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NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics are given on this page. Students can check the Thermodynamics Class 11 NCERT solutions for free. Class 11 Thermodynamics NCERT solutions are designed by our subject experts. Those who are preparing for entrance exam as well as board exams are advised to learn the Thermodynamics NCERT solutions. Class 11 Chemistry Thermodynamics NCERT solutions are given in step by step method so that students can understand easily.
Thermodynamics means study of heat and its properties. Thermodynamics consists of two words, “Thermo” meaning heat and “Dynamics” meaning flow or motion. The various laws of Thermodynamics demonstrate the energy changes in a system under equilibrium state and when there’s a heat flow from one equilibrium state to another. In this chapter, students study various properties of heat under motion in various conditions and laws & principles governing them. Students will also learn about heat and work relation, Enthalpy and Gibb’s law of energy change and equilibrium. NCERT Class 11 Chemistry Thermodynamics is a very important chapter as it lays the foundation of a very complex subject which students learn in engineering at graduate and postgraduate level. This chapter is also very important as many questions are asked in exam like JEE Main from this particular topic. Thermodynamics class 11 Chemistry NCERT solutions are available in PDF format.
Students can check here NCERT class 11 Chemistry solution Thermodynamics which will help them in understanding this chapter better and scoring good marks in the exam. Students preparing for entrance exams can refer NCERT Solutions Chemistry class 12.
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics – Important Topics
Candidates can check here the list of important topics that are covered in the Thermodynamics chapter of NCERT class 11 Chemistry subject.
- Thermodynamic Terms
- The System and the Surroundings
- Types of Thermodynamic Systems
- State of the System
- Internal Energy as a State Function
- Applications
- Work
- Enthalpy, H
- Measurement of ΔU and ΔH: Calorimetry
- Enthalpy Change and Reaction Enthalpy
- Enthalpies for Different Types of Reactions
- Spontaneity
- Gibbs Energy Change and Equilibrium
Thermodynamics Additional Question
Assertion and Reason:
Directions:
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.
6.1. Assertion: Variables like pressure (p), volume (V), temperature (T) etc. are called state variables or state functions.
Reason: Their values depend only on the state of the system and not on how it is reached
Answer: (a) The state of a thermodynamic system is described by its measurable or macroscopic (bulk) properties. We can describe the state of a gas by quoting its pressure (p), volume (V), temperature (T), amount (n) etc. Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached. Example: Volume of water in a pond, is a state function, because change in volume of its water is independent of the route by which water is filled in the pond, either by rain or by tube well or by both.
6.2. Assertion: In an exothermic reaction, heat is evolved, and system gains heat from the surroundings.
Reason: qp and ΔrH will also be negative.
Answer: (d) In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, qp will be negative and ΔrH will also be negative. Similarly in an endothermic reaction, heat is absorbed, qp is positive and ΔrH will be positive.
6.3. Assertion: Entropy decreases when a liquid crystallizes into solid.
Reason: Molecules tend to be in a less ordered state.
Answer:(c) When a liquid crystallizes into solid or after freezing, the molecules attain an ordered state and therefore, entropy decreases.
6.4. Assertion: When temperature of a crystalline solid is increased, the degree of randomness of its constituents increases.
Reason:The entropy of any pure crystalline substance approaches zero as the temperature approaches absolute zero.
Answer: (b) A system at higher temperature has greater randomness in it than one at lower temperature. Thus, temperature is the measure of average chaotic motion of particles in the system.
6.5. Assertion: It requires less heat to vaporise 1 mol of acetone than it does to vaporize 1 mol of water.
Reason: For an organic liquid, such as acetone, the intermolecular dipole-dipole interactions are significantly weaker.
Answer:(a) The magnitude of the enthalpy change depends on the strength of the intermolecular interactions in the substance undergoing the phase transformations. For example, the strong hydrogen bonds between water molecules hold them tightly in liquid phase. For an organic liquid, such as acetone, the intermolecular dipole-dipole interactions are significantly weaker. Thus, it requires less heat to vaporise 1 mol of acetone than it does to vaporize 1 mol of water.
MCQs:
6.1. Reaction spontaneous at all temperatures when:
(a) ΔrHϴ is negative, ΔrSϴ positive, ΔrGϴ is negative.
(a) ΔrHϴ is positive, ΔrSϴ positive, ΔrGϴ is negative.
(a) ΔrHϴ is negative, ΔrSϴ positive, ΔrGϴ is positive.
(a) ΔrHϴ is positive, ΔrSϴ positive, ΔrGϴ is positive.
Answer: (a) ΔrHϴ is negative, ΔrSϴ positive, ΔrGϴ is negative.
6.2. Which of the following statements is incorrect?
(a) Entropy is a measure of disorder or randomness.
(b) First law of thermodynamics does not guide us about the direction of chemical reactions
(c) Variables like p, V, T are called state variables or state functions.
(d) The state of the surroundings can always be completely specified.
Answer: (d) is an incorrect statement.
6.3. Thermodynamics is applicable to
(a) macroscopic system only (b) microsopic system only
(c) homogeneous system only (d) heterogeneous system only
Answer: (a) macroscopic system only
6.4. An isochoric and isobaric process takes place at constant
(a) temperature and pressure respectively
(b) pressure and concentration respectively
(c) volume and pressure respectively
(d) concentration and temperature respectively
Answer: (c) volume and pressure respectively
6.5. Which of the following always has a negative value?
(a) heat of reaction (b) heat of solution
(c) heat of combustion (d) heat of formation
Answer: (c) heat of combustion
6.6. Which one is the correct unit for entropy?
(a) KJ mol (b)JK mol
(c)JK mol-1 (d) KJ mol-1
Answer: (c) JK mol-1
Questions and Answers:
6.1. When is bond energy equal to bond dissociation energy?
Answer: For diatomic molecules e.g. H2, O2, Cl2 etc. both energies are equal.
6.2. What is the enthalpy of formation of the most stable form of an element in its standard state?
Answer: It is zero.
6.3. Out of diamond and graphite, which has greater entropy? At what temperature entropy of a substance is zero?
Answer: Graphite has greater entropy since it is loosely packed. At absolute zero the entropy of a substance is zero.
6.4. State Hess’s law.
Answer: Hess law states that “If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.”
Or it can be stated as “The change of enthalpy of a reaction remains same whether the reaction is carried out in one step or several steps.”
∆H = ∆H1 + ∆H2 + ∆H3……………
6.5 Predict the sign of ∆S for the following reaction heat
CaCO3 (s) →CaO(s) + CO2(g)
Answer: Since, there is an increase in the number of moles on the product side so, entropy increases and ∆S is positive.
6.6. What are extensive and intensive properties?
Answer: In thermodynamics, a distinction is made between extensive properties and intensive properties. An extensive property is a property whose value depends on the
quantity or size of matter present in the system. For example, mass, volume, internal energy, enthalpy, heat capacity, etc. are extensive properties.
Those properties which do not depend on the quantity or size of matter present are known as intensive properties. For example, temperature, density, pressure etc. are intensive properties.
6.7. Define the following:
(i) First law of thermodynamics.
(ii) Standard molar enthalpy of formation.
Answer: (i) First law of thermodynamics: It states that energy can neither be created nor be destroyed. The energy of an isolated system is constant.
∆u = q + w
(ii) The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference state is called Standard Molar Enthalpy of Formation. Its symbol is Δf Hϴ, where the subscript ‘ f ’ indicates that one mole of the compound in question has been formed in its standard state from its elements in their most stable states of aggregation.
6.8. Define standard enthalpy of Vapourization.
Answer: Amount of heat required to vapourize one mole of a liquid at constant temperature and under standard pressure (1bar) is called its standard enthalpy of vapourization or molar enthalpy of vapourization, ΔvapHϴ.
6.9. What is Gibbs Helmholtz equation?
Answer: ∆G = ∆H – T∆S
Where ∆G = free energy change.
∆H = enthalpy change.
∆S = entropy change.
6.10. What is a spontaneous change? Give one example.
Answer: A process which can take place of its own or initiate under some condition.
For example: Common salt dissolves in water of its own.
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics and FAQs
Q 6.1. Choose the correct answer: A thermodynamic state junction is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure volume work (iv) whose value depends on temperature only.
A 6.1 (ii) whose value is independent of path
Q 6.2. For the process to occur under adiabatic conditions, the correct condition is: (i) ∆T= 0 (ii) ∆p = 0 (iii) q = 0 (iv) w = 0
A 6.2 (iii) q = 0
Q 6.3. The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) < 0 (iv) different for each element
A 6.3 (ii) zero
Q 6.4. ΔU⸰ of combustion of methane is – X kJ mol^–1. The value of ΔH⸰ is (i) = ΔU⸰ (ii) > ΔU⸰ (iii) < ΔU⸰ (iv) = 0
- A 6.4
- The chemical equation for the combustion reaction is:
- CH4 (g) + 2O2 (g) →CO2 (g) + 2H2O (l)
- Δng= 1 – 3 = -2
- ΔH⸰= ΔU⸰ + ΔngRT = ΔU⸰- 2RT
- Therefore, ΔH⸰ <ΔU⸰
- i.e. option (iii) is correct.
Q 6.5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are -890.3 KJ mol^-1, – 393.5 KJ mol^-1 and – 285.8 KJ mol^-1 respectively. Enthalpy of formation of CH_4(g) will be (i) – 74.8 KJ mol^-1 (ii) – 52.27 KJ mol^-1 (iii) + 74.8 KJ mol^-1 (iv) + 52.26 KJ mol^-1
A 6.5 As per question:
- (i) CH4+2O2⟶CO2+2H2O , ΔH1=−890kJmol−1
- (ii) C+O2⟶CO2 , ΔH2=−393.5kJmol−1
- (iii) 2H2+O2⟶2H2O , ΔH3=2× (−285.8)kJmol−1
- Required reaction is
- C+2H2⟶CH4(g); ΔHf =?
From equations (ii)-(i)+(iii) - ΔHf = (−393.5) + 890.3 + 2(−285.8)
- = −74kJmol−1
- Hence, option (i) is correct.
Q 6.6. A reaction, A + B—>C + D + q is found to have a positive entropy change. The reaction will be (i) possible at high temperature (ii) possible only at low temperature (iii) not possible at any temperature (iv) possible at any temperature
A 6.6 (iv) possible at any temperature
Q 6.7 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
- A 6.7 As per the first law of thermodynamics,
- ∆U= q + W
- Heat absorbed by the system, q = 701 J
- Work done by the system, W = – 394 J
- Change in internal energy, ∆U = q + w = 701 – 394 = 307 J
Q 6.8. The reaction of cyanamide,NH2CN(s) with dioxygen was carried out in a bomb calorimeter and ∆U was found to be -742,7 KJ^-1 mol^-1 at 298 K. Calculate the enthalpy change for the reaction at 298 K. NH2CN (S) + 3/2O2(g) → N2(g) + CO2(g) + H2O(l)
A 6.8 Given, ∆U = – 742.7 KJ-1 mol-1 ∆ng = 2 – 3/2 = + 1/2 mol.
R = 8.314 x 10-3KJ-1 mol-1 T = 298 K According to the relation, ∆H = ∆U+ ∆ng RT
∆H = (- 742.7 KJ) + (1/2 mol) x (8.314 x10-3 KJ-1 mol-1 ) x (298 K)
= – 742.7 KJ + 1.239 KJ = – 741.5 KJ.
Q 6.9. Calculate the number of KJ of heat necessary to raise the temperature of 60 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.
A 6.9 No. of moles of Al, n = (60g)/ (27 g mol-1) = 2.22 mol
Molar heat capacity (C) = 24 J mol-1 K-1.
Rise in temperature (∆T) = 55 – 35 = 20°C = 20 K
Heat evolved (q) = C x m x T = (24 J mol-1 K-1) x (2.22 mol) x (20 K)
= 1065.6 J = 1.067 KJ.
Q 6.10 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. A,
ΔfusH = 6.03 KJ mol-1 at 0°C.
Cp [H2O(l)] = 75.3 J mol-1 K-1;
Cp [H2O(s)] = 36.8 J mol-1 K-1.
A 6.10 Total enthalpy change involved in the transformation is the sum of the following changes: (i) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C ΔH= Cp [H2O (l)] ΔT (ii) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of ice at 0°C ΔHfreezing (iii) Energy change involved in the transformation of 1 mol of ice at 10°C to 1 mol of ice at 0°C ΔH= Cp [H2O (s)] ΔT ΔH= Cp [H2O (l)] ΔT + ΔHfreezing + Cp [H2O (s)] ΔT = 75.3 J mol-1 K-1 (0-10 K) + 6.03 x 10-3 J mol-1 + 36.8 J mol-1 K-1 (-10-0 K) = -7533 mol-1 -6030 mol-1 – 368 J = -7151 J mol-1 = -7.151 kJ mol-1
Q 6.11. Enthalpy of combustion of carbon to CO2 is – 393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.
A 6.11 The combustion equation is:
C(s) + O2 (g) → CO2 (g); ∆H = – 393.5 KJ mol-1
Heat released in the formation of 44g of CO2 = 393.5 kJ
Heat released in the formation of 35.2 g of CO2 = (393.5 KJ) x (35.2g)/ (44g) = 314.8 kJ
Q 6.12. Find the question below:
Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ΔrH for the reaction:
N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)
A 6.12 Given, ∆fH of CO (g) = – 110 kJ mol-1;
∆fH of CO2(g) = – 393 kJ mol-1
∆fH of N2O(g) = 81 kJ mot-1;
∆fH of N2O4(g) = 9.7 kJ mol-1
Enthalpy of reaction (∆rH) = [81 + 3 (- 393)] – [9.7 + 3 (- 110)]
= [81 – 1179] – [9.7 – 330]
= – 778 kJ mol-1
Q 6.13. Find the question below:
Given
N2(g) + 3H2(g) → 2NH3(g); ∆r H– = -92.4 kJ mol-1
What is the standard enthalpy of formation of NH3 gas?
A 6.13 N2 (g) + 3H2 (g) → 2NH3 (g); ∆rH= -92.4 kJ/mol Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of of a substance in its standard form from its constituent elements in their standard state. Re-writing the given equation for 1 mol of NH3 1/2N2 (g) + 3/2H2 (g) → NH3 (g); Standard enthalpy of formation of NH3 = ½ ∆rH ∆fH (g) = – (92.4)/2 = – 46.2 kJ mol-1
Q 6.14. Calculate the standard enthalpy of formation of CH3OH from the following data:
(i) CH3OH(l) + 3/2 O2 (g) → CO2 (g) + 2H2O (l); ∆rH⸰= – 726kJ mol-1
(ii) C(graphite) + O2(g) → CO2 (g); ∆cH⸰ = – 393 kJ mol-1
(iii) H2(g) + 1/2O2(g) → H2O (l); ∆fH⸰ = -286 kJ mol-1
A 6.14 The reaction for the formation of CH3OH can be obtained by: C(s) + 2H2(g) + l/2O2(g) → CH3OH (l) This can be obtained by the algebraic calculations of the reactions: Equation (ii) + 2 x equation (iii) – equation (i) ∆Hf [CH3OH] = ∆cH⸰ + 2 x ∆fH⸰ - ∆rH⸰ = (– 393 kJ mol-1) + 2 (- 286 kJ mol-1) – (– 726 kJ mol-1) = (– 393 – 572 + 726) kJ mol-1 = 239 kJ mol-1
Q 6.15. Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g).
ΔvapHϴ(CCl4) = 30.5 kJ mol–1.
ΔfHϴ (CCl4) = –135.5 kJ mol–1.
ΔaHϴ (C) = 715.0 kJ mol–1, where ΔaHϴ is enthalpy of atomisation
ΔaHϴ (Cl2) = 242 kJ mol–1
A 6.15 According to the question:
- 1. CCl4 (l) → CCl4 (g), ∆vapH⸰ = 30.5 kJ mol–1
- 2. C(s) + 2Cl2(g) → CCl4 (l), ∆fH⸰ = –135.5 kJ mol–1
- 3. C(s) → C (g), ∆aH⸰ = 715.0 kJ mol–1
- 4. Cl2 (g) → C(g) + 4 Cl(g), ∆aH⸰ = 242 kJ mol–1
- The equation we want is: CCl4 (g) → C (g) + 4Cl (g), ∆H =?
- Equation (iii) + {2 x Equation (iv)} – Equation (ii), gives the required equation with ∆H = 715.0 + 2 x (242) – 30.5 – (-135.5) kJ mol-1
- = 1340 kJ mol-1
- Bond enthalpy of C-Cl in CCl4 = 1304 / 4 = 326 kJ mol-1
Q 6.16. For an isolated system ∆U = 0; what will be ∆S?
A 6.16 Change in internal energy (∆U) for an isolated system is zero for it does not exchange any energy with the surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore, ∆S > 0 or positive.
Q 6.17. For the reaction at 298 K, 2A + B → C ΔH = 400 kJ mol^–1 and ΔS = 0.2 kJ K^–1 mol^–1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
A 6.17 As per the Gibbs Helmholtz’s equation: ΔG = Δ H - TΔ S
- For ΔG=0; ΔH=TΔS Or
- T=ΔH/ΔS
- T = (400 KJ mol-1)/ (0.2 KJ K-1 mol-1)
- = 2000 k
- Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.
Q 6.18. For the reaction; 2Cl (g) → Cl2 (g); what will be the signs of ∆H and ∆S?
A 6.18
∆H: negative (-ve) because energy is released in bond formation
∆S: negative (-ve) because entropy decreases when atoms combine to form molecules.
Q 6.19. For the reaction 2 A(g) + B(g) → 2D(g) ΔU° = – 10.5 kJ and ΔS⸰ = – 44.1 JK^–1. Calculate ΔG⸰ for the reaction, and predict whether the reaction may occur spontaneously.
- A 6.19 ΔH° = ΔU° + Δng RT
- ΔU° = -10.5 kJ, Δng = 2-3 = -1 mol, R = 8.314 x 10-3 kJ mol-1, T = 298 K
- ΔH° = (- 10.5 kJ) + [(- 1 mol) x (8.314 x 10-3 kJ mol-1) x (298 K)]
- = -10.5 kJ – 2.478 kJ
- = -12.978 kJ
- According to Gibbs Helmholtz equation:
- ΔG° = ΔH° - TΔS°
- = (- 12.978 kJ) – (298 K) x (- 0.0441 kJ K-1)
- = -12.978 + 13.142
- = 0.164 kJ
- Since the value of ΔG° is positive, the reaction is a non-spontaneous reaction
6.20. The equilibrium constant for a reaction is 10. What will be the value of ΔG°? R = 8.314 JK^–1 mol^–1, T = 300 K.
- A 6.20 ΔG° = -RT ln K = - 2.303 RT log K
- Putting the values of
- R = 8.314 J K-1 mol-1,
- T = 300 K and
- K =10; we get
- ΔG° = - 2.303 x 8.314 J K-1 mol-1 x 300 K and K x log 10
- = - 5527 J mol-1
- = -5.527 kJ mol-1
Q 6.21. Comment on the thermodynamic stability of NO (g), given
½ N2 (g) + ½ O2 (g) → NO (g); ΔrHϴ = 90 kJ mol–1
NO (g) + ½ O2 (g) → NO2 (g); ΔrHϴ = –74 kJ mol–1
A 6.21 For NO (g), ΔrH° is a +ve value. So, it is unstable in nature. For NO2 (g), ΔrH° is a -ve value. So, it is stable in nature
Q 6.22. Calculate the entropy change in surroundings when 1.00 mol of H2O (l) is formed under standard conditions. ΔfH⸰ = – 286 kJ mol^–1.
- A 6.22 qrev = (-ΔfH⸰) = - (- 286kJ mol-1) = 286 x 103 Jmol-1 = 286000 J mol-1
- ΔS (surroundings) = qrev/ T = 286000 J mol-1 / 298 K
- = 959.73 J mol-1 K-1
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