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Thermodynamics — NEET & JEE MCQs with answers
Unit 5 introduces thermodynamic concepts: systems, state functions, internal energy, work, and heat. The First Law of Thermodynamics ($\Delta U = q + w$) relates energy changes to heat and work. Enthalpy is defined as $H = U + pV$; at constant pressure the heat absorbed by the system equals the change in enthalpy, $qp = \Delta H$. The unit covers heat capacity, Hess's law, enthalpy of reactions, bond enthalpies, and the spontaneity criteria introduced via entropy and the Second Law.
Topics in this chapter
- 5.1 Thermodynamic Terms
- 5.2 Applications of Thermodynamic Concepts
- 5.3 Measurement of ΔU and ΔH: Calorimetry
- 5.4 Enthalpy Change, ΔrH of a Reaction - Reaction Enthalpy
- 5.5 Enthalpies for Different Types of Reactions
- 5.6 Spontaneity
- 5.7 Gibbs Energy Change and Equilibrium
Free practice questions
The enthalpy of sublimation of dry ice ($\text{CO}_2$) is 25.2 kJ/mol at 195 K. To sublime 0.5 mol of dry ice, the heat required is:
- A. 50.4 kJ
- B. 25.2 kJ
- C. 6.3 kJ
- D. 12.6 kJ
Show answer
Answer: D — Heat = $n \times \Delta_{sub}H = 0.5 \times 25.2 = 12.6$ kJ. Sublimation (solid → gas) is endothermic.
NCERT Class 11 Chemistry · Ch 5 · 5.2 Applications of Thermodynamic Concepts
Strongly exothermic reactions may be expected to have a large $K$. From this the text concludes that such reactions:
- A. are unlikely to form much product
- B. are nonspontaneous at all temperatures
- C. have a free energy that is maximum at equilibrium
- D. can go to near completion
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Answer: D — NCERT p.162: 'We may expect strongly exothermic reactions to have a large $K$, and hence can go to near completion.' Forming little product is said of strongly endothermic reactions, and the free energy is minimum (not maximum) at equilibrium.
NCERT Class 11 Chemistry · Ch 5 Thermodynamics · 5.7 Gibbs Energy Change and Equilibrium · p.162
The presence of reactants in a closed vessel made of copper or steel is an example of:
- A. A closed system
- B. An open system
- C. A system with imaginary boundary
- D. An isolated system
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Answer: A — NCERT: 'The presence of reactants in a closed vessel made of conducting material e.g., copper or steel is an example of a closed system.' Energy (heat) can pass through the conducting wall but matter cannot.
NCERT Class 11 Chemistry · Ch 5 · 5.1 Thermodynamic Terms
In the stepwise atomization of methane ($+427$, $+439$, $+452$, $+347\ \text{kJ mol}^{-1}$), the largest step enthalpy exceeds the smallest by:
- A. $80\ \text{kJ mol}^{-1}$
- B. $92\ \text{kJ mol}^{-1}$
- C. $25\ \text{kJ mol}^{-1}$
- D. $105\ \text{kJ mol}^{-1}$
Show answer
Answer: D — Largest $= 452$ (third step), smallest $= 347$ (fourth step); difference $= 105\ \text{kJ mol}^{-1}$ (p.154). $25$ is $452 - 427$, $92$ is $439 - 347$ and $80$ is $427 - 347$.
NCERT Class 11 Chemistry · Ch 5 Thermodynamics · 5.5(c) Bond Enthalpy · p.154
Which of the following reactions, quoted in NCERT with its $\Delta_rH^\ominus$, is endothermic and yet spontaneous?
- A. $H_2(g) + \tfrac12O_2(g) \rightarrow H_2O(l)$
- B. $\tfrac12N_2(g) + O_2(g) \rightarrow NO_2(g)$
- C. Formation of $NH_3(g)$ from $N_2(g)$ and $H_2(g)$
- D. $\tfrac12H_2(g) + \tfrac12Cl_2(g) \rightarrow HCl(g)$
Show answer
Answer: B — p.157: $\tfrac12N_2 + O_2 \rightarrow NO_2$ has $\Delta_rH^\ominus = +33.2\ kJ\,mol^{-1}$ and p.158 says such reactions 'though endothermic, are spontaneous'. The HCl ($-92.32$), $H_2O(l)$ ($-285.8$) and $NH_3$ ($-46.1\ kJ\,mol^{-1}$) reactions are all exothermic.
NCERT Class 11 Chemistry, Ch 5 Thermodynamics, 5.6 Spontaneity (a) Is Decrease in Enthalpy a Criterion for Spontaneity?, p.157
For an endothermic reaction carried out at constant pressure:
- A. heat is absorbed, $q_p$ is negative and $\Delta_r H$ is negative
- B. heat is evolved, $q_p$ is positive and $\Delta_r H$ is positive
- C. heat is evolved, $q_p$ is negative and $\Delta_r H$ is negative
- D. heat is absorbed, $q_p$ is positive and $\Delta_r H$ is positive
Show answer
Answer: D — Page 146: 'in an endothermic reaction, heat is absorbed, $q_p$ is positive and $\Delta_r H$ will be positive'. Heat evolved with negative signs is the exothermic case; the other two options mix the direction of heat flow with the wrong sign.
NCERT Class 11 Chemistry · Ch 5 · 5.3 Measurement of ΔU and ΔH: Calorimetry · p.146
Table 5.1 gives $\Delta_{vap}H^\ominus = 29.1\ kJ\,mol^{-1}$ and $\Delta_{fus}H^\ominus = 5.72\ kJ\,mol^{-1}$ for acetone ($CH_3COCH_3$). The heat required to vaporise 2.5 mol of liquid acetone at its boiling point is:
- A. $14.30$ kJ
- B. $29.10$ kJ
- C. $72.75$ kJ
- D. $101.98$ kJ
Show answer
Answer: C — Heat $= n\,\Delta_{vap}H^\ominus = 2.5\ mol \times 29.1\ kJ\,mol^{-1} = 72.75$ kJ (Table 5.1, p.147). 14.30 kJ uses the fusion value 5.72 instead; 29.10 kJ is for 1 mol only; 101.98 kJ (101.975 rounded) uses water's 40.79 instead of acetone's value.
NCERT Class 11 Chemistry · Ch 5 Thermodynamics · §5.4 (b) Enthalpy Changes during Phase Transformations · Table 5.1 · p.147
In an isothermal irreversible expansion of an ideal gas, the work done is $w = -810\,\text{J}$. The heat absorbed $q$ is:
- A. $+810\,\text{J}$
- B. $-810\,\text{J}$
- C. $+1620\,\text{J}$
- D. $0$
Show answer
Answer: A — For an isothermal process of an ideal gas $\Delta U = 0$, so $q = -w = -(-810) = +810\,\text{J}$.
NCERT Class 11 Chemistry · Ch 5 · 5.1 Thermodynamic Terms
The energy our body obtains from food (e.g. glucose) is released by essentially the same overall process as:
- A. the combustion (oxidation) of the food in oxygen
- B. the neutralisation of acids in the stomach
- C. the formation of the food from its elements
- D. the sublimation of solid food
Show answer
Answer: A — The body extracts energy from food through oxidation, which is the same overall combustion process as burning the substance; hence the standard enthalpy of combustion measures the energy available.
NCERT Class 11 Chemistry · Ch 5 · 5.2 Applications of Thermodynamic Concepts
For $N_2O_4(g) \rightleftharpoons 2NO_2(g)$ with $N_2O_4$ 50 per cent dissociated, the mole fraction of $NO_2$ in the worked solution is given by:
- A. $\dfrac{1 - 0.5}{1 + 0.5}$
- B. $\dfrac{2 \times 0.5}{1 - 0.5}$
- C. $\dfrac{2 \times 0.5}{1 + 0.5}$
- D. $\dfrac{0.5}{1 + 0.5}$
Show answer
Answer: C — NCERT p.163, Problem 5.14: $x_{NO_2} = \dfrac{2 \times 0.5}{1 + 0.5} = \dfrac{1}{1.5}$. The expression $\dfrac{1-0.5}{1+0.5}$ is the mole fraction of $N_2O_4$; the factor 2 is needed because each $N_2O_4$ gives two $NO_2$, and the total is $1 + 0.5$, not $1 - 0.5$.
NCERT Class 11 Chemistry · Ch 5 Thermodynamics · 5.7 Gibbs Energy Change and Equilibrium · p.163
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