Questions on entropy and Gibbs free-energy change

Question 1

Explain how the feasibility of the reaction 2NO₂(g) → N₂O₄(g) depends on temperature, using the data in the table. Include a calculation of the temperature at which ΔG = 0.

Accepted Answer

Indicative content

ΔS = 304 − (2 × 240) = −176 J K⁻¹ mol⁻¹, negative because two moles of gas become one and the system becomes more ordered
ΔH is negative and favours the reaction, while the −TΔS term is positive and works against it, more strongly the higher the temperature
at low temperature TΔS is small compared with ΔH, so ΔG is negative and the dimerisation of NO₂ is feasible
ΔG = 0 when T = ΔH ÷ ΔS = −57.2 ÷ −0.176 = 325 K, with ΔS first converted to kJ K⁻¹ mol⁻¹; allow 325 to 326 K
above about 325 K the −TΔS term outweighs ΔH, so ΔG is positive and the reverse change, N₂O₄ splitting into NO₂, is the feasible one
the calculation assumes ΔH and ΔS do not change with temperature, so 325 K is an estimate; this is the case of negative ΔH and negative ΔS, feasible only below ΔH/ΔS

Levels

Level 3 (5–6 marks) ΔS is calculated with its sign explained, the competition between ΔH and TΔS is set out, the temperature at which ΔG = 0 is calculated with the unit conversion, the feasibility on both sides of it is stated, and the assumption that ΔH and ΔS are constant is noted
Level 2 (3–4 marks) ΔS is calculated and the temperature of ΔG = 0 found; the negative ΔS is not explained, feasibility on either side of that temperature is not stated, or kJ and J are mixed
Level 1 (1–2 marks) ΔG = ΔH − TΔS is quoted or ΔS calculated; the reaction is commonly called feasible at all temperatures because ΔH is negative

Question 2

Calculate the lowest temperature at which the decomposition MgCO₃(s) → MgO(s) + CO₂(g) is feasible, using the data in the table.

Accepted Answer

M1 ΔS = (27 + 214) − 66 = +175 J K⁻¹ mol⁻¹ M1 ΔS is converted to +0.175 kJ K⁻¹ mol⁻¹ and ΔG is set to zero M1 T = ΔH ÷ ΔS = 117 ÷ 0.175 A1 T = 669 K, three significant figures, in kelvin; allow 668 to 669 K

Question 3

Explain, using ΔG = ΔH − TΔS, why the endothermic decomposition of calcium carbonate is not feasible at room temperature but is feasible at high temperature.

Accepted Answer

B1 ΔH is positive, so on its own it makes ΔG positive B1 a solid produces a gas, so ΔS is positive and the term −TΔS is negative B1 at room temperature TΔS is smaller than ΔH, so ΔG is positive and the reaction is not feasible B1 as T rises, TΔS grows until it exceeds ΔH, so above T = ΔH/ΔS the value of ΔG is negative and the reaction is feasible

Question 4

Evaluate this claim about the Haber process: 'ΔG is positive at the temperature the plant uses, so no ammonia can form there, and the process should be run at room temperature, where ΔG is negative.'

Accepted Answer

B1 a positive ΔG means the equilibrium lies on the side of the reactants, not that no ammonia forms; some ammonia is present at equilibrium B1 high pressure moves the equilibrium position towards ammonia, and ammonia is removed as it forms, so a useful yield is obtained at 700 K B1 at room temperature the reaction is feasible but its activation energy is so high that it is far too slow, even with an iron catalyst B1 the claim is therefore wrong: the operating temperature is a compromise between a feasible, high-yield temperature and a fast rate

Question 5

Calculate the entropy change of the surroundings and the total entropy change at 298 K for MgCO₃(s) → MgO(s) + CO₂(g), for which ΔH = +117 kJ mol⁻¹ and ΔS(system) = +175 J K⁻¹ mol⁻¹.

Accepted Answer

M1 ΔH is converted to joules: +117 kJ mol⁻¹ = +117000 J mol⁻¹ M1 ΔS(surroundings) = −ΔH ÷ T = −117000 ÷ 298 A1 ΔS(surroundings) = −393 J K⁻¹ mol⁻¹, three significant figures, negative sign required A1 ΔS(total) = +175 + (−393) = −218 J K⁻¹ mol⁻¹, error carried forward from the previous mark; negative, so the decomposition does not take place at 298 K

Question 6

Calculate the entropy change for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), using the standard entropies in the table.

Accepted Answer

M1 products: 214 + 2 × 70 = 354 J K⁻¹ mol⁻¹ M1 reactants: 186 + 2 × 205 = 596 J K⁻¹ mol⁻¹; the oxygen must be included, since elements do not have zero entropy A1 ΔS = 354 − 596 = −242 J K⁻¹ mol⁻¹, three significant figures, negative sign required

Question 7

Calculate ΔG at 298 K for 2H₂(g) + O₂(g) → 2H₂O(l), for which ΔH = −572 kJ mol⁻¹ and ΔS = −327 J K⁻¹ mol⁻¹.

Accepted Answer

M1 ΔS is converted to kJ: −327 ÷ 1000 = −0.327 kJ K⁻¹ mol⁻¹ M1 ΔG = −572 − (298 × −0.327) = −572 + 97.4 A1 ΔG = −475 kJ mol⁻¹, three significant figures, negative sign required; an answer near +96 900, from ΔS left in joules, is not credited

Question 8

Calculate the lowest temperature at which carbon reduces zinc oxide, ZnO(s) + C(s) → Zn(s) + CO(g), for which ΔH = +237 kJ mol⁻¹, using the standard entropies in the table.

Accepted Answer

M1 ΔS = (42 + 198) − (44 + 6) = +190 J K⁻¹ mol⁻¹, or +0.190 kJ K⁻¹ mol⁻¹ M1 at the lowest feasible temperature ΔG = 0, so T = 237 ÷ 0.190 A1 T = 1250 K to three significant figures (1247 K), in kelvin

Question 9

Calculate ΔH and ΔS for a reaction from its graph of ΔG against T.

Accepted Answer

M1 the intercept on the ΔG axis is ΔH, so ΔH = +50 kJ mol⁻¹ M1 the gradient is −ΔS = (0 − 50) ÷ (400 − 0) = −0.125 kJ K⁻¹ mol⁻¹ A1 ΔS = +0.125 kJ K⁻¹ mol⁻¹ = +125 J K⁻¹ mol⁻¹, three significant figures, positive sign shown

Question 10

Explain why the entropy of a substance increases far more when it boils than when it melts.

Accepted Answer

B1 on melting the particles leave fixed positions but stay in contact, so the disorder increases a little B1 on boiling the particles separate completely and move freely through a much larger volume B1 the number of ways of arranging the particles and their energy rises far more on boiling, so the entropy jump is much larger

Question 11

Explain why diamond does not turn into graphite at room temperature, although ΔG for the change is negative at 298 K.

Accepted Answer

B1 a negative ΔG shows that the change is feasible, but says nothing about its rate B1 the change has a very high activation energy, because strong covalent bonds must be broken B1 at room temperature almost no particles have enough energy to react, so the rate is too slow to observe and diamond is kinetically stable

Question 12

Suggest why solid ammonium chloride heated at the bottom of a test tube decomposes into ammonia and hydrogen chloride, NH₄Cl(s) → NH₃(g) + HCl(g), but a white solid re-forms on the cool upper part of the tube.

Accepted Answer

B1 the decomposition forms two moles of gas from a solid, so ΔS is large and positive, and ΔH is positive B1 at the hot bottom of the tube TΔS exceeds ΔH, so ΔG is negative and the decomposition is feasible B1 at the cooler top TΔS is smaller than ΔH, so ΔG for the decomposition is positive and the reverse reaction, forming solid NH₄Cl, is feasible instead

Question 13

Compare the conditions under which a reaction with negative ΔH and negative ΔS is feasible with those for a reaction with positive ΔH and positive ΔS.

Accepted Answer

B1 the reaction with negative ΔH and negative ΔS is feasible only below the temperature T = ΔH/ΔS B1 the reaction with positive ΔH and positive ΔS is feasible only above the temperature T = ΔH/ΔS B1 in both, the TΔS term grows with temperature: in the first it works against feasibility and in the second it makes the reaction feasible

Question 14

Describe how the entropy of one mole of water changes as it is heated from 0 K to 450 K.

Accepted Answer

B1 the entropy starts at zero at 0 K and rises gradually through the solid as the particles vibrate more B1 it rises vertically at 273 K as the ice melts, then rises gradually again through the liquid B1 it rises vertically by a much larger amount at 373 K as the water boils, then rises gradually through the gas

Question 15

Explain why an exothermic reaction increases the entropy of its surroundings, and why the increase is larger when the surroundings are cold.

Accepted Answer

B1 the heat released by the reaction passes into the surroundings and is spread among their particles B1 this increases the number of ways the energy of the surroundings can be arranged, so ΔS(surroundings) = −ΔH/T is positive when ΔH is negative B1 a given quantity of heat makes a larger difference to cold surroundings, which hold less energy to begin with, so the increase is larger at low T, as the division by T shows

Question 16

Calculate the equilibrium constant, K, at 298 K for 2NO₂(g) ⇌ N₂O₄(g), for which ΔG = −4.7 kJ mol⁻¹. Use ΔG = −RT ln K and R = 8.31 J K⁻¹ mol⁻¹.

Accepted Answer

M1 ln K = −ΔG ÷ RT, with ΔG converted to −4700 J mol⁻¹ M1 ln K = 4700 ÷ (8.31 × 298) = 1.90 A1 K = exp(1.90) = 6.7, two significant figures, with no units

Question 17

Suggest why a reaction with ΔG = +5 kJ mol⁻¹ at 298 K still gives an equilibrium mixture containing a measurable amount of product. R = 8.31 J K⁻¹ mol⁻¹.

Accepted Answer

B1 ln K = −ΔG ÷ RT = −5000 ÷ (8.31 × 298) = −2.0 B1 K = exp(−2.0) = 0.13, less than 1 but far from zero B1 the equilibrium lies towards the reactants but contains products in appreciable amounts: a positive ΔG means the reaction does not go to completion, not that no product forms

Question 18

Show that the thermal decomposition CaCO₃(s) → CaO(s) + CO₂(g) becomes feasible at about 1110 K. ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹, both taken as constant.

Accepted Answer

M1 the reaction becomes feasible when ΔG = ΔH − TΔS falls to zero, so T = ΔH ÷ ΔS M1 the units must match: ΔH = 178 000 J mol⁻¹, so T = 178 000 ÷ 161 A1 T = 1106 K, about 1110 K; above this temperature TΔS exceeds ΔH, ΔG is negative and the decomposition is feasible

Question 19

State what is meant by entropy, and give its units.

Accepted Answer

B1 entropy is a measure of disorder, the number of ways the particles and their energy can be arranged B1 its units are J K⁻¹ mol⁻¹

Question 20

Identify which of these reactions has the largest positive entropy change, and give a reason: A, H₂(g) + Cl₂(g) → 2HCl(g); B, CaCO₃(s) → CaO(s) + CO₂(g); C, 2SO₂(g) + O₂(g) → 2SO₃(g).

Accepted Answer

B1 reaction B has the largest positive entropy change B1 a solid produces a mole of gas, whereas A has no change in the moles of gas and C has fewer moles of gas in the products

Question 21

Explain why a graph of ΔG against T for a reaction changes gradient at the boiling point of one of the reactants.

Accepted Answer

B1 the gradient of the line is −ΔS B1 when the reactant boils its entropy jumps, so ΔS for the reaction changes and so does the gradient

Question 22

Suggest why a value of ΔG at 1500 K calculated from standard values of ΔH and ΔS measured at 298 K may be inaccurate.

Accepted Answer

B1 ΔH and ΔS both change with temperature, but the calculation assumes they stay at their 298 K values B1 a substance in the reaction may melt or boil between 298 K and 1500 K, changing its entropy and so ΔS

Question 23

State the condition for a reaction to be feasible in terms of ΔG.

Accepted Answer

B1 a reaction is feasible when ΔG is zero or negative

Quantity	Value
ΔH for 2NO₂(g) → N₂O₄(g)	−57.2 kJ mol⁻¹
Standard entropy of NO₂(g)	240 J K⁻¹ mol⁻¹
Standard entropy of N₂O₄(g)	304 J K⁻¹ mol⁻¹

Quantity	Value
ΔH for the decomposition	+117 kJ mol⁻¹
Standard entropy of MgCO₃(s)	66 J K⁻¹ mol⁻¹
Standard entropy of MgO(s)	27 J K⁻¹ mol⁻¹
Standard entropy of CO₂(g)	214 J K⁻¹ mol⁻¹