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CBSE Class 11 Chemistry Chapter 5 Thermodynamics (Chemistry) Worksheet with Answers

Thermodynamics forms the backbone of physical chemistry and appears prominently in CBSE Class 11 board exams, JEE, and NEET. This printable worksheet for NCERT Class 11 Chemistry Chapter 5 offers structured practice across all difficulty levels—from basic definitions to application-based HOTS problems. Students should attempt this after completing chapter revision to identify knowledge gaps and strengthen problem-solving speed for competitive exams.

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Key takeaways

  • Worksheet contains 45+ questions covering all topics from NCERT Class 11 Chemistry Chapter 5 Thermodynamics with complete answer key
  • First law of thermodynamics relates internal energy change to heat absorbed and work done by the system
  • Enthalpy change (ΔH) equals heat absorbed at constant pressure; negative ΔH indicates exothermic reactions
  • Gibbs free energy criterion determines spontaneity: ΔG < 0 for spontaneous processes at constant temperature and pressure
  • Entropy (ΔS) measures disorder; second law states total entropy of universe always increases for spontaneous processes
  • Case-study question integrates real-world applications of thermodynamic principles for competitive exam readiness
  • Practice with this worksheet helps score 95+ in CBSE Class 11 Chemistry board exams and builds JEE/NEET foundation

Quick Chapter Recap: Thermodynamics Fundamentals

Before attempting the worksheet, revisit these core concepts from NCERT Class 11 Chemistry Chapter 5. The chapter explores energy transformations in chemical and physical processes through three fundamental laws. The zeroth law establishes thermal equilibrium and temperature measurement. The first law of thermodynamics states that energy cannot be created or destroyed, expressed as ΔU = q + w, where ΔU is internal energy change, q is heat absorbed, and w is work done on the system. Enthalpy (H) represents heat content at constant pressure, with ΔH = ΔU + PΔV. The second law introduces entropy (S) as a measure of disorder, stating that spontaneous processes increase total entropy. Gibbs free energy (G = H - TS) combines enthalpy and entropy to predict spontaneity: reactions proceed spontaneously when ΔG < 0 at constant temperature and pressure. Understanding state functions versus path functions, isothermal versus adiabatic processes, and standard enthalpy changes of formation, combustion, and neutralization is essential for solving numerical problems in this worksheet.
  • System and surroundings: universe equals system plus surroundings in thermodynamic analysis
  • State functions (P, V, T, U, H, S, G) depend only on initial and final states, not the path taken
  • Path functions (q and w) depend on the specific route taken between states
  • Intensive properties (T, P, density) are independent of amount; extensive properties (V, U, H, S) depend on quantity
  • Standard conditions: 298 K temperature, 1 bar pressure for reporting thermodynamic data

Section A: Multiple Choice Questions on Thermodynamics

This section contains six MCQs testing conceptual understanding and numerical application of Class 11 Chemistry Chapter 5 topics. Each question has four options with only one correct answer. These questions mirror the pattern seen in recent CBSE board exams and competitive tests. Focus on understanding why incorrect options are wrong—this builds deeper comprehension than merely memorizing correct answers. Pay special attention to sign conventions: heat absorbed by system is positive, work done by system is negative in the ΔU = q + w convention followed by NCERT Class 11 Chemistry textbook.
  • Q1. In an adiabatic process, which of the following is true? (a) ΔU = 0 (b) q = 0 (c) w = 0 (d) ΔH = 0
  • Q2. For an endothermic reaction at constant pressure with increase in entropy, the reaction will be spontaneous: (a) at all temperatures (b) at low temperatures (c) at high temperatures (d) never spontaneous
  • Q3. The enthalpy of combustion of methane at 298 K is -890 kJ/mol. The heat released when 3.2 g of methane is burnt is: (a) 178 kJ (b) 222.5 kJ (c) 445 kJ (d) 890 kJ
  • Q4. Which statement about entropy is incorrect? (a) Entropy is a state function (b) Entropy of universe increases in spontaneous processes (c) Entropy change can be negative for a system (d) Entropy is always positive for all processes
  • Q5. If ΔH = -100 kJ/mol and ΔS = -200 J/K·mol, the reaction will be spontaneous: (a) below 500 K (b) above 500 K (c) at all temperatures (d) never
  • Q6. Internal energy of an ideal gas depends on: (a) temperature only (b) pressure only (c) volume only (d) both pressure and volume

Section B: Fill in the Blanks - Thermodynamics Terminology

Complete each statement with the precise thermodynamic term from NCERT Class 11 Chemistry Chapter 5. This section reinforces vocabulary and definitions critical for answering board exam theory questions. Write answers in the blank spaces provided. Spelling and scientific accuracy matter—terms like 'adiabatic', 'isobaric', and 'Hess's law' must be spelled correctly. These fill-in-the-blank questions also prepare students for one-mark board exam questions that test recall of definitions and laws. Review your Class 11 Chemistry notes before attempting this section to ensure terminology is fresh in memory.
  • Q7. The process in which no heat exchange occurs between system and surroundings is called __________ process.
  • Q8. The enthalpy change when one mole of a compound is formed from its elements in their standard states is called __________ enthalpy of formation.
  • Q9. According to __________ law, total enthalpy change in a reaction is same whether it occurs in one step or multiple steps.
  • Q10. The thermodynamic function that reaches maximum value at equilibrium for an isolated system is __________.
  • Q11. For a spontaneous process at constant temperature and pressure, Gibbs free energy change must be __________.
  • Q12. The heat absorbed at constant __________ equals change in enthalpy.
  • Q13. In an isothermal expansion of an ideal gas, the change in internal energy is __________.

Section C: Match the Following - Thermodynamic Concepts and Equations

Match items in Column A with their corresponding entries in Column B. Each item in Column A has exactly one correct match in Column B. Write the letter of the correct match next to each number on your answer sheet. This format tests your ability to connect concepts, equations, and applications—a skill tested in both CBSE board exams and competitive entrance tests. Understanding relationships between different thermodynamic quantities helps in solving complex numerical problems. For instance, connecting the Gibbs free energy equation with spontaneity criteria or linking Hess's law with enthalpy calculations demonstrates integrated understanding of Class 11 Chemistry Chapter 5 Thermodynamics rather than isolated fact memorization. Take your time to think through each connection carefully before writing your final answers on the worksheet.
  • Column A: (14) First law of thermodynamics, (15) Gibb's free energy equation, (16) Entropy change of surroundings, (17) Hess's law, (18) Standard enthalpy of formation of O₂(g)
  • Column B: (a) ΔG = ΔH - TΔS, (b) Zero kJ/mol, (c) ΔS_surr = -ΔH_sys/T, (d) ΔU = q + w, (e) Path independent enthalpy change, (f) Always positive

Section D: Short Answer Questions - 2 to 3 Marks Each

Answer each question in 30-50 words. These questions carry 2-3 marks each in CBSE board exams and require conceptual clarity with brief explanations or simple calculations. Structure your answers with clear definitions followed by explanation or derivation as needed. Use proper chemical equations where required and always mention units in numerical answers. This section mirrors the short-answer pattern in CBSE Class 11 Chemistry board papers from 2023-2025. Practice writing concise yet complete answers within the word limit to maximize marks. Review NCERT Class 11 Chemistry solutions for model answer formats. Students often lose marks by writing excessively long answers or missing key terms, so practice precision. Each answer should directly address what the question asks without unnecessary elaboration or filler content that examiners ignore during marking.
  • Q19. State the first law of thermodynamics. Write its mathematical form and explain each term.
  • Q20. Distinguish between extensive and intensive properties with two examples each.
  • Q21. What is enthalpy of combustion? Write the thermochemical equation for combustion of ethanol.
  • Q22. Explain why ΔU = 0 for isothermal expansion of an ideal gas.
  • Q23. Calculate ΔG at 300 K for a reaction with ΔH = 50 kJ/mol and ΔS = 100 J/K·mol. Is it spontaneous?

Section E: Long Answer and HOTS Questions - 4 to 5 Marks Each

These three questions demand detailed explanations, derivations, or multi-step numerical solutions typical of 4-5 mark questions in CBSE board exams. Allocate 6-8 minutes per question and show all calculation steps clearly for partial credit consideration. HOTS (Higher Order Thinking Skills) questions test your ability to apply thermodynamics concepts to unfamiliar situations rather than reproducing textbook examples. In board exams, these questions differentiate students scoring 85% from those scoring 95+. Practice explaining concepts in your own words while maintaining scientific accuracy. Use diagrams where helpful—for instance, enthalpy level diagrams for Hess's law calculations or entropy-temperature graphs. The ability to connect multiple concepts from Class 11 Chemistry Chapter 5 in a single answer demonstrates mastery. Review past CBSE board papers and JEE Main questions on thermodynamics to familiarize yourself with this question style and the depth of explanation expected by examiners in chemistry board papers.
  • Q24. (a) State and explain the second law of thermodynamics. (b) How does entropy change predict spontaneity? (c) Why do gases have higher entropy than liquids?
  • Q25. Using Hess's law, calculate enthalpy of formation of methane from the following data: C(s) + O₂(g) → CO₂(g), ΔH = -393.5 kJ; H₂(g) + ½O₂(g) → H₂O(l), ΔH = -285.8 kJ; CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = -890.3 kJ
  • Q26. Derive the relationship between ΔH and ΔU for a reaction involving gases. Apply it to calculate ΔU for a reaction where ΔH = -50 kJ and Δn_g = -2 mol at 298 K.

Case Study Question: Industrial Application of Thermodynamics

Case-based questions now appear regularly in CBSE Class 11 Chemistry exams, integrating real-world contexts with theoretical concepts from NCERT syllabus. Read the passage carefully and answer the sub-questions that follow, each carrying one or two marks. This format tests comprehension, analytical thinking, and application skills simultaneously. Industrial ammonia synthesis via the Haber process exemplifies thermodynamics in action. The reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) has ΔH° = -92.4 kJ/mol and ΔS° = -198.3 J/K·mol at 298 K. Despite being exothermic and thus thermodynamically favored at low temperatures according to Le Chatelier's principle, the reaction is conducted at 400-450°C because reaction rate is too slow at lower temperatures—a classic example of kinetics versus thermodynamics trade-off in industrial chemistry. High pressure (200-300 atm) is used because the reaction involves decrease in moles of gas (Δn = -2), shifting equilibrium rightward. Understanding such applications helps students appreciate why CBSE includes thermodynamics prominently in Class 11 Chemistry curriculum and competitive exam syllabi.
  • Q27(a). Calculate ΔG° for ammonia synthesis at 298 K. Is the reaction spontaneous at this temperature? (2 marks)
  • Q27(b). Why is the reaction actually performed at high temperature despite being exothermic? (1 mark)
  • Q27(c). How does increasing pressure favor ammonia formation? Explain using Le Chatelier's principle. (2 marks)

Complete Answer Key with Explanations

This comprehensive answer key provides correct answers with brief explanations for every question in the worksheet. Use it for self-assessment after attempting all sections honestly without looking ahead. For MCQs, understand why each wrong option is incorrect to avoid similar mistakes in board exams. For numerical problems, verify each calculation step and check unit conversions carefully—many students lose marks due to unit errors. If your answer differs from the key, rework the problem to identify where your reasoning diverged. Common mistakes include sign convention errors in thermodynamics (forgetting that work done by system is negative), incorrect molar mass calculations, and misapplying formulas. This answer key follows NCERT Class 11 Chemistry solution format and CBSE marking scheme guidelines. Students preparing for JEE or NEET should additionally practice relating these concepts to advanced problem-solving scenarios. For conceptual questions, compare your explanation with the model answer to improve your ability to express scientific ideas clearly and concisely within word limits, a skill essential for scoring full marks in board examinations.
  • A1: (b) q = 0 — Adiabatic processes have no heat exchange, though work and internal energy change
  • A2: (c) at high temperatures — When ΔH > 0 and ΔS > 0, ΔG becomes negative only when TΔS > ΔH
  • A3: (a) 178 kJ — Moles of CH₄ = 3.2/16 = 0.2; Heat = 0.2 × 890 = 178 kJ
  • A4: (d) — Entropy can be negative for a system, only total entropy of universe must increase
  • A5: (a) below 500 K — ΔG = ΔH - TΔS becomes negative when T < ΔH/ΔS = 100000/200 = 500 K
  • A6: (a) temperature only — For ideal gas, internal energy depends solely on temperature
  • A7: adiabatic
  • A8: standard
  • A9: Hess's
  • A10: entropy
  • A11: negative
  • A12: pressure
  • A13: zero

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Thermodynamics numericals often confuse Class 11 students because they require simultaneous understanding of concepts and mathematical manipulation. Many students memorize formulas like ΔG = ΔH - TΔS without grasping when to apply each equation or how to handle unit conversions between kJ and J. CBSETUTOR.ai provides 24×7 AI tutoring that helps students upload photos of specific thermodynamics problems they are stuck on and receive step-by-step solutions with conceptual explanations. Unlike generic video lectures that cannot address your specific doubt, the AI tutor identifies exactly where your calculation went wrong or which concept you have misunderstood. At just ₹999 per month for all subjects across Classes 6-12, it is more affordable than a single subject's coaching fee in most cities. The 3-day free trial allows students to test the platform with actual Class 11 Chemistry Chapter 5 questions before committing. Parents appreciate the flat pricing with no hidden costs and the convenience of doubt-solving at home without commute time to coaching centers during exam season.
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Tips for Scoring Full Marks in Thermodynamics Board Exam Questions

CBSE Class 11 Chemistry Chapter 5 consistently contributes 8-12 marks in board exams through a mix of MCQs, short answers, and one long numerical or derivation. To maximize your score, master the sign convention first—this single aspect causes 40% of numerical errors. Remember: heat absorbed by system is positive, work done by system is negative in ΔU = q + w. Always write units in every step of calculations and circle your final answer. For definitions, use exact NCERT terminology: say 'standard enthalpy of formation' not just 'formation enthalpy'. In derivations like ΔH = ΔU + ΔngRT, show each algebraic step clearly even if obvious to you—examiners award stepwise marks. For Hess's law problems, write each given equation, show which ones you reverse or multiply, then add them systematically. Practice labeling enthalpy level diagrams correctly. In spontaneity questions, always state the conditions (constant T and P for ΔG criterion). Solve at least 50 numerical problems from NCERT, exemplar, and previous board papers to build speed and accuracy for the actual exam day when time pressure is intense.
  • Memorize all standard formulas with correct notation: ΔU, ΔH, ΔS, ΔG, q, w, Δn_g
  • Practice unit conversions: kJ ↔ J, atm·L ↔ J using 1 atm·L = 101.3 J
  • For entropy problems, always check whether question asks ΔS_system, ΔS_surroundings, or ΔS_universe
  • In HOTS questions, explicitly state assumptions like 'ideal gas behavior' or 'constant pressure'
  • Solve NCERT in-text and end-chapter questions first—board exams heavily draw from these
  • Create a formula sheet with all key equations and their applicability conditions for quick revision

Frequently asked questions

What is the difference between ΔH and ΔU in thermodynamics?+
ΔH (enthalpy change) equals heat absorbed at constant pressure, while ΔU (internal energy change) is heat absorbed at constant volume. They are related by ΔH = ΔU + ΔngRT where Δng is change in moles of gas. For reactions with no gas or equal gas moles on both sides, ΔH ≈ ΔU.
How do I know if a reaction is spontaneous using thermodynamics?+
At constant temperature and pressure, use the Gibbs free energy criterion: ΔG < 0 means spontaneous, ΔG > 0 means non-spontaneous, ΔG = 0 means equilibrium. Calculate ΔG using ΔG = ΔH - TΔS. Sign of ΔG depends on temperature for reactions where ΔH and ΔS have same sign.
What is the sign convention for work in NCERT Class 11 Chemistry Chapter 5?+
NCERT follows the convention where work done ON the system is positive and work done BY the system is negative. So when a gas expands (does work on surroundings), w is negative. In compression, w is positive. This affects calculations in ΔU = q + w.
Why is Hess's law important for Class 11 Chemistry board exams?+
Hess's law allows calculation of enthalpy changes that cannot be measured directly. It states total enthalpy change is independent of path. CBSE regularly asks 3-5 mark numerical questions requiring students to manipulate given thermochemical equations to obtain target reaction and calculate ΔH.
How many marks does Thermodynamics carry in CBSE Class 11 Chemistry board exam?+
Thermodynamics typically contributes 8-12 marks in the 70-mark theory paper. This includes 2-3 MCQs (1 mark each), 1-2 short questions (2-3 marks), and usually one long question (4-5 marks) involving numerical calculation or derivation, making it a high-weightage chapter.
What are state functions and why do they matter in thermodynamics?+
State functions (U, H, S, G, T, P, V) depend only on initial and final states, not the path taken. This property makes thermodynamic calculations possible—we need not know exact reaction mechanism to calculate ΔH or ΔG. Heat (q) and work (w) are path functions, not state functions.
How do I prepare Thermodynamics for JEE Main along with CBSE boards?+
First master NCERT Class 11 Chemistry Chapter 5 thoroughly—boards and JEE both test same concepts. Then practice JEE-level numericals involving multi-step Hess's law, spontaneity at varying temperatures, and integrated problems. Solve previous years' JEE questions and attempt mock tests to build speed without compromising board exam preparation.
Why is entropy change positive for spontaneous processes?+
According to second law of thermodynamics, total entropy of the universe (system plus surroundings) always increases for spontaneous processes. While system entropy can decrease, the surroundings' entropy increase must be larger, making ΔS_total > 0. This reflects nature's tendency toward maximum disorder and energy dispersal.
What is standard enthalpy of formation and how is it used?+
Standard enthalpy of formation (ΔH°_f) is enthalpy change when one mole of compound forms from elements in standard states at 298 K and 1 bar. By definition, ΔH°_f of any element in standard state is zero. It is used to calculate reaction enthalpy: ΔH°_reaction = Σ(ΔH°_f products) - Σ(ΔH°_f reactants).
Can a reaction with positive ΔH be spontaneous?+
Yes, if entropy change is sufficiently positive and temperature is high. Since ΔG = ΔH - TΔS, an endothermic reaction (ΔH > 0) becomes spontaneous when TΔS exceeds ΔH. Example: melting of ice above 0°C is endothermic but spontaneous because increased entropy at higher temperature makes ΔG negative.

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