How to Use This Page
This page provides complete exercise solutions for Chapter 7 — Thermodynamics of the Samacheer Kalvi Class 11 Chemistry textbook. Follow this approach:
1. Key Concepts — revise all formulas and definitions first. 2. MCQs through Match the Following — test your recall and conceptual clarity. 3. Brief and Detailed Answers — practice writing structured, exam-ready responses. 4. Worked Numericals — follow each step; practise the method, not just the answer. 5. Common Mistakes & Exam Checklist — final review before your exam.
Key Concepts (Quick Recap)
| Concept | Formula / Statement |
|---|
| First Law | ΔU = q + w (energy conserved) |
| Work (PV work) | w = −PΔV (expansion work at constant pressure) |
| Enthalpy | H = U + PV; ΔH = ΔU + PΔV (at constant P) |
| ΔH at constant P | ΔH = q_p (heat absorbed at constant pressure) |
| ΔU at constant V | ΔU = q_v (heat absorbed at constant volume) |
| Relation ΔH and ΔU | ΔH = ΔU + Δn_g RT (for gaseous reactions) |
| Heat Capacity | C_p = ΔH/ΔT (at constant P); C_v = ΔU/ΔT (at constant V) |
| C_p − C_v | C_p − C_v = R (for 1 mole of ideal gas) |
| Hess's Law | ΔH_total = ΔH₁ + ΔH₂ + ΔH₃ + ... (path independent) |
| Standard Enthalpy of Formation | ΔH°f of elements in standard state = 0 |
| Bond Enthalpy | ΔH_reaction = Σ(bond energies of reactants) − Σ(bond energies of products) |
| Entropy | ΔS = q_rev/T; ΔS_total = ΔS_system + ΔS_surroundings |
| Second Law | For a spontaneous process: ΔS_total > 0 |
| Gibbs Free Energy | G = H − TS; ΔG = ΔH − TΔS |
| Spontaneity Conditions | ΔG < 0 → spontaneous; ΔG > 0 → non-spontaneous; ΔG = 0 → equilibrium |
Spontaneity Summary Table:
| ΔH | ΔS | ΔG = ΔH − TΔS | Spontaneity |
|---|
| − (exothermic) | + | Always − | Spontaneous at all T |
| − (exothermic) | − | − at low T, + at high T | Spontaneous at low T |
| + (endothermic) | + | + at low T, − at high T | Spontaneous at high T |
| + (endothermic) | − | Always + | Non-spontaneous at all T |
I. Choose the Best Answer (MCQs)
1. Which of the following is a state function?
Answer
(a) Heat (q) (b) Work (w) (c) Enthalpy (H) (d) Path-dependent quantity
Answer(c) Enthalpy (H) depends only on the initial and final states, not on the path taken. Heat and work are path functions — their values depend on how the change is carried out.
2. For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), Δn_g is:
Answer
(a) +2 (b) −2 (c) +1 (d) −1
Answer(b)
Δn_g = n_products(g) − n_reactants(g)
Δn_g = 2 − (1 + 3) = 2 − 4 = −2
3. If ΔH = −92 kJ and ΔS = −199 J/K for the reaction N₂ + 3H₂ → 2NH₃, the reaction is spontaneous:
Answer
(a) At all temperatures (b) At low temperatures (c) At high temperatures (d) Never spontaneous
Answer(b)
| ΔH is negative and ΔS is negative. ΔG = ΔH − TΔS. At low temperatures, | ΔH | > | TΔS | , so ΔG < 0 (spontaneous). At high temperatures, | TΔS | > | ΔH | , making ΔG > 0 (non-spontaneous). |
|---|
4. The standard enthalpy of formation of CO₂(g) is −393.5 kJ/mol. This means:
Answer
(a) Burning CO₂ releases 393.5 kJ/mol (b) Forming 1 mol of CO₂ from C(s) and O₂(g) releases 393.5 kJ (c) CO₂ decomposes absorbing 393.5 kJ (d) The enthalpy of CO₂ is −393.5 kJ
Answer(b) The standard enthalpy of formation is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states. The negative sign indicates heat is released (exothermic).
C(s) + O₂(g) → CO₂(g) ΔH°f = −393.5 kJ/mol
5. For an ideal gas expanding isothermally, which is true?
Answer
(a) ΔU > 0 (b) ΔU < 0 (c) ΔU = 0 (d) ΔH ≠ 0
Answer(c) For an ideal gas, internal energy (U) depends only on temperature. In an isothermal process, T is constant, so ΔU = 0. Since H = U + PV and both U and PV (= nRT = constant) don't change, ΔH = 0 as well.
6. The work done when 2 moles of an ideal gas expand from 5 L to 10 L against a constant external pressure of 1 atm at 300 K is:
Answer
(a) +5 L·atm (b) −5 L·atm (c) +10 L·atm (d) −10 L·atm
Answer(b)
w = −P_ext × ΔV = −P_ext × (V₂ − V₁)
w = −1 × (10 − 5) = −5 L·atm
The negative sign indicates work is done BY the system (expansion).
7. Which of the following has the highest standard entropy?
Answer
(a) Diamond (b) Graphite (c) Liquid water (d) Water vapour
Answer(d) Entropy increases with molecular randomness. Diamond (rigid 3D network) has the lowest entropy, followed by graphite (layered), then liquid water (moderate freedom), and water vapour (maximum disorder) has the highest.
II. Fill in the Blanks
1. The first law of thermodynamics is mathematically expressed as ΔU = q + w. 2. At constant volume, the work done by the system is zero because ΔV = 0. 3. The enthalpy change for the formation of 1 mole of a compound from its elements in standard states is called standard enthalpy of formation. 4. For a spontaneous process, the total entropy change ΔS_total must be greater than zero (positive). 5. The relationship between ΔH and ΔU for a gaseous reaction is ΔH = ΔU + Δn_g RT. 6. Hess's law is a consequence of enthalpy being a state function. 7. The SI unit of entropy is J K⁻¹ mol⁻¹. 8. When ΔH < 0 and ΔS > 0, the reaction is spontaneous at all temperatures. 9. The bond dissociation enthalpy of H₂ is 436 kJ/mol. 10. For an isothermal expansion of an ideal gas, ΔU = 0. 11. The standard state pressure is 1 bar (or 1 atm). 12. If q = 100 J and w = −30 J, then ΔU = 70 J. 13. The heat capacity at constant pressure is related to enthalpy by C_p = (∂H/∂T)_p. 14. The condition for equilibrium is ΔG = 0. 15. When a solid melts, the entropy change is positive (disorder increases).
III. True or False (with Corrections)
1. Heat and work are state functions.
False. Heat and work are path functions — they depend on how the process is carried out, not just on the initial and final states. State functions include U, H, S, G, P, V, T.
2. For an exothermic reaction, ΔH is positive.
False. For an exothermic reaction, heat is released by the system, so ΔH is negative. For an endothermic reaction, heat is absorbed and ΔH is positive.
3. The enthalpy of an ideal gas depends only on temperature.
True. For an ideal gas, H = U + PV = U + nRT. Since U depends only on T and nRT is also a function of T only, enthalpy depends only on temperature for an ideal gas.
4. A reaction with ΔS > 0 is always spontaneous.
False. Spontaneity depends on ΔG = ΔH − TΔS, not on ΔS alone. A reaction with ΔS > 0 could still be non-spontaneous if ΔH is very positive and T is low enough that ΔG > 0.
5. Hess's law applies only to enthalpy changes, not to entropy changes.
False. Since entropy is also a state function, Hess's law applies to entropy changes as well. The total entropy change is the same regardless of the path taken.
IV. Match the Following
| Column A | Column B |
|---|
| 1. Isothermal process | (a) ΔU = q (no PV work) |
| 2. Isobaric process | (b) ΔH = q_p |
| 3. Isochoric process | (c) ΔT = 0 |
| 4. Adiabatic process | (d) q = 0 |
| 5. Cyclic process | (e) ΔU = 0, ΔH = 0 |
| 6. Spontaneous process | (f) ΔG < 0 |
Answer
1 → (c) Isothermal: ΔT = 0 (constant temperature)
2 → (b) Isobaric: ΔH = q_p (constant pressure, heat = enthalpy change)
3 → (a) Isochoric: ΔU = q (constant volume, no PV work)
4 → (d) Adiabatic: q = 0 (no heat exchange)
5 → (e) Cyclic: ΔU = 0, ΔH = 0 (returns to initial state)
6 → (f) Spontaneous: ΔG < 0 (Gibbs energy decreases)
V. Answer Briefly (2-mark)
1. Distinguish between intensive and extensive properties with two examples each.
Answer
Intensive properties are independent of the amount of matter: temperature, pressure, density, molar heat capacity. Extensive properties depend on the amount of matter: volume, mass, internal energy, enthalpy, entropy.
2. Define standard enthalpy of formation. Give one example.
Answer
The standard enthalpy of formation (ΔH°f) is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states (1 atm, 298 K).
Example: C(s) + O₂(g) → CO₂(g) ΔH°f = −393.5 kJ/mol
3. What is the relationship between ΔH and ΔU for a reaction involving gases?
Answer
For a reaction involving gases:
ΔH = ΔU + Δn_g RT
where Δn_g = (moles of gaseous products) − (moles of gaseous reactants)
R = 8.314 J mol⁻¹ K⁻¹
T = temperature in Kelvin
If Δn_g = 0, then ΔH = ΔU. If Δn_g > 0, ΔH > ΔU. If Δn_g < 0, ΔH < ΔU.
4. State the second law of thermodynamics.
Answer
The second law states that in any spontaneous process, the total entropy of the system and surroundings always increases. A process is spontaneous only if ΔS_total = ΔS_system + ΔS_surroundings > 0. For a reversible process, ΔS_total = 0.
VI. Answer in Detail (5-mark)
1. Explain the first law of thermodynamics. A system absorbs 500 J of heat and does 200 J of work on the surroundings. Calculate ΔU. If the same change were brought about at constant volume, how much heat would be absorbed?
Answer
The first law of thermodynamics states that energy can neither be created nor destroyed, though it can change from one form to another. The total energy of the universe remains constant.
Mathematical form: ΔU = q + w
Where:
ΔU = change in internal energy
q = heat absorbed by the system
w = work done ON the system
Convention:
- q > 0: heat absorbed by system
- q < 0: heat released by system
- w > 0: work done on system
- w < 0: work done by system
Numerical calculation:
Given: q = +500 J (heat absorbed), work done BY system = 200 J
So: w = −200 J (work done ON the system is negative when system expands)
ΔU = q + w = 500 + (−200) = +300 J
The internal energy increases by 300 J.
At constant volume (ΔV = 0):
w = −PΔV = 0
ΔU = q_v = 300 J
At constant volume, all heat goes into changing internal energy.
The system would absorb only 300 J (instead of 500 J) to achieve
the same ΔU, because no energy is spent doing PV work.
2. Explain Hess's law with an example. Calculate the enthalpy of formation of CH₄ using the following data.
Answer
Hess's law states that the total enthalpy change for a chemical reaction is the same regardless of whether it occurs in one step or in multiple steps. This is a direct consequence of enthalpy being a state function — the enthalpy change depends only on the initial and final states, not the path.
Example — Formation of CO₂:
Path 1 (Direct):
C(s) + O₂(g) → CO₂(g) ΔH = −393.5 kJ
Path 2 (Two steps):
C(s) + ½O₂(g) → CO(g) ΔH₁ = −110.5 kJ
CO(g) + ½O₂(g) → CO₂(g) ΔH₂ = −283.0 kJ
Total: ΔH₁ + ΔH₂ = −110.5 + (−283.0) = −393.5 kJ
Both paths give the same result, confirming Hess's law.
Numerical — Enthalpy of formation of CH₄:
Given data:
(1) C(s) + O₂(g) → CO₂(g) ΔH = −393.5 kJ
(2) H₂(g) + ½O₂(g) → H₂O(l) ΔH = −285.8 kJ
(3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890.3 kJ
Target reaction:
C(s) + 2H₂(g) → CH₄(g) ΔH = ?
Using Hess's law, rearrange the given equations:
Eq(1): C(s) + O₂(g) → CO₂(g) ΔH₁ = −393.5 kJ
2 × Eq(2): 2H₂(g) + O₂(g) → 2H₂O(l) ΔH₂ = 2(−285.8) = −571.6 kJ
Reverse Eq(3): CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g) ΔH₃ = +890.3 kJ
Adding all three:
C(s) + 2H₂(g) → CH₄(g)
ΔH = (−393.5) + (−571.6) + (890.3) = −74.8 kJ
The standard enthalpy of formation of CH₄ is −74.8 kJ/mol.
3. Explain Gibbs free energy and its role in determining spontaneity. Calculate ΔG at 298 K for a reaction with ΔH = −92 kJ and ΔS = −199 J/K. Is the reaction spontaneous? At what temperature does it become non-spontaneous?
Answer
Gibbs free energy (G) is defined as G = H − TS. It represents the maximum useful work obtainable from a process at constant temperature and pressure. The change in Gibbs energy is:
ΔG = ΔH − TΔS
This equation combines enthalpy (energy content) and entropy (disorder)
into a single criterion for spontaneity at constant T and P:
- ΔG < 0: Spontaneous (feasible)
- ΔG > 0: Non-spontaneous
- ΔG = 0: System at equilibrium
Numerical calculation:
Given: ΔH = −92 kJ = −92000 J, ΔS = −199 J/K, T = 298 K
Step 1: Calculate ΔG
ΔG = ΔH − TΔS
ΔG = (−92000) − (298 × (−199))
ΔG = −92000 + 59302
ΔG = −32698 J = −32.698 kJ
Since ΔG < 0, the reaction IS spontaneous at 298 K.
Step 2: Find the temperature at which it becomes non-spontaneous
Set ΔG = 0:
0 = ΔH − TΔS
T = ΔH/ΔS = (−92000)/(−199) = 462.3 K
Below 462.3 K: ΔG < 0 (spontaneous)
Above 462.3 K: ΔG > 0 (non-spontaneous)
The reaction N₂ + 3H₂ → 2NH₃ is spontaneous at room temperature
but becomes non-spontaneous above approximately 462 K (189 °C).
Common Mistakes to Avoid
1. Confusing sign conventions for q and w — heat absorbed by system is positive (q > 0); work done ON the system is positive (w > 0). Work done BY the system (expansion) is negative.
2. Using Δn_g incorrectly — always calculate Δn_g as (moles of gaseous products) minus (moles of gaseous reactants). Only count gaseous species, not solids or liquids.
3. Forgetting to convert units in ΔH = ΔU + Δn_gRT — ΔH and ΔU are typically in kJ, but R = 8.314 J/mol·K. Convert Δn_gRT to kJ by dividing by 1000.
4. Assuming ΔS < 0 means non-spontaneous — spontaneity depends on ΔG, not just ΔS. A reaction can be spontaneous even with ΔS < 0 if ΔH is sufficiently negative.
5. Confusing C_p and C_v — C_p is heat capacity at constant pressure (used with enthalpy); C_v is at constant volume (used with internal energy). For an ideal gas, C_p − C_v = R.
6. Not reversing equations correctly in Hess's law — when you reverse an equation, you must reverse the sign of ΔH. When you multiply an equation by a factor, multiply ΔH by the same factor.
7. Forgetting that ΔH°f of elements is zero — the standard enthalpy of formation of an element in its standard state (e.g., O₂(g), N₂(g), C(s, graphite)) is zero by definition.
8. Confusing isothermal, isobaric, isochoric, and adiabatic processes — isothermal (ΔT = 0), isobaric (ΔP = 0), isochoric (ΔV = 0), adiabatic (q = 0). Each has different implications for q, w, ΔU, and ΔH.
Exam Checklist
- [ ] Can you state and apply the first law of thermodynamics (ΔU = q + w)?
- [ ] Can you calculate work done during expansion (w = −PΔV)?
- [ ] Do you know the relationship ΔH = ΔU + Δn_gRT and can you apply it?
- [ ] Can you calculate ΔH of a reaction using Hess's law and given thermochemical equations?
- [ ] Can you calculate ΔH using standard enthalpies of formation (ΔH°f)?
- [ ] Can you determine spontaneity using ΔG = ΔH − TΔS?
- [ ] Can you find the temperature at which a reaction becomes spontaneous (T = ΔH/ΔS)?
- [ ] Do you know the difference between state functions and path functions?
- [ ] Can you explain the second law of thermodynamics in terms of entropy?
- [ ] Can you classify processes as isothermal, isobaric, isochoric, or adiabatic?
