Level 3 — ProductionThermodynamics (Chemical)

Thermodynamics (Chemical)

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level: 3 (Derivations from scratch, explain-out-loud reasoning, quantitative synthesis) Time limit: 45 minutes Total marks: 60

Use $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $\ln 10 = 2.303$ . Show all reasoning. State every assumption "out loud."

Question 1 — Derive the reversible isothermal work (10 marks)

(a) Starting from the definition of expansion work $\delta w = -P_{ext}\,dV$ , derive from scratch the expression for the work done on an ideal gas during a reversible isothermal expansion from $V_1$ to $V_2$ . State explicitly at each step why you may make the substitutions you make. (6)

(b) Hence explain out loud why $w_{rev}$ (isothermal) is always more negative than $w_{irrev}$ against a constant external pressure for the same expansion, and why work is a path function while $\Delta U$ is a state function. (4)

Question 2 — First law numerics (8 marks)

$2.00\ \text{mol}$ of an ideal gas expands isothermally and reversibly at $300\ \text{K}$ from $5.00\ \text{L}$ to $20.0\ \text{L}$ .

(a) Calculate $w$ , $q$ , and $\Delta U$ for the process (chemist sign convention). (6) (b) Justify your value of $\Delta U$ in one sentence referencing the nature of $U$ for an ideal gas. (2)

Question 3 — $C_p - C_v = nR$ from first principles (10 marks)

(a) Derive the relationship $C_p - C_v = nR$ for one mole of an ideal gas, beginning from $H = U + PV$ and the definitions $C_v = \left(\frac{\partial U}{\partial T}\right)_V$ , $C_p = \left(\frac{\partial H}{\partial T}\right)_P$ . State the ideal-gas fact you invoke. (6)

(b) For an ideal monatomic gas $C_v = \tfrac{3}{2}R$ . Compute $\gamma = C_p/C_v$ and comment on why this differs for a diatomic gas. (4)

Question 4 — Born–Haber cycle from memory (12 marks)

Construct the Born–Haber cycle for solid $\text{NaCl}$ and use it to calculate the lattice enthalpy $\Delta H_{latt}$ (defined as the enthalpy of $\text{Na}^+(g) + \text{Cl}^-(g) \to \text{NaCl}(s)$ ).

Data (kJ mol⁻¹):

Quantity	Value
$\Delta H_f^\circ(\text{NaCl},s)$	$-411$
Sublimation of Na	$+108$
Ionization energy of Na	$+496$
Bond dissociation $\tfrac{1}{2}\text{Cl}_2$	$+122$
Electron affinity of Cl	$-349$

(a) Draw/describe the cycle labelling each step. (5) (b) Calculate $\Delta H_{latt}$ . (5) (c) Explain why lattice enthalpies cannot be measured directly and must be obtained this way. (2)

Question 5 — Gibbs, entropy and equilibrium (12 marks)

For the reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ at $298\ \text{K}$ : $\Delta H^\circ = +57.2\ \text{kJ mol}^{-1}$ , $\Delta S^\circ = +176\ \text{J K}^{-1}\text{mol}^{-1}$ .

(a) Calculate $\Delta G^\circ$ at $298\ \text{K}$ and state whether the reaction is spontaneous under standard conditions. (3) (b) Derive/state the temperature at which the reaction becomes spontaneous, explaining out loud how the sign of $\Delta G$ flips. (3) (c) Using $\Delta G^\circ = -RT\ln K$ , calculate $K$ at $298\ \text{K}$ . (4) (d) Explain the physical meaning of your $K$ value in one sentence. (2)

Question 6 — Coupling reactions (8 marks)

The reaction $\text{A} \to \text{B}$ has $\Delta G_1^\circ = +25\ \text{kJ mol}^{-1}$ (non-spontaneous). The hydrolysis of ATP has $\Delta G_2^\circ = -31\ \text{kJ mol}^{-1}$ .

(a) Explain, using the state-function property of $G$ , how coupling these reactions drives $\text{A}\to\text{B}$ . (3) (b) Calculate $\Delta G^\circ$ for the coupled process and its equilibrium constant at $310\ \text{K}$ . (5)

Answer keyMark scheme & solutions

Question 1 (10)

(a) Start: $\delta w = -P_{ext}\,dV$ . (1) Reversible ⇒ system is always in mechanical equilibrium with surroundings, so $P_{ext} = P_{gas} = P$ at every instant. (1, the key substitution justification) Ideal gas: $P = \dfrac{nRT}{V}$ . (1) $w = -\int_{V_1}^{V_2} \frac{nRT}{V}\,dV$ Isothermal ⇒ $T$ constant, pull $nRT$ out. (1) $w = -nRT\int_{V_1}^{V_2}\frac{dV}{V} = -nRT\ln\frac{V_2}{V_1}$ (2)

(b) For expansion ( $V_2>V_1$ ), $P_{ext}$ in the reversible case is always the (higher) instantaneous gas pressure whereas irreversibly it is a fixed lower value $P_{ext}=P_2$ ; the reversible path integrates over larger pressures at every stage, giving maximum (most negative) work. (2) $w$ depends on the path taken between the same endpoints (rev vs irrev give different values), so it is a path function; $\Delta U$ depends only on initial and final states (here $T$ ), so it is a state function. (2)

Question 2 (8)

(a) $w = -nRT\ln(V_2/V_1) = -(2.00)(8.314)(300)\ln(20/5)$ $= -(4988.4)\ln 4 = -(4988.4)(1.3863) = -6915\ \text{J} \approx -6.92\ \text{kJ}$ . (3) Isothermal ideal gas ⇒ $\Delta U = 0$ . (1) First law $\Delta U = q + w \Rightarrow q = -w = +6915\ \text{J} \approx +6.92\ \text{kJ}$ . (2)

(b) $U$ of an ideal gas depends only on $T$ ; since $T$ is constant, $\Delta U = 0$ . (2)

Question 3 (10)

(a) $H = U + PV$ . For ideal gas $PV = nRT$ , so $H = U + nRT$ . (2) Differentiate w.r.t. $T$ : $\dfrac{dH}{dT} = \dfrac{dU}{dT} + nR$ . (2) For an ideal gas $U$ and $H$ depend only on $T$ , so partial derivatives equal total derivatives: $C_p = C_v + nR \Rightarrow C_p - C_v = nR$ . (2)

(b) $C_p = \tfrac{3}{2}R + R = \tfrac{5}{2}R$ ; $\gamma = \dfrac{5/2}{3/2} = 5/3 \approx 1.667$ . (2) Diatomic gases have extra rotational (and vibrational) degrees of freedom raising $C_v$ to $\tfrac52 R$ , giving $\gamma = 7/5 = 1.4$ ; more ways to store energy ⇒ lower $\gamma$ . (2)

Question 4 (12)

(a) Cycle (all in kJ mol⁻¹): $\text{Na}(s)+\tfrac12\text{Cl}_2(g) \xrightarrow{\Delta H_f=-411} \text{NaCl}(s)$ Alternative route: sublimation (+108) → ionization (+496) → ½ bond dissoc. (+122) → electron affinity (−349) → lattice enthalpy $(\Delta H_{latt})$ . (5)

(b) Hess's law around the cycle: $\Delta H_f = \Delta H_{sub} + IE + \tfrac12 D + EA + \Delta H_{latt}$ $-411 = 108 + 496 + 122 - 349 + \Delta H_{latt}$ $-411 = 377 + \Delta H_{latt} \Rightarrow \Delta H_{latt} = -788\ \text{kJ mol}^{-1}$ (5)

(c) Free gaseous ions combining directly into a crystal is not experimentally realisable in isolation; the value is inferred indirectly via the cycle using measurable quantities. (2)

Question 5 (12)

(a) $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 57200 - 298(176) = 57200 - 52448 = 4752\ \text{J} \approx +4.75\ \text{kJ}$ . Positive ⇒ non-spontaneous under standard conditions. (3)

(b) Spontaneous when $\Delta G < 0 \Rightarrow T > \Delta H/\Delta S = 57200/176 = 325\ \text{K}$ . Above this $T$ , the $T\Delta S$ term outweighs $\Delta H$ , flipping $\Delta G$ negative. (3)

(c) $\ln K = -\dfrac{\Delta G^\circ}{RT} = -\dfrac{4752}{8.314\times298} = -1.918$ ; $K = e^{-1.918} = 0.147$ . (4)

(d) $K<1$ ⇒ at equilibrium reactant $\text{N}_2\text{O}_4$ predominates over $\text{NO}_2$ at 298 K. (2)

Question 6 (8)

(a) $G$ is a state function, so for the summed (coupled) reaction $\Delta G^\circ_{total} = \Delta G_1^\circ + \Delta G_2^\circ$ . If the large negative $\Delta G_2^\circ$ outweighs the positive $\Delta G_1^\circ$ , the overall process becomes spontaneous. (3)

(b) $\Delta G^\circ = +25 + (-31) = -6\ \text{kJ mol}^{-1}$ . (2) $\ln K = -\dfrac{-6000}{8.314\times310} = \dfrac{6000}{2577.3} = 2.328$ ; $K = e^{2.328} = 10.3$ . (3)

[
  {"claim":"Q2 reversible work ≈ -6915 J","code":"import math; w=-2.00*8.314*300*math.log(20/5); result = abs(w-(-6915))<2"},
  {"claim":"Q4 lattice enthalpy = -788 kJ/mol","code":"latt=-411-(108+496+122-349); result = latt==-788"},
  {"claim":"Q5 dG = +4752 J and K ≈ 0.147","code":"import math; dG=57200-298*176; K=math.exp(-dG/(8.314*298)); result = abs(dG-4752)<1 and abs(K-0.147)<0.005"},
  {"claim":"Q5 crossover T = 325 K","code":"T=57200/176; result = abs(T-325)<1"},
  {"claim":"Q6 coupled K ≈ 10.3 at 310 K","code":"import math; dG=-6000; K=math.exp(-dG/(8.314*310)); result = abs(K-10.3)<0.3"}
]