Level 4 — ApplicationThermodynamics (Chemical)

Thermodynamics (Chemical)

50 marksprintable — key stays hidden on paper

Level: 4 (Application — novel problems, no hints) Time: 60 minutes Total marks: 50

Use $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ unless a value is supplied. Assume ideal gas behaviour where relevant.

Question 1 (10 marks)

Two moles of an ideal monatomic gas ( $C_{v,m} = \tfrac{3}{2}R$ ) at $300\ \text{K}$ occupy $10.0\ \text{L}$ . The gas is expanded isothermally to $30.0\ \text{L}$ .

(a) Calculate $q$ , $w$ , $\Delta U$ and $\Delta H$ if the expansion is carried out reversibly. (5)

(b) The same initial-to-final change is instead carried out irreversibly against a constant external pressure equal to the final gas pressure. Calculate $w$ and $q$ for this path. (3)

(c) State, with a one-line justification, which quantities in (a) and (b) differ and which are identical, relating this to state vs path functions. (2)

Question 2 (10 marks)

Consider the reaction at $298\ \text{K}$ and constant pressure:

$\text{C}_2\text{H}_4(g) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)$

(a) Using the bond enthalpies below, estimate $\Delta H_{rxn}$ : C=C $612$ , C–C $348$ , C–H $412$ , H–H $436$ (all $\text{kJ mol}^{-1}$ ). (4)

(b) The value from standard enthalpies of formation is $-137\ \text{kJ mol}^{-1}$ . Account for the discrepancy in one or two sentences. (2)

(d) Predict, without a numerical entropy calculation, the sign of $\Delta S_{rxn}$ and hence comment on how $\Delta G$ behaves as temperature rises. (2)

Question 3 (10 marks)

Construct a Born–Haber cycle to find the lattice enthalpy $\Delta H_L$ (defined as $\text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{NaCl}(s)$ ) from:

$\Delta H^\circ_f[\text{NaCl}(s)] = -411\ \text{kJ mol}^{-1}$
Sublimation of Na $= +108$
$\tfrac{1}{2}$ bond dissociation of $\text{Cl}_2 = +121$
Ionisation energy of Na $= +496$
Electron affinity of Cl $= -349$ (all $\text{kJ mol}^{-1}$ )

(a) Draw/describe the cycle and calculate $\Delta H_L$ . (6)

(b) A student proposes that replacing Na by K (larger cation) would make $\Delta H_L$ more exothermic. Evaluate this claim. (2)

Question 4 (12 marks)

For the reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\,\text{NO}_2(g)$ :

$\Delta H^\circ = +57.2\ \text{kJ mol}^{-1}, \qquad \Delta S^\circ = +175.8\ \text{J K}^{-1}\text{mol}^{-1}$

(a) Calculate $\Delta G^\circ$ at $298\ \text{K}$ and state whether the forward reaction is spontaneous under standard conditions. (3)

(b) Determine the temperature above which the forward reaction becomes spontaneous at standard conditions. (3)

(d) Estimate $K$ at $400\ \text{K}$ assuming $\Delta H^\circ$ and $\Delta S^\circ$ are temperature-independent, and comment on the direction of shift. (3)

Question 5 (8 marks)

At $298\ \text{K}$ the reaction

$\text{ATP}^{4-} + \text{H}_2\text{O} \rightarrow \text{ADP}^{3-} + \text{HPO}_4^{2-} + \text{H}^+ \qquad \Delta G^\circ = -30.5\ \text{kJ mol}^{-1}$

is used to drive the phosphorylation of glucose:

$\text{glucose} + \text{HPO}_4^{2-} + \text{H}^+ \rightarrow \text{glucose-6-phosphate} + \text{H}_2\text{O} \qquad \Delta G^\circ = +13.8\ \text{kJ mol}^{-1}$

(a) Write the coupled overall reaction and calculate its $\Delta G^\circ$ . (3)

(b) Determine the equilibrium constant of the coupled reaction at $298\ \text{K}$ . (3)

Answer keyMark scheme & solutions

Question 1 (10)

(a) Reversible isothermal, 2 mol, 300 K, $10 \to 30$ L

$w = -nRT\ln(V_2/V_1) = -(2)(8.314)(300)\ln 3$ $\ln 3 = 1.0986$ , so $w = -4988.4 \times 1.0986 = -5480\ \text{J} \approx -5.48\ \text{kJ}$ (1) Isothermal ideal gas ⇒ $\Delta U = 0$ (1), $\Delta H = 0$ (1) (both depend only on $T$ ). First law $\Delta U = q + w = 0 \Rightarrow q = -w = +5.48\ \text{kJ}$ (1) Sign logic: gas does work on surroundings ( $w<0$ ), absorbs equal heat. (1)

(b) Irreversible against $P_{ext}=P_{final}$

$P_{final} = \dfrac{nRT}{V_2} = \dfrac{2 \times 8.314 \times 300}{30\times10^{-3}} = 1.663\times10^{5}\ \text{Pa}$ (1) $w = -P_{ext}\Delta V = -1.663\times10^{5}(30-10)\times10^{-3} = -3326\ \text{J} \approx -3.33\ \text{kJ}$ (1) $\Delta U = 0$ (state function, same endpoints) ⇒ $q = +3.33\ \text{kJ}$ (1)

(c) $\Delta U$ and $\Delta H$ are identical in both (state functions, path-independent). $q$ and $w$ differ (path functions); the reversible path extracts maximum work. (2)

Question 2 (10)

(a) Bonds broken (reactants): C=C + 4(C–H) + H–H $= 612 + 4(412) + 436 = 2696$ Bonds formed (product ethane): C–C + 6(C–H) $= 348 + 6(412) = 2820$ (2) $\Delta H = \sum\text{broken} - \sum\text{formed} = 2696 - 2820 = -124\ \text{kJ mol}^{-1}$ (2)

(b) Bond enthalpies are averaged over many molecules, not exact for these species; hence $-124$ vs true $-137\ \text{kJ mol}^{-1}$ (≈13 kJ discrepancy). (2)

(c) $\Delta n_g = 1 - 2 = -1$ . $\Delta U = \Delta H - \Delta n_g RT = -137000 - (-1)(8.314)(298) = -137000 + 2478 = -134522\ \text{J} \approx -134.5\ \text{kJ mol}^{-1}$ (2)

(d) $\Delta n_g = -1$ (2 gas mol → 1) ⇒ $\Delta S_{rxn} < 0$ . (1) With $\Delta H<0$ , $\Delta S<0$ : $\Delta G=\Delta H-T\Delta S$ becomes less negative as $T$ rises, so reaction becomes less favourable (eventually non-spontaneous) at high $T$ . (1)

Question 3 (10)

(a) Cycle (Hess): $\Delta H_f = \Delta H_{sub} + \tfrac12 D(\text{Cl}_2) + IE + EA + \Delta H_L$ (2) Solve for $\Delta H_L$ : $\Delta H_L = \Delta H_f - (\Delta H_{sub} + \tfrac12 D + IE + EA)$ $= -411 - (108 + 121 + 496 - 349)$ $= -411 - 376 = -787\ \text{kJ mol}^{-1}$ (4)

(b) False as stated. Larger K⁺ increases interionic distance, so $|\Delta H_L|$ decreases (less exothermic), by $\Delta H_L \propto 1/r$ . (2)

(c) Sign is negative (exothermic) — consistent with a stable ionic lattice releasing energy on formation from gaseous ions. (2)

Question 4 (12)

(a) $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 57200 - 298(175.8) = 57200 - 52388 = 4812\ \text{J} \approx +4.81\ \text{kJ mol}^{-1}$ (2) $\Delta G^\circ>0$ ⇒ forward reaction non-spontaneous at standard conditions, 298 K. (1)

(b) Spontaneous when $\Delta G^\circ<0$ : $T > \dfrac{\Delta H^\circ}{\Delta S^\circ} = \dfrac{57200}{175.8} = 325.4\ \text{K}$ (3)

(c) $\Delta G^\circ = -RT\ln K \Rightarrow \ln K = -\dfrac{4812}{8.314\times298} = -1.943$ $K = e^{-1.943} = 0.143$ (3)

(d) At 400 K: $\Delta G^\circ = 57200 - 400(175.8) = 57200 - 70320 = -13120\ \text{J}$ $\ln K = -\dfrac{-13120}{8.314\times400} = 3.946 \Rightarrow K = e^{3.946} = 51.7$ (2) $K$ rises sharply; equilibrium shifts toward $\text{NO}_2$ (endothermic favoured by heating — Le Chatelier consistent). (1)

Question 5 (8)

(a) Adding the two reactions (HPO₄²⁻, H⁺, H₂O cancel appropriately): $\text{ATP}^{4-} + \text{glucose} \rightarrow \text{ADP}^{3-} + \text{glucose-6-phosphate}$ (2) $\Delta G^\circ_{total} = -30.5 + 13.8 = -16.7\ \text{kJ mol}^{-1}$ (1)

(b) $\ln K = -\dfrac{\Delta G^\circ}{RT} = -\dfrac{-16700}{8.314\times298} = 6.741$ $K = e^{6.741} = 846$ (3)

(c) $\Delta G$ is a state function and additive; a large negative $\Delta G^\circ$ (ATP hydrolysis) coupled to a positive one makes the sum negative, so the overall process is spontaneous even though one step alone is not. (2)

[
  {"claim":"Q1a reversible work ≈ -5480 J","code":"import sympy as sp\nw=-2*8.314*300*sp.log(3)\nresult = abs(float(w)+5480) < 5"},
  {"claim":"Q2a bond-energy ΔH = -124 kJ/mol","code":"broken=612+4*412+436\nformed=348+6*412\nresult = (broken-formed)==-124"},
  {"claim":"Q3 lattice enthalpy = -787 kJ/mol","code":"dHL=-411-(108+121+496-349)\nresult = dHL==-787"},
  {"claim":"Q4b crossover T ≈ 325.4 K","code":"T=57200/175.8\nresult = abs(T-325.4)<0.5"},
  {"claim":"Q4c K(298) ≈ 0.143","code":"import sympy as sp\nK=sp.exp(-4812/(8.314*298))\nresult = abs(float(K)-0.143)<0.005"},
  {"claim":"Q5b coupled K ≈ 846","code":"import sympy as sp\nK=sp.exp(16700/(8.314*298))\nresult = abs(float(K)-846)<10"}
]