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NCERT Solutions for Class 11 Chemistry Chapter 5 Chemical Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 5 Chemical Thermodynamics covers the basic thermodynamic terms, the first law of thermodynamics, enthalpy, Hess's Law, and spontaneity of processes (second law).

Overview of Topics & Subtopics (Click to Jump)

1. Basic Concepts of Thermodynamics

Short Note:

• A system is the part of the universe under study; the surroundings is everything else. Systems can be open, closed, or isolated depending on exchange of matter and energy.

• State functions (e.g. internal energy U, enthalpy H) depend only on the initial and final state, not on the path taken; path functions (e.g. heat q, work w) depend on the path.

2. First Law of Thermodynamics

Short Note:

• ΔU = q + w, i.e. energy can neither be created nor destroyed, only converted from one form to another (Law of Conservation of Energy applied to thermodynamics).

3. Enthalpy and Its Types

Short Note:

• Enthalpy (H) = U + pV; ΔH = ΔU + pΔV (at constant pressure), representing the heat absorbed or released in a process at constant pressure.

• Hess's Law of Constant Heat Summation: the total enthalpy change for a reaction is the same, regardless of the number of steps taken, as long as initial and final conditions are the same.

• Types include enthalpy of formation, combustion, neutralisation, atomisation, and bond enthalpy.

4. Spontaneity and Gibbs Energy

Short Note:

• Entropy (S) is a measure of the degree of randomness or disorder of a system; the Second Law states the total entropy of the universe increases in every spontaneous process.

• Gibbs Free Energy: ΔG = ΔH − TΔS. For a spontaneous process at constant T and p, ΔG < 0.

NCERT Textbook Exercise Questions

A gas absorbs 200 J of heat and does 80 J of work on the surroundings. Calculate the change in internal energy.
Answer: ΔU = q + w = 200 + (−80) = 120 J (work done by the system on surroundings is taken as negative in the convention ΔU=q+w).
Calculate the standard enthalpy of formation of CH4(g) given: C(graphite) + 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.
Answer: Using Hess's Law: ΔHf(CH4) = ΔHf(CO2) + 2ΔHf(H2O) − ΔHcombustion(CH4) = (−393.5) + 2(−285.8) − (−890.3) = −393.5 − 571.6 + 890.3 = −74.8 kJ/mol.
For the reaction 2A(g) + B(g) → 2D(g), ΔU = −10.5 kJ and Δng = −1 mol. Calculate ΔH.
Answer: ΔH = ΔU + ΔngRT = −10.5 kJ + (−1)(8.314 × 10−3 kJ/K/mol)(298 K) = −10.5 − 2.478 = −12.98 kJ (approx).
Predict the sign of ΔS for the reaction: 2NH3(g) → N2(g) + 3H2(g).
Answer: The reaction converts 2 moles of gas into 4 moles of gas, increasing the number of gaseous particles and hence the randomness/disorder of the system, so ΔS is positive.
For a reaction, ΔH = 40.63 kJ/mol and ΔS = 108.8 J/K/mol at 373 K. Is the reaction spontaneous at this temperature?
Answer: ΔG = ΔH − TΔS = 40.63 − (373)(0.1088) = 40.63 − 40.58 = 0.05 kJ/mol, which is approximately zero, indicating the system is nearly at equilibrium at this temperature (373 K corresponds closely to the boiling point of water for such vaporisation-type processes).
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Class 11 › Chemistry › Chemical Thermodynamics
Chapter notesRead the chapterPractice MCQsCheck what you rememberChapter test10 questions on one pageOnline testTest 1, 2 and 3, one question at a timePYQPrevious year questionsStudy materialPDF notes
Topics in this chapter: System, Surroundings, State Functions & First Law of ThermodynamicsEnthalpy Changes (ΔH), Heat Capacities & Hess's LawSpontaneity, Entropy (S) & Second Law of ThermodynamicsGibbs Free Energy (ΔG) & Relationship with Equilibrium Constant