻⁴) = −9.21.
−55398.8 kJ mol⁻¹ — A student who adds RT ln Q in J mol⁻¹ to ΔG⦵ in kJ mol⁻¹ gets −61.9 + (−55 336.9) = −55398.8. With Q = 1.0 × 10⁻⁴, RT ln Q must be converted to −55.3 kJ mol⁻¹ before it is added.
−117.2 kJ mol⁻¹ — T = 450 + 273 = 723 K and Q = (0.0025)²/((0.50)²(0.25)) = 1.0 × 10⁻⁴. RT = 8.31 × 723 = 6008 J mol⁻¹ = 6.008 kJ mol⁻¹. ΔG = ΔG⦵ + RT ln Q = −61.9 + 6.008 × ln(1.0 × 10⁻⁴) = −61.9 + 6.008(−9.210) = −61.9 − 55.3 = −117.2 kJ mol⁻¹. Because Q is less than 1, RT ln Q is negative and ΔG is more negative than ΔG⦵; because Q is far below K (about 3 × 10⁴ at 723 K), ΔG is negative and the forward reaction proceeds.
−6.6 kJ mol⁻¹ — A student who writes Q the wrong way up, as [SO₂]²[O₂]/[SO₃]², gets Q = 1.0 × 10⁴ and ΔG = −61.9 + 6.008 × 9.210 = −61.9 + 55.3 = −6.6 kJ mol⁻¹. For the reaction as written, Q = [SO₃]²/([SO₂]²[O₂]) = 1.0 × 10⁻⁴.
4 For the decomposition of ammonia, 2NH₃(g) ⇌ N₂(g) + 3H₂(g), ΔG⦵ = +32.9 kJ mol⁻¹ at 25 °C. What is the value of the equilibrium constant, K, for this reaction at 25 °C? (R = 8.31 J K⁻¹ mol⁻¹) HL
Answer and reasoning
5.9 × 10⁵ — A student who drops the minus sign and uses ln K = ΔG⦵/RT gets ln K = +13.29 and K = e^13.29 = 5.9 × 10⁵. From ΔG⦵ = −RT ln K, ln K = −ΔG⦵/RT, which is negative when ΔG⦵ is positive, so K < 1.
5.2 × 10⁻¹⁴ — A student who treats the logarithm as log₁₀ takes −13.29 as log₁₀ K and gets K = 10^−13.29 = 5.2 × 10⁻¹⁴. The equation uses the natural logarithm, so K = e^−13.29 = 1.7 × 10⁻⁶.
1.67 × 10⁻⁶⁹ — A student who substitutes the Celsius temperature gets ln K = −32 900/(8.31 × 25) = −158.4 and K = e^−158.4 = 1.67 × 10⁻⁶⁹. T must be the absolute temperature, 298 K.
1.7 × 10⁻⁶ — T = 298 K. ln K = −ΔG⦵/RT = −32 900/(8.31 × 298) = −13.29, so K = e^−13.29 = 1.7 × 10⁻⁶. A positive ΔG⦵ gives a negative ln K and K < 1: at 298 K the reactant, ammonia, is favoured at equilibrium.
Working T = 25 + 273 = 298 K. ΔG⦵ = −RT ln K, so ln K = −ΔG⦵/RT = −(32 900 J mol⁻¹)/((8.31 J K⁻¹ mol⁻¹)(298 K)) = −13.29. K = e^−13.29 = 1.7 × 10⁻⁶.
That was your twenty minutes. Real practice on R1.4 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
DP Science Cafe