Model Test Questions XI Chemistry on Thermochemistry (Chapter # 11)

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Welcome to Inam Jazbi’s ultimate guide to XI Chemistry – Thermochemistry (Chapter # 11)! 🔥 Whether you're preparing for board exams or competitive tests like MDCAT / ECAT, this blog is packed with everything you need to ace Thermochemistry.

• 🔥 Chapter 11 Test Questions to practice core concepts of thermochemistry
• 🔥 Detailed solutions for numericals and problems
• 🔥 MCQs to test your understanding and boost exam confidence

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🧪 Class 11 Chemistry – Chapter 11: Thermochemistry ✨

Preparing for Class 11 Chemistry exams and worried about Chapter 11: Thermochemistry ? 😟 You’re in the right place! ✅

In this blog, Inam Jazbi’s smart, board-oriented model test questions are carefully designed to sharpen your concepts, boost your confidence 💪, and help you score maximum marks 📈.

These questions strictly follow the latest exam pattern, covering important numericals, MCQs, short and long questions, with special focus on the areas examiners love to repeat 🔁.

If you want easy understanding + high scores 🏆, this guide is a must-read before your chemistry paper! 📘🔥

✏️ Model Test Questions XI Chemistry Chapter # 11………… Thermochemistry ✏️

✏️ Short Questions of Thermochemistry ✏️

Q1. State precisely the meaning of each of the following terms:
internal energy, Enthalpy, system, surrounding, state, state of system, macroscopic properties, internal energy, enthalpy, thermochemical reaction, heat of neutralization, Heat of combustion, Thermodynamics, Thermochemistry, thermodynamical standard state

Q2. What is meant by internal energy change (∆E) and enthalpy change (∆H)? Under what conditions are ∆E and ∆H equal?

Q3. How can you define standard enthalpy of formation and standard enthalpy of reaction.

Q4. Differentiate between
(i) Intensive and Extensive properties
(ii) Exothermic and Endothermic reactions

Q5. Give 5 general examples of each of exothermic and endothermic changes in tabular form

Q6. Give 5 examples of each of extensive and intensive properties in tabular form

Q7. Categories the following into exothermic and endothermic changes in tabular form:
Photosynthesis, Evaporation, Combustion, Freezing, Sublimation, Cracking, Bond cleavage, Electrolysis, condensation, fermentation, bond formation, melting, hydration.

Q8. Categories the following into intensive and extensive properties in tabular form:
Area, Enthalpy, Refractive index, Density, Internal energy, Volume, Temperature, Entropy

✏️ Descriptive Questions ✏️

Q9. State and explain first law of thermodynamics. Derive pressure-volume work of a system.

Q10. Discuss the applications of first law of thermodynamics at constant pressure and constant volume. (Prove that: i) ∆H = qₚ by deriving pressure volume work ii) ∆E = qᵥ)

Q11. State, explain and prove Hess’s law of enthalpy summation. Discuss its applications.

Q12. Explain exothermic and endothermic reactions with the help of the energy diagram.

✏️ Numericals on Thermochemistry✏️

Q1. A gaseous chemical reaction is carried out in a cylinder under a constant external pressure of 1 atm. If during reaction, volume increases from 3 dm³ to 5 dm³ by moving the piston upward, calculate the work done and express in kJ. (Example 11.1, Page # 225)

Answer: −0.20265 kJ

Q2. Convert the values of following quantities (Self-assessment, Page # 225)
(i) 20 calories energy into joule

Answer: 83.68 J

(ii) 3.5 atm dm³ work into kJ

Answer: 0.355 kJ

Q3. Burning of petrol in an automobile engine gives carbon dioxide and water vapours. If the gases do 675 J work in pushing piston outward and the system loses 435 J heat to the surrounding, calculate the internal energy change in kJ. [Example 11.2, Page # 226]

Answer: ∆E in J = −1110 J, ∆E in kJ = −1.110 kJ

Q4. A thermochemical process is carried out at constant pressure of 8.52 atm. If it absorbs 15.5 kJ energy from the surrounding, due to which an expansion in the volume of 4.7dm³ is occurred. Calculate its change in internal energy. [Exercise Q1, Page # 237]

Answer: Work = −40.04 atm dm³ = −4057.45 J = −4.05745 kJ, ∆E in kJ = −11.44 kJ

Q5. Calculate the enthalpy of combustion of propane at 25°C by the given information: [Example 11.3, Page # 230]
C₃H₈ (g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g) ∆H° = ?
∆Hf° of C₃H₈ (g) = −103.9 kJ/mol
∆Hf° of CO₂(g) = −393.5 kJ/mol
∆Hf° of H₂O(g) = −285.8 kJ/mol

Answer: ∆Hreaction° = −2219.8 kJ/mol

Q6. Using data in following table, calculate the standard enthalpy change for each of the following reactions:
2H₂S₍g₎ + 3O₂(g) → 2H₂O₍ₗ₎ + 2SO₂₍g₎
Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
∆Hf° of H₂S₍g₎ = −20.17 kJ/mol, ∆Hf° of Fe₂O₃₍ₛ₎ = −824.2 kJ/mol
∆Hf° of O₂₍g₎ = 0 kJ/mol, ∆Hf° of CO₍g₎ = −110.5 kJ/mol
∆Hf° of H₂O₍g₎ = −285.8 kJ/mol, ∆Hf° of 2Fe(s) = 0 kJ/mol
∆Hf° of SO₂₍g₎ = −296.8 kJ/mol, ∆Hf° of CO₂(g) = −393.5 kJ/mol
[Exercise Q2, Page # 237]

Answer: ∆Hreaction° = −1124.86 kJ/mol, −24.8 kJ/mol

Q7. In the manufacturing of HNO₃ by the Ostwald process, one of the most important exothermic reactions is the oxidation of ammonia. [Exercise Q3, Page # 238]
4NH₃ (g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
Determine standard heat of reaction (ΔH°) from the following given data:
ΔHf° of NH₃ = −46.91 kJ/mol
ΔHf° of NO = 90.25 kJ/mol
ΔHf° of H₂O = −285.8 kJ/mol

Answer: ∆H°reaction = −1166.16 kJ/mol

Q8. Calculate the standard heat of formation of acetylene (C₂H₂) using the data of the following thermochemical equations: [Example 11.4, Page # 231]
C(s) + O₂(g) → CO₂(g) ∆H = −393.5 kJ
H₂(g) + ½O₂(g) → H₂O₍ₗ₎ ∆H = −285.8 kJ
C₂H₂(g) + 5/2O₂(g) → 2CO₂(g) + H₂O₍ₗ₎ ∆H = −1299.5 kJ
2C(s) + H₂(g) → C₂H₂(g) ∆Hf = ?

Answer: +226.7 kJ

Q9. Iso octane (C₈H₁₈) combustion in engine: [Exercise Q4, Page # 238]
C₈H₁₈₍ₗ₎ + 12 ½ O₂ → 8CO₂(g) + 9H₂O₍ₗ₎
Given that
ΔHf° of CO₂ = −393.5 kJ/mol,
ΔHf° of H₂O = −285.8 kJ/mol,
ΔHf° of C₈H₁₈ = −223.8 kJ/mol

Answer: ΔHf° = +5496.4 kJ/mol

Q10. Glycerol (C₃H₈O₃) is a well-known organic compound due to its versatile uses. Calculate the standard enthalpy of formation of Glycerol from the data given below. [Exercise Q5, Page # 238]
3C(s) + 4H₂(g) + 3/2 O₂ → C₃H₈O₃(l) (ΔHf° = ?)
C(s) + O₂(g) → CO₂(g) ΔH° = −393.5 kJ/mol
H₂(g) + ½O₂(g) → H₂O₍ₗ₎ ΔH° = −285.8 kJ/mol
C₃H₈O₃(l) + 31/2 O₂(g) → 3CO₂(g) + 4H₂O₍ₗ₎ ΔH° = −1654.1 kJ/mol

Answer: ΔHf° = −669.6 kJ/mol

Q11. Use the data provided below for the formation of RbCl(S), write thermochemical equations for all the steps involved in the Born Haber cycle and determine the enthalpy of formation of RbCl(s).
Sublimation energy of Rb₍ₛ₎ = 82 kJ/mol
Ionization energy of Rb(g) = 403 kJ/mol
Dissociation energy of Cl₂(g) = 242 kJ/mol
Electron affinity of Cl₂(g) = −348.5 kJ/mol
Lattice energy of RbCl₍ₛ₎ = −689 kJ/mol

Answer: ∆Hf°(RbCl(s)) = −431.5 kJ/mol

Q12. Draw a fully labeled Born Haber cycle for Rubidium chloride (RbCl) and determine the lattice energy by using the following values. (all in kJ/mol) [Exercise Q6, Page # 238]
I.P 1st of Rb = 403 kJ/mol
Electron affinity of Cl = −349 kJ/mol
Bond energy of Cl₂ = 242 kJ/mol
Sublimation energy of Rb = 86.5 kJ/mol
Heat of formation of RbCl = −430.5 kJ/mol

Answer: ∆Hf° = −430.5 kJ/mol

Extra Numericals from Past Papers

Q1. When 5400 J of heat is added to a system of gas at a constant pressure of 2.0 x 10⁵ N/m², its internal energy increases by 1000 J. Calculate change in volume of the system.

Answer: 0.022 m³

Q2. In a thermodynamical process, 500 kJ work is done on the system and its internal energy increases by 273 kJ. State whether the process is endothermic or exothermic and what is the heat exchange between the system and the surroundings.

Answer: Endothermic process, Heat absorbed = 227 kJ

Q3. A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01 m³. A constant pressure process gives 54 kJ of work out. Find the final volume of the air.

Answer: Vfinal = 0.1 m³

Q4. 4800 kJ heat is entered in a gas cylinder, calculate the value of ΔE:
(a) If the volume is kept constant.
(b) If the volume is not constant, work of +1800 kJ was performed.
(c) If the gas is allowed to expand, the value of work is 8800 kJ.

(a) ΔE = 4800 kJ, (b) ΔE = 3000 kJ, (c) ΔE = -4000 kJ

Q5. When 6800 J of heat is added to a system of gas at a constant pressure of 2.0 x 10⁵ N/m², its internal energy increases by 2000 J. Calculate change in volume of the system.

Answer: ΔV = 0.024 m³

Q6. Calculate the enthalpy of formation of butane (C₄H₁₀) from the following data:
Enthalpy of combustion of graphite = –393.6 kJ/mol
Enthalpy of combustion of hydrogen = –285.9 kJ/mol
Enthalpy of combustion of butane = –2877.1 kJ/mol

Answer: ΔHf° of C₄H₁₀ = −125.7 kJ/mol

Q7. Calculate heat of formation of Ethanol (C₂H₅OH) from following data: [K.B. – 2006]
(i) C + O₂ → CO₂(g) ΔH = –394 kJ
(ii) H₂ + ½ O₂ → H₂O(g) ΔH = –285.8 kJ
(iii) C₂H₅OH₍ₗ₎ + 3O₂ → 2CO₂ + 3H₂O ΔH = –1402.14 kJ
(iv) 2C + 3H₂ + ½ O₂ → C₂H₅OH₍ₗ₎ ΔHf = ?

Answer: ΔHf° = −243.3 kJ/mol

Q8. Calculate heat of formation of benzene (C₆H₆) from the following data: [KB 2017]
(i) C + O₂ → CO₂(g) ΔH = –394 kJ
(ii) H₂ + ½ O₂ → H₂O(g) ΔH = –286 kJ
(iii) C₆H₆ + 7.5O₂ → 6CO₂ + 3H₂O ΔH = –3267 kJ
(iv) 6C + 3H₂ → C₆H₆ ΔHf = ?

Answer: ΔHf° = +45 kJ/mol

Q9. Calculate heat of formation of Ethyne (C₂H₂) from the following data: [K.B. – 2002]
(i) C + O₂ → CO₂(g) ΔH = –394 kJ
(ii) H₂ + ½ O₂ → H₂O(g) ΔH = –284 kJ
(iii) C₂H₂ + 5/2 O₂ → 2CO₂ + H₂O ΔH = –1296 kJ
(iv) 2C(s) + H₂(g) → C₂H₂(g) ΔHf = ?

Answer: ΔHf° = +224 kJ/mol

Q10. Calculate the standard heat of formation of methyl alcohol (CH₃OH) from the following data: [K.B. – 2015]
(i) C(s) + ½ O₂(g) → CO(g) ΔH = –111 kJ/mol
(ii) H₂(g) + ½ O₂ → H₂O(g) ΔH = –286 kJ/mol
(iii) CH₃OH₍ₗ₎ + O₂(g) → CO(g) + 2H₂O(l) ΔH = –561 kJ/mol
(iv) C(s) + 2H₂(g) + ½ O₂(g) → CH₃OH₍ₗ₎ ΔHf = ?

Answer: ΔHf° = −122 kJ/mol

✏️ Text Book MCQs on Thermochemistry with Explanatory Answers ✏️

Thermochemistry MCQs

Q1. Least entropy found in which of the following state of water:

(a) Steam at 100°C

(b) Liquid water at 25°C

(c) Liquid water at 4°C

(d) Ice at 0°C

Answer: (d) Ice at 0°C – Solids have least entropy compared to liquids and gases.

Q2. The energy corresponds to the given thermochemical process: Li⁺ (g) + Cl⁻ (g) →LiCl₍ₛ₎

(a) Ionization energy

(b) Enthalpy of atomization

(c) Enthalpy of combustion

(d) Lattice energy

Answer: (d) Lattice energy – Energy released when ions form a solid lattice.

Q3. Which of the following change is not an endothermic reaction:

(a) Cracking of alkanes

(b) Decomposition of lime

(c) Combustion of butane

(d) Photosynthesis

Answer: (c) Combustion of butane – Combustion releases heat, so it's exothermic.

Q4. Heat transfer cannot be feasible across the boundary of a:

(a) Open system

(b) Thermo permeable system

(c) Isolated system

(d) Close system

Answer: (c) Isolated system – No energy or matter can cross its boundaries.

Q5. In a thermochemical process, no work is done if the system is kept at:

(a) Constant temperature

(b) Constant pressure

(c) Constant volume

(d) Constant mass

Answer: (c) Constant volume – No expansion or compression means no work done (w = PΔV = 0).

Q6. Standard enthalpy of formation of all of the following elements at 25°C and 1 atm pressure are zero except:

(a) C(diamond)

(b) C(graphite)

(c) O₂

(d) N₂

Answer: (a) C(diamond) – Graphite is standard state; diamond has nonzero enthalpy of formation.

Q7. In the equation of First law of thermodynamics (ΔE=q + w), the property(s) which depends upon initial and final state is (are):

(a) ΔE

(b) q

(c) W

(d) Both q and W

Answer: (a) ΔE – Internal energy is a state function; q and W depend on the path.

Q8. Volume is a:

(a) State function

(b) Colligative property

(c) Intensive properties

(d) Path function

Answer: (a) State function – Its value depends only on the state, not path.

Q9. Which statement is incorrect:

(a) For constant pressure process, ΔH = ΔE + PΔV

(b) For constant volume process, ΔE = q

(c) For exothermic reactions, ΔH > 0

(d) For Hess law ΣΔH°(cycle) = 0

Answer: (c) For exothermic reactions, ΔH < 0 – Exothermic reactions release heat.

Q10. Which of the following enthalpy change is always negative:

(a) Enthalpy of formation

(b) Enthalpy of decomposition

(c) Enthalpy of combustion

(d) Enthalpy of reaction

Answer: (c) Enthalpy of combustion – Combustion reactions always release heat.

✏️ Extra MCQs from Past Papers on Thermochemistry with Explanatory Answers ✏️

Q11. The heat released during an experiment is 94 kJmol⁻¹. The heat absorbed by the surrounding is:

(a) -94 kJmol⁻¹

(b) 47 kJmol⁻¹

(c) +94 kJmol⁻¹

(d) 188 kJmol⁻¹

Answer: (c) +94 kJmol⁻¹ – Heat released by the system is absorbed by surroundings (equal magnitude, opposite sign convention).

Q12. In an endothermic reaction, the product has:

(a) Lower energy

(b) Equal energy

(c) Higher energy

(d) Lesser energy

Answer: (c) Higher energy – Endothermic reactions absorb energy, so products are at higher energy than reactants.

Q13. The vigorous reaction taking place between Na metal and water is:

(a) Isolated

(b) Endothermic

(c) Exothermic

(d) Isothermal

Answer: (c) Exothermic – Sodium reacts with water releasing heat and hydrogen gas.

Q14. The reaction between carbon doxide and water in the presence of sunlight is:

(a) Isolated

(b) Exothermic

(c) Endothermic

(d) Isothermal

Answer: (c) Endothermic – Photosynthesis is endothermic and a photochemical reaction.

Q15. The substance undergoing a physical or chemical change forms a ………..

(a) Physical system

(b) Chemical system

(c) Microscopic system

(d) Isolated system

Answer: (b) Chemical system – Any reaction creates a chemical system where substances interact.

Q16. Spontaneous reactions are:

(a) Endothermic

(b) Reversible

(c) Not irreversible

(d) Irreversible

Answer: (d) Irreversible – Spontaneous reactions occur naturally and typically cannot reverse without external input.

Q17. Most of the reactions which give stable products are:

(a) Isothermal

(b) Endothermic

(c) Exothermic

(d) None of them

Answer: (c) Exothermic – Formation of stable products usually releases energy.

Q18. At constant volume, qv is equal to:

(a) ΔH

(b) ΔP

(c) ΔE

(d) ΔV

Answer: (c) ΔE – At constant volume, heat exchanged equals change in internal energy.

Q19. Which of the following processes has always ΔH = +ve?

(a) Formation of compound

(b) Atomization

(c) Dilution of a solution

(d) Combustion

Answer: (b) Atomization – always absorsb heat during separation of atoms (endothermic).

Q20. Under what conditions ∆H is equal to ∆E?

(a) The reaction does not involve any gaseous substance

(b) The reaction does not involve any change in volume (∆V = 0)

(c) No change in the number of moles of gases (∆n = 0)

(d) All of these

Answer: (d) All of these – ΔH = ΔE when no PV work occurs, i.e., no gaseous change or volume change.

Q21. Which of the following processes is endothermic?

(a) Freezing

(b) Condensation

(c) Electrolysis

(d) Bond formation

Answer: (c) Electrolysis – It requires input of energy to split compounds.

Q22. Which of the following processes is endothermic?

(a) Photosynthesis

(b) Condensation

(c) Respiration

(d) Fermentation

Answer: (a) Photosynthesis – It absorbs sunlight to form glucose.

Q23. A substance Z releases energy. When the amount of Z is doubled energy released will be:

(a) Halved

(b) Quadrupled

(c) Doubled

(d) Remains same

Answer: (c) Doubled – Energy released is proportional to the amount of substance.

Q24. A substance Z releases energy. When the amount of Z is halved energy released will be:

(a) Halved

(b) Doubled

(c) Quadrupled

(d) Remains same

Answer: (a) Halved – Energy released is directly proportional to the amount of substance.

Q25. Combustion of CH₄ gives off 890 kJmol⁻¹ heat. How much heat would be released if 64 g of methane is combusted?

(a) 1780 kJmol⁻¹

(b) 890 kJmol⁻¹

(c) 3560 kJmol⁻¹

(d) 2670 kJmol⁻¹

Answer: (c) 1780 kJ – 64 g is 4 moles of CH₄ (16 g/mol), so heat released = 4 × 890 = 3560 kJ.

Q26. Combustion of CH₄ gives off 890 kJmol⁻¹ heat. How much heat would be released if 32 g of methane is combusted?

(a) 3560 kJmol⁻¹

(b) 1780 kJmol⁻¹

(c) 890 kJmol⁻¹

(d) 2670 kJmol⁻¹

Answer: (b) 890 kJ – 32 g is 2 moles of CH₄ (16 g/mol), heat released = 2 × 890 = 1780 kJ.

Q27. A gas undergoes constant-temperature expansion from 2 liter to 6 liter. The work done in joule by the gas if it expands against vacuum is:

(a) Zero

(b) 4

(c) -4

(d) None of these

Answer: (a) Zero – Expansion against vacuum does no work (w = PΔV = 0).

Q28. Which one of the following is intensive property?

(a) Enthalpy

(b) Density

(c) Internal energy

(d) Entropy

Answer: (b) Density – Intensive properties do not depend on the amount of substance.

Q29. Which one of the following is intensive property?

(a) Viscosity

(b) Enthalpy

(c) Internal energy

(d) Entropy

Answer: (a) Viscosity – Intensive property independent of mass/volume.

Q30. As ice at 0°C changes to water at 0°C, the average kinetic energy of the ice molecules:

(a) Increases

(b) Decreases

(c) Remains the same

(d) Cannot be predicted

Answer: (c) Remains the same – Temperature is constant at 0°C, so average kinetic energy does not change during melting.

Q31. A leaky thermos would be an example of a (an)

(a) Isolated system

(b) Closed system

(c) Open system

(d) Insulated system

Answer: (c) Open system – Both matter and energy can escape from a leaky thermos.

Q32. If ΔH value is greater than zero then the reaction will be

(a) Exothermic

(b) Endothermic

(c) Isothermic

(d) May or may not be Exothermic or Endothermic

Answer: (b) Endothermic – Positive ΔH indicates heat is absorbed by the system.

Q33. Most of thermodynamic parameters are

(a) System

(b) Surrounding

(c) State functions

(d) Phase

Answer: (c) State functions – Parameters like ΔE, ΔH, ΔS depend only on initial and final states.

Q34. Which one of the following is not a state function?

(a) Temperature

(b) Heat

(c) Pressure

(d) Volume

Answer: (b) Heat – Heat depends on the path, not just initial and final states.

Q35. In a bomb calorimeter, the reactions are carried out at

(a) Constant volume

(b) Constant pressure

(c) Constant temperature

(d) All of them

Answer: (a) Constant volume – Bomb calorimeter measures ΔE at constant volume.

Q36. When coefficients of chemical equation are doubled, ΔH

(a) Doubles

(b) Halves

(c) Remains same

(d) No correlation

Answer: (a) Doubles – Enthalpy change is proportional to the amount of reactants/products.

Q37. The enthalpy of combustion is

(a) Positive

(b) Negative

(c) Either positive or negative

(d) No correlation

Answer: (b) Negative – Combustion is exothermic, releasing heat.

Q38. Heat of neutralization of a strong acid and strong base is

(a) 13.7 kcal

(b) 57 kJ

(c) 5.7 x 10⁴ J

(d) All of them

Answer: (d) All of them – All values are equivalent expressions of heat released (~57 kJ).

Q39. Mole is an example of ………………..property.

(a) Intensive

(b) Extensive

(c) Non-additive

(d) None of these

Answer: (b) Extensive – Mole depends on the amount of substance.

Q40. Pressure is an example of ……………. property.

(a) Intensive

(b) Extensive

(c) Non-additive

(d) None of these

Answer: (a) Intensive – Pressure does not depend on the amount of substance.

Q41. Bond formation is an example of ……………. reaction.

(a) Spontaneous

(b) Exothermic

(c) Endothermic

(d) Both a and b

Answer: (d) Both a and b – Bond formation releases energy and occurs spontaneously.

Q42. Enthalpy change during reaction depends upon?

(a) Physical state

(b) Amount

(c) Temperature

(d) All of them

Answer: (d) All of them – ΔH depends on the state, quantity, and temperature of reactants/products.

Q43. Which will double if amount of substance doubled?

(a) Internal energy

(b) Mole

(c) Enthalpy

(d) All of them

Answer: (d) All of them – Extensive properties like ΔE, ΔH, and moles double with the amount of substance.

Q44. The property of a system in bulk is termed as ……………. property:

(a) Macroscopic

(b) Microscopic

(c) Intensive

(d) Extensive

Answer: (a) Macroscopic – Bulk properties observed without looking at molecules.

Q45. A system which allows only the exchange of energy is called:

(a) Closed system

(b) Isolated system

(c) Open system

(d) Chemical system

Answer: (a) Closed system – Energy can be exchanged but matter cannot.

Q46. Which one of the following processes is exothermic?

(a) Condensation

(b) Sublimation

(c) Decomposition

(d) Evaporation

Answer: (a) Condensation – Heat is released when gas turns into liquid.

Q47. Which change is not endothermic?

(a) Freezing

(b) Electrolysis

(c) Melting

(d) All of them

Answer: (a) Freezing – Freezing releases heat, so it is exothermic.

Q48. An endothermic reaction is one in which the temperature of surrounding:

(a) Decreases

(b) Increases

(c) Remains the same

(d) None of them

Answer: (a) Decreases – Endothermic reactions absorb heat from surroundings, lowering their temperature.

Q49. In an exothermic reaction, which has the highest energy?

(a) Product

(b) Reactant

(c) Both of them

(d) None of them

Answer: (b) Reactant – Reactants have higher energy than products in exothermic reactions.

Q50. Which one of the reactions has H₂ greater than H₁?

(a) Combustion

(b) Endothermic

(c) Exothermic

(d) Neutralization

Answer: (b) Endothermic – H₂ > H₁ because the products have higher enthalpy than reactants.

✏️ Smart Answers of Model Test Questions XI Chemistry on Thermochemistry Chapter # 11 Test # 15✏️

✏️ Smart Answers of Short-Answer Questions on Thermochemistry ✏️

Q1. State precisely the meaning of each of the following terms:
Define internal energy, Enthalpy, system, surrounding, state, state of system, macroscopic properties, internal energy, enthalpy, thermochemical reaction, heat of neutralization, Heat of combustion, Thermodynamics, Thermochemistry, thermodynamical standard state

Answer

🔥 Thermodynamics (Greek; means ‘Heat flow’)
Thermodynamics is the branch of science that deals with the study of interconversion of heat, work, and other forms of energy during physical and chemical processes, based on the law of conservation of energy. It is the study of ‘energy transformations or energy changes’ or flow of heat ⚖️.

🧪 Thermochemistry
Thermochemistry is the branch of chemistry that studies the heat energy changes associated with chemical and physical reactions 🌡️. Thermochemistry is the study of relationship between heat energy and chemical energy.

♨️ Thermochemical Reaction
A thermochemical reaction is a chemical reaction that is accompanied by absorption or evolution of heat energy with the material changes 🔥❄️.

📏 Macroscopic Properties (Measurable Properties)
Macroscopic properties are the bulk properties of a system that can be easily measured, such as temperature, pressure, volume, mass, chemical composition and entropy 📊.

🔬 System
A system is the specified part of the universe under study, separated from its surroundings by a real or imaginary boundary, in which physical or chemical changes occur ⚗️.

🌍 Surroundings
Surroundings are the rest of the universe except the system, which can exchange energy or matter with the system 🌐.

🚧 Boundary
A boundary is a real or imaginary closed surface that separates a system from its surroundings. It may allow (adiabatic boundary) or prevent heat (diathermal boundary) and matter exchange 🚪.

🧭 State of a System
The state of a system is defined completely by its macroscopic properties like temperature, pressure, volume, and composition at a given time ⏱️.

📌 Thermodynamic Standard State
The thermodynamic standard state is the most stable form of a pure substance at 1 atm pressure, 25°C (298 K), and 1 M concentration for solutions (standard conditions) ✅.

🔁 State Function
A state function is a property or thermodynamic parameter whose value depends only on the initial and final states of the system, not on the path followed 🔄. e.g. internal energy, enthalpy, temperature, volume etc.

⚡ Internal Energy (U or E)
Internal energy is the total energy of a system, equal to the sum of kinetic and potential energies of all particles present ⚛️.

🔥 Enthalpy (H)
Enthalpy ((meaning to warm) is the total heat content of a system at constant pressure and it is the algebraic product of pressure and volume (PV energy or mechanical work), given by H = E + PV ♨️.

🧾 Standard Enthalpy of Reaction (ΔH°_rxn) or Heat of Reaction (ΔH°_r)
It is the enthalpy change that occurs when a reaction takes place in the standard state under standard conditions 📉📈. It may be + or –. It is equal to the heat exchanged, q_p (heat flow into or out of the system) at constant pressure.

🧱 Standard Enthalpy of Formation (ΔH_f°)/ Heat of Formation
It is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions 🧩. It may be + or –.

💧 Heat of Neutralization (ΔH°_{neutralization})
The heat of neutralization is the enthalpy change when one mole of water is formed by the reaction of an acid and a base; it is always negative ❄️.

🔥 Heat of Combustion/ Enthalpy of Combustion
The heat of combustion (ΔH_c°) is the enthalpy change when one mole of a substance in its standard state under standard conditions undergoes complete combustion in oxygen; it is always negative 🔥⬇️.

📈 Entropy (S)
Entropy is a measure of the degree of disorder or randomness of a system and increases with temperature 🔀.

2. What is meant by internal energy change (∆E) and enthalpy change (∆H)? Under what conditions are ∆E and ∆H equal?

Answer

✨Internal Energy Change (ΔE) 🔋
➡️Internal energy change (ΔE) is the difference between the final and initial internal energies of a system.
➡️ΔE = E₂−E₁
➡️It is a state function, as it depends only on the initial and final states of the system and not on the path followed 🔄.

✨Enthalpy Change (ΔH) ♨️
➡️Enthalpy change (ΔH) is the change in heat content of a system at constant pressure. Enthalpy change (ΔH) is the difference between the final and initial enthalpies of a system.
➡️It is also a state function and depends only on the initial and final states of the system.
➡️Exothermic reactions: ΔH is negative (heat released) 🔻
➡️Endothermic reactions: ΔH is positive (heat absorbed) 🔺

✨Condition when ΔE = ΔH ⚖️
For reactions involving solids and liquids, the volume change (ΔV) is negligible, so the term PΔV ≈ 0.
According to the first law of thermodynamics:
ΔH = ΔE + PΔV
Since PΔV≈0
📌 ΔH = ΔE (enthalpy change will be equal to internal energy change)

Q3. How can you define standard enthalpy of formation and standard enthalpy of reaction?

Answer

🔥 Standard Enthalpy of Reaction (ΔH°_rxn) or Heat of Reaction
The standard enthalpy of reaction is the enthalpy change that occurs when a chemical reaction takes place with reactants and products in their standard states under standard conditions (1 atm pressure and 25°C / 298 K).
It is equal to the heat exchanged at constant pressure (qₚ) and may be positive or negative 📉📈.

🧱 Standard Enthalpy of Formation (ΔH°_f)/ Heat of Formation
The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions (1 atm pressure and 25°C / 298 K).
Its value may be positive or negative and is usually expressed in kJ mol⁻¹ ⚛️.

Q4. Differentiate between

(i) Intensive and Extensive properties
(ii) Exothermic and Endothermic reactions

Answer

Exothermic Reactions Vs Endothermic Reactions

S. #	Exothermic Reactions 🔥	Endothermic Reactions ❄️
1.	Heat is released to surroundings ♨️	Heat is absorbed from surroundings ❄️
2.	Surroundings (container) become hot 🌍🔥	Surroundings become cold 🌍❄️
3.	Causes an increase in temperature 🌡️⬆️	Causes decrease in temperature 🌡️⬇️
4.	Products have lower enthalpy than reactants 📉	Products have higher enthalpy than reactants 📈
5.	ΔH (enthalpy change) is negative (–) i.e. ΔH<0 ⬇️	ΔH (enthalpy change) is positive (+) i.e. ΔH>0 ⬆️
6.	Bond formation energy > bond breaking energy 🔗	Bond breaking energy > bond formation energy ✂️
7.	Usually spontaneous 🔁	Usually non-spontaneous ❌
8.	Examples: combustion, neutralization	Examples: photosynthesis, melting of ice

Difference between Intensive and Extensive Properties

S. #	Intensive Properties ⚖️	Extensive Properties 📦
1.	Independent of amount of substance ⚖️	Dependent on amount of substance 📦
2.	Do not depend on size of system 📏	Depend on size of system 🏗️
3.	Help in identification of substances 🔍	Cannot identify substances ❓
4.	Non-additive in nature 🚫➕	Additive in nature ➕
5.	Examples: Density, specific gravity, pressure, temperature, etc.	Examples: Length, Distance, mass, Time, volume, internal energy, enthalpy, entropy etc.

Q5. Give 5 general examples of each of exothermic and endothermic changes in text format

Answer

✨Exothermic Changes 🔥 (Heat Released)

➡️ Combustion reactions (release heat to the surroundings 🔥)
➡️ Respiration (exothermic process that produces energy 🫁)
➡️ Neutralization reactions (evolve heat ⚗️)
➡️ Fermentation (releases heat during chemical change 🍞)
➡️ Bond formation (always an exothermic process 🔗)
➡️ Metal–water reactions (liberate heat 💧)
➡️ Most oxidation reactions (are exothermic ⚙️)

✨Endothermic Changes ❄️ (Heat Absorbed)

➡️ Decomposition reactions (absorb heat ❄️)
➡️ Electrolysis (requires energy input 🔌)
➡️ Photosynthesis (absorbs solar energy 🌿)
➡️ Bond breaking (always an endothermic process 🔓)
➡️ Molecular addition reactions (absorb heat ➕)
➡️ Water gas formation (absorbs heat 💨)
➡️ Melting of ice (absorbs heat 🧊)

Q6. Give 5 examples of each of extensive and intensive properties in text format

Answer

✨Extensive Properties 📦

➡️ Density, specific gravity, pressure, temperature, viscosity, vapour pressure, surface tension, melting point, boiling point, freezing point, refractive index, concentration, velocity, chemical properties, chemical potential, specific heat capacity, specific volume, specific energy, electrical resistivity, hardness, ductility, elasticity, malleability, etc.

✨Intensive Properties ⚖️

➡️ Length, distance, mass, time, volume, area, weight, momentum, mole numbers, particle number, electric charge, work, enthalpy, entropy, internal energy, Gibb’s free energy, energy, heat capacity, magnetic moment.

Q7. Categorize the following into exothermic and endothermic changes in tabular form:

Photosynthesis, Evaporation, Combustion, Freezing, Sublimation, Cracking, Bond cleavage, Electrolysis, condensation, fermentation, bond formation, melting, hydration.

Answer

🔥 Exothermic

Combustion

Condensation

Freezing

Fermentation

Bond Formation

Hydration

❄️ Endothermic

Photosynthesis

Evaporation

Sublimation

Cracking

Bond Cleavage

Electrolysis

Melting

Q8. Categorize the following into intensive and extensive properties in tabular form:

Area, Enthalpy, Refractive index, Density, Internal energy, Volume, Temperature, Entropy

Answer

⚖️ Intensive

Refractive index

Density

Temperature

📦 Extensive

Area

Enthalpy

Internal energy

Volume

Entropy

Q9. State and explain the first law of thermodynamics. Derive pressure-volume work of a system.

Answer

📚 Introduction
➡️The concept was enunciated by J.R. Mayer (1841), formulated by Helmholtz (1847).
➡️The first law of thermodynamics relates energy, heat, and work.
➡️It is an adaptation of the law of conservation of energy: energy cannot be created or destroyed, only transferred between a system and its surroundings as heat (q) or work (W).

📌 Statement of the First Law
In any physical or chemical process:
A system cannot create or destroy energy, but may exchange it with surroundings as heat or work, keeping the total energy of system + surroundings (or the universe) constant.
OR
The change in internal energy of the system is the sum of heat exchanged (heat lost or gained) and work done on or by the system. i.e.
Change in internal energy = Heat exchanged + Work done on or by system

🧮 Mathematical Formulation of the Law
ΔE = q + W
Where:
ΔE = change in internal energy of the system 🔋
q = Heat transfer or heat absorbed (+) or released (–) 🌡️
W = work done on (+) or by (–) the system ⚙️

✨ Other Forms:
➡️ ΔE = q, if no work is done
➡️ ΔE = W, if no heat is exchanged
➡️ ΔE = q – PΔV (substituting W = –PΔV)
➡️ q = ΔE + PΔV

⚖️ Sign Conventions for work, heat and energy
➡️ W: Negative for thermochemical expansion, positive for compression
➡️ q: Positive when heat is absorbed, negative when released
➡️ ΔE: Positive if system gains energy, negative if it loses energy

🔧 Derivation of Pressure-Volume Work
Pressure-volume (PV) work
It is the mechanical work that occurs when a system expands or compresses at constant pressure.

Basic Assumptions
Consider a gas in a cylinder with a movable frictionless piston of cross-area “A” at a constant external pressure. If the force exerted by the gas on the inner wall of the piston is greater than external pressure, the piston moves upward from a height h₁ to h₂ and hence does some work. Insertion of a negative sign indicates that external pressure opposes the expansion of gas.

✨ Formula of Work
Work done by gas: W = − Force × Displacement = – F (h₂ – h₁) = −FΔh

✨ Pressure Calculation of Piston:
P = FA  ⟹  F = PA

✨ Volume Change Calculation of Gas:
V = A⋅h  ⟹  ΔV = V₂ – V₁ ⇒ ΔV = Ah₂ – Ah₁ ⇒ ΔV = A (h₂–h₁) ⇒ ΔV = A⋅Δh [∵(h₂–h₁)= Δh]

✨ Calculation of Work Done by the Expansion of the Gas at Constant Pressure:
(Substituting F and ΔV) W = – PA⋅Δh ⇒ W = –P (A⋅Δh) ⇒W = – P⋅ΔV [∵A⋅Δh = ΔV] ⇒ W = –P (V₂ – V₁)
The negative sign indicates that the work is done by the system on the surrounding.

✨ Interpretation:
🔼Expansion (V₂ > V₁) → W negative (work done by system)
🔽Compression (V₁ > V₂) → W positive (work done on system)

💡 Units of Pressure-Volume (PV) Work
SI unit: Joule (J), with P in pascal and V in m³
Other unit: atm·dm³
Conversion: 1 atm·dm³ = 101.325 J

✅ Summary
➡️ First law: ΔE = q + W
➡️ P–V work: W = –PΔV
Sign conventions:
➡️ W < 0 → expansion
➡️ W > 0 → compression
➡️ q > 0 → heat absorbed
➡️ q < 0 → heat released

Q10. Discuss the applications of the first law of thermodynamics at constant pressure and constant volume. (Prove that: i) ∆H = qₚ by deriving Pressure volume work ii) ∆E = qᵥ)

Answer

🔥 Heat Exchanged at Constant Volume (qᵥ) 📦

When a reaction is carried out at constant volume (ΔV = 0), no work is done on or by the system. Consider a closed system in which q_v heat (the subscript “v” specifies constant volume process) is absorbed or evolved when volume is constant. In this situation, internal energy change must equal to the heat change of the reaction. The equation of the first law of thermodynamics may be reformed as:

At constant volume (ΔV = 0), no PV work is done (W = PΔV = 0) ⚙️
ΔV = 0
W = PΔV ⇒ W = P × 0 ⇒ W = 0

By applying the First Law of Thermodynamics:
ΔE = q + W  ⇒  ΔE = qᵥ + 0  ⇒  ΔE = qᵥ [qᵥ = Heat absorbed or evolved at constant volume.]

✅ Conclusion: At constant volume, the heat absorbed or evolved equals the change in internal energy of the system.
Heat absorbed → increases internal energy of the system 🔋
Heat evolved → decreases internal energy 🌡️

🔥 Heat Exchanged at Constant Pressure (qₚ) 💨

Most reactions occur in open vessels at constant pressure, allowing PV work (expansion/compression) 🔥❄️
First Law at constant pressure:
ΔE = qₚ + W
W = –PΔV (At constant pressure, Pressure-Volume work, W = –PΔV)
Therefore: ΔE = qₚ – PΔV (is the heat flow to or from the system at constant pressure)
OR
qₚ = ΔE + P ΔV

Derivation of ΔH = qₚ

Internal energy: ΔE = E₂ – E₁
Volume change: ΔV = V₂ – V₁
Therefore: qₚ = (E₂ – E₁) + P(V₂ – V₁)
qₚ = E₂ – E₁ + PV₂ – PV₁
qₚ = (E₂ + PV₂) – (E₁ + PV₁) (By arranging this equation)

Define Enthalpy: H = E + PV (E + PV is called Enthalpy, denoted by H)
For initial state (1): H₁ = E₁ + PV₁
For final state (2): H₂ = E₂ + PV₂
Substituting the values in equation (1), we get:
qₚ = H₂ – H₁
qₚ = ΔH (Heat Exchanged = Change in enthalpy) (ΔH = H₂ – H₁)

✅ Conclusion: At constant pressure, enthalpy change ΔH equals heat exchanged or energy flow qₚ.

⚖️ Relation Between ΔH and ΔE

Substitute the value of qₚ into the qₚ equation:
qₚ = ΔE + P ΔV
ΔH = qₚ = ΔE + P ΔV (for gaseous reaction only)
For reactions involving solids or liquids, ΔV ≈ 0 → PΔV ≈ 0
ΔH = ΔE (for reactions involving solids or liquids, PΔV = 0)

✅ Conclusion: For reactions of solids/liquids, enthalpy change equals internal energy change.

Q11. State, explain and prove Hess’s law of enthalpy summation. Discuss its applications.

Answer

✨Discovered by: G. H. Hess (1840)
✨Dealing Variable: to compute the enthalpy changes for cyclic thermochemical changes involving two or more
✨ Other name. Second Law of Thermochemistry/ Hess’s law of enthalpy summation⚛️

📜 Statement
If a chemical reaction can occur by more than one pathway, the total enthalpy change (ΔH) is the same, provided the initial and final states are identical.
In simple words:
The heat evolved or absorbed (i.e. total enthalpy change) in a reaction is independent of the number of steps and depends only on initial and final states.

📐 Mathematical Form
∆H = ∆H₁ + ∆H₂ + ∆H₃ OR q = q₁ + q₂ + q₃ [∆H = qₚ]

💡 Explanation with Example
⚡Consider reactant A → Z
⚡Direct path: A → Z (ΔH)
⚡Indirect path: A → X → Y → Z (ΔH₁ + ΔH₂ + ΔH₃)
⚡According to Hess’s Law, enthalpy change in direct step will be equal to the sum of the enthalpies changes of indirect steps. Thus,
∆H = ∆H₁ + ∆H₂ + ∆H₃
✅ Total enthalpy change is independent of the path

🔬 Enthalpy Change Calculation
The change in enthalpy of some reactions are measured by calorimetric method. However, experimentally it is difficult to determine change in enthalpy of very slow and very fast reactions.
Example: Combustion of carbon vs carbon monoxide
C₍ₛ₎ + O₂₍g₎ → CO₂₍g₎; …………………. ΔH° =−393.5 kJ/mol --------------- (i)
CO₍g₎ + ½ O₂₍g₎ → CO₂₍g₎;…………… ΔH° = −283.5 kJ/mol --------------- (ii)

Contradictorily, the measurement of ∆H° in the combustion of carbon to carbon monoxide is quite difficult because some of the carbon monoxide is further oxidized to carbon dioxide (subtracting equation ‘ii’ form ‘i’)
C₍ₛ₎ + ½O₂₍g₎ → CO₍g₎ ∆H° = ?
ΔH° for CO formation can be calculated indirectly using Hess’s law (considering the energy cycle of the above three reactions).

🧪 Applications of Hess’s Law
➡️Calculation of heat of reaction (ΔH)
➡️Determination of heat of formation of compounds
➡️Measurement of heat of combustion
➡️Estimation of lattice energy of substances
➡️Useful in indirect calorimetric measurements ⚡

Q12. Explain exothermic and endothermic reactions with the help of the energy diagram.

Answer

🔥 Exothermic Reactions
✨Definition: Reactions in which heat is released from the system to the surroundings, making surroundings warmer.
✨Enthalpy Change: Products have lower enthalpy than reactants → ΔH negative (–ve)
✨Nature: Energetically favorable, often spontaneous; may require initial activation energy.
✨Reaction Form: Reactants → Products + Heat (ΔH=−ve)
✨Examples:
➡️C + O₂ → CO₂, ΔH = –ve (–393.7 kJ-mol⁻¹)
➡️CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = –ve (–890 kJ-mol⁻¹)

✨General Examples:
➡️Combustion of fuels 🔥 (burning of fuel and coal, oxidation of sui gas)
➡️Neutralization ⚗️ (acid + base → salt + water)
➡️Addition or synthesis reactions ➕
➡️Bond formation 🔗

✨Energy Diagram:
➡️Reactants start at a higher energy level than products
➡️Energy released = difference between reactants and products
➡️Activation energy is small hill to start reaction

❄️ Endothermic Reactions
✨Definition: Reactions in which heat is absorbed from the surroundings into the system, making surroundings colder.
✨Enthalpy Change: Products have higher enthalpy than reactants → ΔH positive (+ve)
✨Nature: Non-spontaneous; energy is required to break bonds in reactants.
✨Reaction Form: Reactants + Heat → Products (ΔH=+ve)
✨Examples:
➡️6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂, ΔH = +ve (+2816 kJ-mol⁻¹)
➡️CH₄ → C + 2H₂, ΔH = +ve (+75 kJ-mol⁻¹)

✨General Examples:
➡️Decomposition reactions ⚡
➡️Photosynthesis 🌿
➡️Bond breaking 🔓

✨Energy Diagram:
➡️Reactants start at lower energy than products
➡️Energy absorbed = difference between products and reactants
➡️Requires activation energy to initiate reaction

✏️ Solution of Numericals of Thermochemistry from Textbook and Past Papers✏️

Q1. A gaseous chemical reaction is carried out in a cylinder under a constant external pressure of 1 atm. If during reaction, volume increases from 3 dm³ to 5 dm³ by moving the piston upward, calculate the work done and express in kJ. (Example 11.1, Page # 225]

Solution

✨Given:
➡️External pressure: P = 1 atm ⚖️
➡️Initial volume: V₁ = 3 dm³
➡️Final volume: V₂ = 5 dm³
➡️Formula: Work = −P∆V or Work = −P(V₂ – V₁) [Work done in an expansion process has negative sign]
🔥Note: Negative sign indicates work done by the system on the surroundings 🔥

✨Calculate Work in atm·dm³
➡️Work = −1 atm (5 dm³ – 3 dm³) ⇒ Work = −1 atm (2 dm³) ⇒ Work = −2 atm dm³⚙️✅

✨Convert Work to Joules
➡️Work in J = work in atm dm³ × 101.325 J = −2 × 101.325 J = −202.65 J ✅ [💡 1 atm dm³ = 101.325 J]

✨Convert Work to kJ
➡️Work in kJ = work in J ÷ 1000 = −202.65 ÷ 1000 = −0.20265 kJ or −0.203 kJ💥✅

Q2. Convert the values of following quantities (Self-assessment, Page # 225]

Answer

🔥 (i) Convert 20 calories to Joules
Given: 1 cal = 4.184 J 💡
Formula:
Conversion Factor: Multiply by 4.184 for cal → J
Energy (J) = Energy (cal) × 4.184
Calculation: 20 cal = 20 × 4.184 = 83.68 J⚡
✅ Answer: 83.68 J

⚙️ (ii) Convert 3.5 atm·dm³ to kJ
Given: 1 atm·dm³ = 101.325 J = 0.101325 kJ 🔹
Conversion Factor: Multiply by 0.101325 for atm·dm³ → kJ
Formula:
Work (kJ) = Work (atm·dm³) × 0.101325
Calculation: 3.5 × 0.101325 = 0.355 kJ 💡 ✅

Q3. Burning of petrol in an automobile engine gives carbon dioxide and water vapours. If the gases do 675 J work in pushing piston outward and the system loses 435 J heat to the surrounding, calculate the internal energy change in kJ. [Example 11.2, Page # 226]

Solution

✨Given:
‘q’ = −435 J (Heat is given out, q is −)
W = −675 J (Work is done by the system, W is −)
ΔE in kJ = ?

📐 Step 1: Apply First Law of Thermodynamics
ΔE (in J) = q + W
ΔE (in J) = (−435) + (−675)

📐 Step 2: Simplify
ΔE (in J) = −1110 J

📐 Step 3: Convert to kJ
ΔE (in kJ) = −1110 ÷ 1000 = −1.110 kJ 💡 ✅

Q4. A thermochemical process is carried out at constant pressure of 8.52 atm. If it absorbs 15.5 kJ energy from the surrounding, due to which an expansion in the volume of 4.7 dm³ is occurred. Calculate its change in internal energy. [Ans: 11.34 kJ] [Exercise Q1, Page # 237]

🧮 Solution

📌 Given:
⚡Pressure: P = 8.52 atm ⚖
⚡Volume change: ΔV = 4.7 dm³ 📏
⚡Heat absorbed: q = +15.5 kJ 🔥 (Heat is absorbed, q is +)
🔍 Find: ΔE

⚙️ Step 1: Calculation of Work Done (W)
⚡Formula for work in expansion: W = −PΔV [Work done in an expansion process has negative sign]
⚡Substitute values: Work = −8.52 atm × 4.7 dm³ = −40.044 atm·dm³ 💡 ✅

✨ Convert atm·dm³ → Joules
[1 atm·dm³ = 101.325 J 💡]
⚡W in J = −40.044 × 101.325 J = −4057.45 J 💡 ✅

✨ Convert Joules → kJ
⚡W in kJ = −4057.45 ÷ 1000 = −4.057 kJ ≈ −4.058 kJ 💡 ✅

⚙️ Step 2: Calculation of ΔE using First Law of Thermodynamics
⚡ΔE = q + W = (+15.5) + (−4.058) = 11.442 kJ ≈ 11.34 kJ 💡 ✅

Q5. Calculate the enthalpy of combustion of propane at 25°C by the given information: C₃H₈ (g) + 5O₂₍g₎ → 3CO₂(g) + 4H₂O(g) ∆H° = ? [Example 11.3, Page # 230]

🧮 Solution

📌 Given Reaction:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

📌 Given Data:
ΔHᵳ°(C₃H₈) = −103.9 kJ/mol
ΔHᵳ°(CO₂) = −393.5 kJ/mol
ΔHᵳ°(H₂O) = −285.8 kJ/mol

⚙️ Formula:
ΔH°ᵣₓₙ = [∑(nₚ × ΔH°ᵳ of products)] − [∑(nᵣ × ΔH°ᵳ of reactants)] 🧪

📐 Substitution:
ΔH°ᴄомᴃ = [(3 × −393.5) + (4 × −285.8)] − [(1 × −103.9) + (5 × 0)]

📐 Simplify Products:
−1180.5 + (−1143.2) = −2323.7 kJ

📐 Simplify Reactants:
(−103.9) + (0) = −103.9 kJ

💡 Calculate ΔH:
ΔH°ᴄомᴃ = −2323.7 − (−103.9) = −2323.7 + 103.9 = −2219.8 kJ/mol 💥 ✅

Q6. Using data in following table, calculate the standard enthalpy change for each of the following reactions. [Exercise Q2, Page # 237]

🧮 Solution

📌 (i) Given Reaction:
2H₂S₍g₎ + 3O₂₍g₎ → 2H₂O₍ₗ₎ + 2SO₂₍g₎

📌 Given Data:
ΔHᵳ°(H₂S) = −20.17 kJ/mol
ΔHᵳ°(O₂) = 0 kJ/mol
ΔHᵳ°(H₂Oₗ) = −285.8 kJ/mol
ΔHᵳ°(SO₂) = −296.8 kJ/mol

⚙️ Step 1: Calculate Σ ΔHᵳ° (Products)
(2 × −285.8) + (2 × −296.8) = −571.6 − 593.6 = −1165.2 kJ

⚙️ Step 2: Calculate Σ ΔHᵳ° (Reactants)
(2 × −20.17) + (3 × 0) = −40.34 + 0 = −40.34 kJ

📐 Formula:
ΔH°ᵣₓₙ = [∑(nₚ × ΔHᵳ° of products)] − [∑(nᵣ × ΔHᵳ° of reactants)] 🧪

📐 Substitution:
ΔH°ᵣₓₙ = −1165.2 − (−40.34) = −1165.2 + 40.34 = −1124.86 kJ/mol 💥 ✅

---

📌 (ii) Given Reaction:
Fe₂O₃₍ₛ₎ + 3CO₍g₎ → 2Fe₍ₛ₎ + 3CO₂₍g₎

📌 Given Data:
ΔHᵳ°(Fe₂O₃) = −824.2 kJ/mol
ΔHᵳ°(CO) = −110.5 kJ/mol
ΔHᵳ°(Fe) = 0 kJ/mol
ΔHᵳ°(CO₂) = −393.5 kJ/mol

⚙️ Step 1: Σ ΔHᵳ° (Products)
(2 × 0) + (3 × −393.5) = 0 − 1180.5 = −1180.5 kJ

⚙️ Step 2: Σ ΔHᵳ° (Reactants)
(−824.2) + (3 × −110.5) = −824.2 − 331.5 = −1155.7 kJ

📐 Formula:
ΔH°ᵣₓₙ = [∑(nₚ × ΔHᵳ° of products)] − [∑(nᵣ × ΔHᵳ° of reactants)] 🧪

📐 Substitution:
ΔH°ᵣₓₙ = −1180.5 − (−1155.7) = −1180.5 + 1155.7 = −24.8 kJ/mol 💥 ✅

Q7. In the manufacturing of HNO₃ by the Ostwald process, one of the most important exothermic reactions is the oxidation of ammonia. [Exercise Q3, Page # 238]

🧮 Solution

📌 Given Reaction:
4NH₃₍g₎ + 5O₂₍g₎ → 4NO₍g₎ + 6H₂O₍g₎ (Ostwald process)

📌 Given Data:
ΔH°ᵳ NH₃ = −46.91 kJ/mol
ΔH°ᵳ NO = 90.25 kJ/mol
ΔH°ᵳ H₂O(g) = −285.8 kJ/mol
ΔH°ᵳ O₂ = 0 kJ/mol

⚙️ Step 1: Calculate Σ ΔH°ᵳ (Products)
(4 × 90.25) + (6 × −285.8) = 361 − 1714.8 = −1353.8 kJ

⚙️ Step 2: Calculate Σ ΔH°ᵳ (Reactants)
(4 × −46.91) + (5 × 0) = −187.64 + 0 = −187.64 kJ

📐 Formula:
ΔH°ᵣₓₙ = [∑(nₚ × ΔH°ᵳ of products)] − [∑(nᵣ × ΔH°ᵳ of reactants)] 🧪

📐 Substitution:
ΔH°ᵣₓₙ = −1353.8 − (−187.64) = −1353.8 + 187.64 = −1166.16 kJ/mol 💥 ✅

Q8. Calculate the standard heat of formation of acetylene (C₂H₂) by using the data of the following thermochemical equations: [Example 11.4, Page # 231]

🧮 Solution

📌 Given Thermochemical Equations:
(i) C₍ₛ₎ + O₂₍g₎ → CO₂₍g₎ ----------------------------- ΔH = −393.5 kJ
(ii) H₂₍g₎ + ½O₂₍g₎ → H₂O₍ₗ₎ -------------------------- ΔH = −285.8 kJ
(iii) C₂H₂₍g₎ + ⁵/₂O₂₍g₎ → 2CO₂₍g₎ + H₂O₍ₗ₎ ----------- ΔH = −1299.5 kJ
(iv) 2C₍ₛ₎ + H₂₍g₎ → C₂H₂₍g₎ -------------------------- ΔHᵳ = ? (Desired equation)

⚙️ Adjust given equations (Hess’s Law):
(i) Multiply by 2 (because target has 2C)… 2C₍ₛ₎ + 2O₂₍g₎ → 2CO₂₍g₎ ΔH = −787.0 kJ
(ii) Keep as it is (as it has 1 H₂)… H₂₍g₎ + ½O₂₍g₎ → H₂O₍ₗ₎ ΔH = −285.8 kJ
(iii) Reverse, change sign of ΔH (C₂H₂ is now product)… 2CO₂₍g₎ + H₂O₍ₗ₎ → C₂H₂₍g₎ + ⁵/₂O₂₍g₎ ΔH = +1299.5 kJ

📐 Final Step:
(iv) Add the three equations with their ΔH (Hess’s Law)…
2C₍ₛ₎ + H₂₍g₎ → C₂H₂₍g₎ ΔH = 226.7 kJ/mol ✅

Q9. Iso octane (C₈H₁₈) is an efficient fuel with a high octane rating. The combustion of (C₈H₁₈) in an international combustion engine is represented in the following thermochemical equation. Find its standard heat of combustion. [Exercise Q4, Page # 238]

🧮 Solution

📌 Given Reaction:
C₈H₁₈₍ₗ₎ + 12½ O₂₍g₎ → 8CO₂₍g₎ + 9H₂O₍ₗ₎ (ΔH° = ?)

📌 Given Data:
ΔHᵳ° CO₂₍g₎ = −393.5 kJ/mol
ΔHᵳ° H₂O₍ₗ₎ = −285.8 kJ/mol
ΔHᵳ° C₈H₁₈₍ₗ₎ = −223.8 kJ/mol

⚙️ Step 1: Calculate Σ ΔH°ᵳ (Products)
(8 × −393.5) + (9 × −285.8) = −3148 + (−2572.2) = −5720.2 kJ

⚙️ Step 2: Calculate Σ ΔH°ᵳ (Reactants)
(−223.8) + (12.5 × 0) = −223.8 kJ

📐 Formula (Hess’s Law):
ΔH° combustion = [∑(nₚ × ΔH°ᵳ of products)] − [∑(nᵣ × ΔH°ᵳ of reactants)] 🧪

📐 Substitution:
ΔH° combustion = −5720.2 − (−223.8) = −5720.2 + 223.8 = −5496.4 kJ/mol 💥 ✅

Q10. Glycerol (C₃H₈O₃) is a well-known organic compound due to its versatile uses. Calculate the standard enthalpy of formation of Glycerol from the data given below.
C₍ₛ₎ + O₂₍g₎ → CO₂₍g₎ (ΔH° = −393.5 kJ/mol)
H₂₍g₎ + ½O₂₍g₎ → H₂O₍ₗ₎ (ΔH° = −285.8 kJ/mol)
C₃H₈O₃₍ₗ₎ + 3½ O₂₍g₎ → 3CO₂₍g₎ + 4H₂O₍ₗ₎ (ΔH° = −1654.1 kJ/mol)
3C₍ₛ₎ + 4H₂₍g₎ + 3/₂O₂ → C₃H₈O₃₍ₗ₎ (ΔHᵳ° = ?)

[Exercise Q5, Page # 238]

🧮 Solution

📌 Target Equation:
3C₍ₛ₎ + 4H₂₍g₎ + 3/₂O₂ → C₃H₈O₃₍ₗ₎ (ΔHᵳ° = ?)

📌 Given Thermochemical Equations:
C₍ₛ₎ + O₂₍g₎ → CO₂₍g₎ (ΔH° = −393.5 kJ/mol)
H₂₍g₎ + ½O₂₍g₎ → H₂O₍ₗ₎ (ΔH° = −285.8 kJ/mol)
C₃H₈O₃₍ₗ₎ + 3½ O₂₍g₎ → 3CO₂₍g₎ + 4H₂O₍ₗ₎ (ΔH° = −1654.1 kJ/mol)

⚙️ Steps (Hess’s Law):
1️⃣ Because the target equation has 3C as a reactant, we multiply the first equation and its ΔH by 3.
2️⃣ Because the target equation has 4H₂ as a reactant, we multiply the second equation and its ΔH by 4.
3️⃣ Because the target equation has C₃H₈O₃ as a product, we reverse the third equation; the sign of ΔH is therefore changed.
4️⃣ We then add the three equations with their enthalpy changes in accordance with Hess’s law.

Q11. Use the data provided below for the formation of RbCl₍ₛ₎, write thermochemical equations for all the steps involved in the Born Haber cycle and determine the enthalpy of formation of RbCl₍ₛ₎.

Sublimation energy of Rb₍ₛ₎ = 82 kJ/mol
Ionization energy of Rb(g) = 403 kJ/mol
Dissociation energy of Cl₂(g) = 242 kJ/mol
Electron affinity of Cl₂(g) = −348.5 kJ/mol
Lattice energy of RbCl₍ₛ₎ = −689 kJ/mol

🧮 Solution

Since, according to Hess’s law overall enthalpy change of a cyclic thermochemical process is zero, the sum of all five steps involved in the formation of solid ionic compound like RbCl(s) will get heat of formation of RbCl₍ₛ₎. Thermochemical equations associated with Born Haber cycle in the formation of RbCl(s) may be written as:

☀️ Sublimation:
Rb₍ₛ₎ → Rb₍g₎...................... ΔH°ᵴᴜᴮ = +82 kJ/mol

⚡ Ionization:
Rb₍g₎ → Rb⁺₍g₎ + ē ............... ΔH°ᵻᴇ = +403 kJ/mol

🧨 Dissociation:
½Cl₂₍g₎ → Cl₍g₎................. ΔH°ᴅ = ½ × +242 kJ/mol

🧲 Electron Affinity:
Cl₍g₎ + ē → Cl⁻₍g₎ ............... ΔH°ᴇѧ = −348.5 kJ/mol

🏗️ Lattice Energy:
Rb⁺₍g₎ + Cl⁻₍g₎ → RbCl₍ₛ₎ .......... ΔH°ʟᴇ = −689 kJ/mol

🔥 Formation of Ionic Solid (Heat of Formation):
Rb₍g₎ + Cl₍g₎ → RbCl₍ₛ₎............. ΔH°ᵳ = −431.5 kJ/mol

📐 Calculation Using Born-Haber Equation:
ΔH°ᵳ = ΔH°ᵴᴜᴮ + IE + ½D + EA + U
Or
ΔH°ᵳ = ΔH°ᵴᴜᴮ + ΔH°ᵻᴇ + ½ΔH°ᴅ + ΔH°ᴇѧ + ΔH°ʟᴇ
ΔH°ᵳ (RbCl) = 82 + 403 + 121 − 348.5 − 689 = −431.5 kJ/mol ✅

Q12. Draw a fully labeled Born Haber cycle for Rubidium chloride (RbCl) and determine the lattice energy by using the following values. (all in kJ/mol)

• I.P1st of Rb = 403 kJ/mol
• Electron affinity of Cl = −349 kJ/mol
• Bond energy of Cl₂ = 242 kJ/mol
• Sublimation energy of Rb = 86.5 kJ/mol
• Heat of formation of RbCl = −430.5 kJ/mol

[Exercise Q6, Page # 238]

🧮 Solution

Since, according to Hess’s law overall enthalpy change of a cyclic thermochemical process is zero, the sum of all five steps involved in the formation of solid ionic compound like RbCl₍ₛ₎ will get heat of formation of RbCl₍ₛ₎. Thermochemical equations associated with Born Haber cycle in the formation of RbCl₍ₛ₎ may be written as:

☀️ Sublimation:
Rb₍ₛ₎ → Rb₍g₎ ................. ΔH°ᵴᴜᴮ = +86.5 kJ/mol

⚡ Ionization:
Rb₍g₎ → Rb⁺₍g₎ + ē ............ ΔH°ᵻᴇ = +403 kJ/mol

🧨 Dissociation:
½Cl₂₍g₎ → Cl₍g₎ ............... ΔH°ᴅ = ½ × 242 = +121 kJ/mol

🧲 Electron Affinity:
Cl₍g₎ + ē → Cl⁻₍g₎............ ΔH°ᴇѧ = −349 kJ/mol

🏗️ Lattice Energy:
Rb⁺₍g₎ + Cl⁻₍g₎ → RbCl₍ₛ₎......... ΔH°ʟᴇ = ?

📐 Calculation Using Born-Haber Equation:
ΔH°ᵳ = ΔH°ᵴᴜᴮ + ΔH°ᵻᴇ + ½ΔH°ᴅ + ΔH°ᴇѧ + ΔH°ʟᴇ
−430.5 = 86.5 + 403 + 121 − 349 + ΔH°ʟᴇ
ΔH°ʟᴇ = −430.5 − (86.5 + 403 + 121 − 349)
ΔH°ʟᴇ = −430.5 − 261 = −692 kJ/mol ✅

Welcome to Inam Jazbi’s ultimate guide to XI Chemistry – Thermochemistry (Chapter # 11)! 🔥 Whether you're preparing for board exams or competitive tests like MDCAT/ ECAT, this blog is packed with model test questions, solved numericals, and MCQs that will help you ace Thermochemistry!

In this blog, you’ll find:

🔥Chapter 11 Test Questions to practice the core concepts of thermochemistry.

🔥Detailed solutions for numericals and problems.

🔥MCQs to test your understanding and boost your exam confidence.

With Inam Jazbi’s expert tips and solutions, you’re all set to master Thermochemistry and score big in your XI Chemistry exams! 🚀 Let’s dive into the heat of learning and make Thermochemistry your strength! 💥

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👀 غور کرنے پہ یاد آتی ہے

🏚️ کون اس گھر کی دیکھ بھال کرے
💔 روز ایک چیز ٹوٹ جاتی ہے

✍️ جونؔ ایلیا

💥 غزل ۔۔۔۔ جونؔ ایلیا 💥

🌟 یہ ملکِ جاں یہ حقیقت نہ رائیگاں جائے
💭 یہ دل کے خواب کی صورت نہ رائیگاں جائے

🏙️ ہے شہر شہر کی محنت یہ منزلِ مقصود
🌌 یہ شہر شہر کی محنت نہ رائیگاں جائے

🤝 یہ رنگ رنگ کے رشتے بکھر نہ جائیں کہیں
💡 یہ خود سے اپنی رفاقت نہ رائیگاں جائے

🌸 سوائے حسنِ طبیعت دھرا بھی اب کیا ہے
🌙 کہیں یہ حسنِ طبیعت نہ رائیگاں جائے

🔊 ہے گوشہ گوشہ یہاں سازشوں کی سرگوشی
💔 ہمارا عہدِ محبت نہ رائیگاں جائے

👥 کہاں کہاں سے یہاں آکے ہم ہوئے ہیں بہم
✨ یہ اجتماع یہ صحبت نہ رائیگاں جائے

⏳ نہ بھولنا کہ یہ مہلت ہے آخری مہلت
🌟 رہے خیال یہ مہلت نہ رائیگاں جائے

💭 مجھے تو اے میرے دل تجھ سے ہے یہی کہنا
🔥 تیرے جنون کی حالت نہ رائیگاں جائے

✍️ جونؔ ایلیا

💥 غزل ۔۔۔۔ جونؔ ایلیا 💥

🌙 نشئہ ماہ و سال ہے، تاحال
💭 شوق اس کا کمال ہے، تاحال

🌸 نکہتِ گل ادھر نہ آئیو تو
😔 جی ہمارا نڈھال ہے، تاحال

💔 میرا سینہ چھلا ہوا ہے مگر
⚡ شوقِ بحث و جدال ہے، تاحال

🌀 اس عبث خانئہ حوادث میں
❓ ہر جواب اک سوال ہے، تاحال

⏳ بڑھ رہا ہوں زوال کی جانب
💔 دل میں زخمِ کمال ہے، تاحال

🏚️ کب کا تاراج ہوچکا ہوں مگر
🌟 ذہن میں اک مثال ہے، تاحال

💔 زخمِ کاری کے باوجود
🌿 ہوسِ اندمال ہے، تاحال

🧣 دامنِ آلودگی کے بعد بھی تو
🌸 آپ اپنی مثال ہے، تاحال

💭 ہے یہ صورت کہ اشتیاق اس کا
💔 بے امیدِ وصال ہے، تاحال

😔 تھا جو شکوہ سو ہے وہ تاایں دم
🌙 وہ جو تھا اک ملال ہے، تاحال

💔 زندگی ہے لہولہان مگر
🎨 رنگ بے خدوخال ہے، تاحال

📖 ہے سوادِ ختن غزل میری
🦌 تو غزل کا غزال ہے، تاحال

🌹 لالہ رویا، شکن شکن مویا
💭 تجھ کو پانا محال ہے، تاحال

🩺 کتنے چارہ گروں نے زحمت کی
😔 پر وہی میرا حال ہے، تاحال

✍️ جونؔ ایلیا

💥 غزل ۔۔۔۔ جونؔ ایلیا 💥

💭 نہیں جذبے کسی بھی قیمت کے
🌙 ہم ہیں حیران اپنی حیرت کے

🤔 اس میں آخر عجب کی بات ہے کیا
💔 تم نہیں تھے مری طبیعت کے

😔 پوچھ مت بے شکایتی کا عذاب
🌟 کیا عجب عیش تھے شکایت کے

💧 یہ جو آنسو ہیں، رخصتی آنسو
🎁 یہ عطیے ہیں دل کی عادت کے

📜 ہم ہی شیعوں کے مجتہد ہیں مغاں!
⚖️ ہم ہی مفتی ہیں اہلسنت کے

💉 ہم تو بس خون تھوکتے ہیں میاں
🛠️ نہیں خوگر کسی مشقت کے

💕 یہ جو لمحے ہیں وصال کے ہیں میاں
🌙 ہیں یہ لمحے تمام ہجرت کے

✨ جونؔ، یزدان و آدم و ابلیس
📖 ہیں عجب معجزے حکایت کے

✍️ جونؔ ایلیا