Level 1 — RecognitionThermodynamics (Chemical)

Thermodynamics (Chemical)

30 marksprintable — key stays hidden on paper

Level 1: Recognition (MCQ + Matching + True/False with Justification)

Time: 20 minutes Total Marks: 30

Use $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ where needed.

Section A — Multiple Choice (1 mark each)

Q1. A sealed rigid metal container that allows heat but not matter to cross its boundary is best described as a: (a) Open system (b) Closed system (c) Isolated system (d) Adiabatic system

Q2. Which of the following is a path function? (a) Enthalpy (b) Internal energy (c) Work (d) Entropy

Q3. In the chemist's sign convention for the first law $\Delta U = q + w$ , work done on the system is: (a) Negative (b) Positive (c) Zero (d) Undefined

Q4. For the reversible isothermal expansion of an ideal gas, the work done is: (a) $w = -P_{ext}\Delta V$ (b) $w = -nRT\ln(V_2/V_1)$ (c) $w = +nRT\ln(V_1/V_2)$ (d) $w = 0$

Q5. For an ideal gas, the relationship between molar heat capacities is: (a) $C_p - C_v = R$ (b) $C_v - C_p = R$ (c) $C_p + C_v = R$ (d) $C_p = C_v$

Q6. The standard enthalpy of formation of $O_2(g)$ at 298 K is: (a) $+249\ \text{kJ mol}^{-1}$ (b) $-286\ \text{kJ mol}^{-1}$ (c) $0$ (d) $+498\ \text{kJ mol}^{-1}$

Q7. Hess's law is a direct consequence of the fact that enthalpy is a: (a) Path function (b) State function (c) Extensive-only property (d) Colligative property

Q8. For a reaction that is spontaneous at all temperatures, the signs of $\Delta H$ and $\Delta S$ must be: (a) $\Delta H > 0,\ \Delta S < 0$ (b) $\Delta H < 0,\ \Delta S > 0$ (c) $\Delta H < 0,\ \Delta S < 0$ (d) $\Delta H > 0,\ \Delta S > 0$

Q9. At equilibrium, $\Delta G^\circ = -RT\ln K$ . If $K > 1$ , then $\Delta G^\circ$ is: (a) Positive (b) Zero (c) Negative (d) Cannot be determined

Q10. Which process is expected to have $\Delta S > 0$ ? (a) Freezing of water (b) $2NO_2(g)\to N_2O_4(g)$ (c) Sublimation of iodine (d) Condensation of steam

Section B — Matching (1 mark each row; 5 marks)

Q11. Match the term (Column I) with its correct description (Column II).

Column I	Column II
(i) Enthalpy of neutralization	(P) Energy released when 1 mol of gaseous ions forms a solid lattice
(ii) Lattice energy	(Q) Heat change when 1 mol substance burns completely in $O_2$
(iii) Enthalpy of combustion	(R) Heat released when acid + base form 1 mol water
(iv) Enthalpy of hydration	(S) $\Delta H = U + PV$ at constant pressure
(v) Enthalpy (definition)	(T) Heat change when 1 mol gaseous ion is dissolved in water

Section C — True/False WITH Justification (2 marks each: 1 T/F + 1 justification; 15 marks)

Q12. "For an ideal gas undergoing free expansion into a vacuum, the work done is zero."

Q13. "The enthalpy of formation of an element in its standard state is always negative."

Q14. " $\Delta H$ for a reaction can be estimated as (sum of bond enthalpies of bonds broken) − (sum of bond enthalpies of bonds formed)."

Q15. "A reaction with positive $\Delta G$ can never be driven forward by any means."

Q16. "In an isolated system, both energy and matter can be exchanged with the surroundings."

Q17. "For an ideal gas, $C_p$ is greater than $C_v$ ."

Q18. "The second law states that for a spontaneous process the total entropy of the universe increases."

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1 — (b) Closed system. Why: Heat crosses (thermal contact) but matter cannot (sealed). Open = matter+energy; isolated = neither.

Q2 — (c) Work. Why: Work depends on the path taken; enthalpy, internal energy and entropy are state functions.

Q3 — (b) Positive. Why: In the chemist convention $\Delta U = q + w$ , work done on the system adds energy, so $w>0$ (compression).

Q4 — (b) $w = -nRT\ln(V_2/V_1)$ . Why: Reversible isothermal work integrates $P=nRT/V$ ; expansion ( $V_2>V_1$ ) gives negative $w$ .

Q5 — (a) $C_p - C_v = R$ . Why: For 1 mol ideal gas, $C_p - C_v = nR = R$ .

Q6 — (c) 0. Why: $\Delta H^\circ_f$ of an element in its most stable form is defined as zero.

Q7 — (b) State function. Why: Enthalpy depends only on initial/final states, so the total $\Delta H$ is path-independent (Hess's law).

Q8 — (b) $\Delta H<0,\ \Delta S>0$ . Why: $\Delta G = \Delta H - T\Delta S < 0$ at all $T$ only when $\Delta H<0$ and $\Delta S>0$ .

Q9 — (c) Negative. Why: $K>1 \Rightarrow \ln K>0 \Rightarrow \Delta G^\circ = -RT\ln K <0$ .

Q10 — (c) Sublimation of iodine. Why: Solid → gas greatly increases disorder ( $\Delta S>0$ ); the other options decrease gas moles/order.

Section B

Q11 (5 marks, 1 each):

(i) → (R) neutralization = acid+base → 1 mol water
(ii) → (P) lattice energy = gaseous ions → solid lattice
(iii) → (Q) combustion = complete burning in $O_2$
(iv) → (T) hydration = gaseous ion dissolved in water
(v) → (S) $H = U + PV$

Section C (2 marks: 1 for T/F, 1 for justification)

Q12 — TRUE. Justification: Free expansion into vacuum means $P_{ext}=0$ , so $w=-P_{ext}\Delta V = 0$ .

Q13 — FALSE. Justification: $\Delta H^\circ_f$ of an element in its standard state is zero, not negative.

Q14 — TRUE. Justification: Breaking bonds absorbs energy (+), forming bonds releases energy (−); net $\Delta H \approx \Sigma(\text{broken}) - \Sigma(\text{formed})$ .

Q15 — FALSE. Justification: An unfavorable ( $\Delta G>0$ ) reaction can be coupled to a strongly favorable reaction; if the summed $\Delta G<0$ the overall process is spontaneous.

Q16 — FALSE. Justification: An isolated system exchanges neither energy nor matter with the surroundings; a closed system exchanges energy only, open exchanges both.

Q17 — TRUE. Justification: $C_p - C_v = R > 0$ , so $C_p > C_v$ (extra energy goes into $PV$ work at constant pressure).

Q18 — TRUE. Justification: Second law: $\Delta S_{universe} = \Delta S_{sys} + \Delta S_{surr} > 0$ for any spontaneous process.

Total = 10 (Sec A) + 5 (Sec B) + 15 (Sec C) = 30 marks

[
  {"claim":"Cp - Cv = R for ideal gas (numeric R value)","code":"R=8.314\nCp=R+20\nCv=20\nresult = abs((Cp-Cv)-R)<1e-9"},
  {"claim":"K>1 implies dG<0 via dG=-RT lnK","code":"R=8.314; T=298; K=10\ndG=-R*T*ln(K)\nresult = dG<0"},
  {"claim":"Free expansion into vacuum: w=-Pext*dV with Pext=0 gives 0","code":"Pext=0; dV=5\nw=-Pext*dV\nresult = w==0"},
  {"claim":"Reversible isothermal expansion V2>V1 gives negative work","code":"n=1;R=8.314;T=300;V1=1;V2=2\nw=-n*R*T*ln(V2/V1)\nresult = w<0"}
]