NCERT Solutions Class 11 Chemistry Chapter 6 – Free PDF Download
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics can be found on this page. Here, students can access detailed and explanative solutions according to the latest term – II CBSE Syllabus 202122 for all the questions present in the NCERT textbook. This chapter contains various subdivisions like closed, open and isolated systems, isothermal and free expansion of gas, internal energy as a state function etc. NCERT Solutions for Class 11 Chemistry contain a brief description of all these important topics to enable a high conceptual knowledge among students.
Chemistry is a branch of Science which deals with molecules, atoms, states of matter, thermodynamics, elements etc. Students have to pay attention to each of these concepts and learn them completely. Many students do not identify the crucial concepts in each chapter. So, following the NCERT Solutions available at BYJU’S will help you attain remarkable grades in the term – II exam. Furthermore, the NCERT Solutions provided on this page can be downloaded as a PDF for free by clicking the download button provided above.
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NCERT Solutions for Class 11 Chemistry Chapter 6: Thermodynamics
“Thermodynamics” is the sixth chapter in the NCERT Class 11 Chemistry textbook. Thermodynamics is a branch of Science that deals with the relationship between heat and other forms of energy. A part of the universe where observations are made is called system. The surrounding of a system is the part of the universe that does not contain the system. Based on the exchange of energy and matter, thermodynamic systems can be classified into 3 types: closed system, open system and an isolated system.
In a closed system, only energy can be exchanged with the surrounding. In open system, energy as well as matter can be exchanged with the surrounding. In an isolated system, both energy and matter cannot be exchanged with the surrounding.
Subtopics included in NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

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
Access the answers of NCERT Solutions for Class 11 Chemistry Chapter 6
Q1: Choose the correct answer. A thermodynamic state function 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
Ans:
(ii) A quantity which is independent of path.
Reason:
Functions like pressure, volume and temperature depends on the state of the system only and not on the path.
Q2: For the process to occur under adiabatic conditions, the correct condition is:
(i) ∆T = 0 (ii) ∆p = 0
(iii) q = 0 (iv) w = 0
Ans:
(iii) q = 0
Reason:
For an adiabatic process heat transfer is zero, i.e. q = 0.
Q3: The enthalpies of all elements in their standard states are:
(i) Unity (ii) Zero
(iii) < 0 (iv) Different for every element
Ans:
(ii) Zero
Q4: ∆U^{0} of combustion of methane is – X kJ mol^{–1}. The value of ∆H^{0} is
(i) = [latex]\Delta U^{\Theta }[/latex]
(ii) > [latex]\Delta U^{\Theta }[/latex]
(iii) < [latex]\Delta U^{\Theta }[/latex]
(iv) 0
Ans:
(iii) < [latex]\Delta U^{\Theta }[/latex]
Reason:
[latex]\Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RT[/latex] ; [latex]\Delta U^{\Theta }[/latex] = – Y [latex]kJ mol^{1}[/latex], [latex]\Delta H^{\Theta } = ( – Y) + \Delta n_{g}RT \Rightarrow \Delta H^{\Theta } < \Delta U^{\Theta }[/latex]
Q5: 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[latex]mol^{1}[/latex]
(ii) 52.27 kJ[latex]mol^{1}[/latex]
(iii) +74.8 kJ[latex]mol^{1}[/latex]
(iv) +52 kJ[latex]mol^{1}[/latex]
Ans:
(i) 74.8kJ[latex]mol^{1}[/latex]
1. CH_{4(g) } + 2O_{2(g)} [latex]\rightarrow[/latex] CO_{2(g)} + 2H_{2}O_{(g)}
[latex]\Delta H = 890.3kJmol^{1}[/latex]2. C_{(s)} + O_{2(g) } [latex]\rightarrow[/latex] CO_{2(g)}
[latex]\Delta H = 393.5kJmol^{1}[/latex]3. 2H_{2(g)} + O_{2(g) }[latex]\rightarrow[/latex] 2H_{2}O_{(g)}
[latex]\Delta H = 285.8kJmol^{1}[/latex]C_{(s)} + 2H_{2(g) }[latex]\rightarrow[/latex] CH_{4(g)}
[latex]\Delta _{f}H_{CH_{4}}[/latex] = [latex]\Delta _{c}H_{c}[/latex] + [latex]2\Delta _{f}H_{H_{2}}[/latex] – [latex]\Delta _{f}H_{CO_{2}}[/latex]= [ 393.5 +2(285.8) – (890.3)] kJ[latex]mol^{1}[/latex]
= 74.8kJ[latex]mol^{1}[/latex]
Q6: 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
Ans:
(iv) possible at any temperature
[latex]\Delta G[/latex] should be –ve, for spontaneous reaction to occur [latex]\Delta G[/latex] = [latex]\Delta H[/latex] – T[latex]\Delta S[/latex]As per given in question,
[latex]\Delta H[/latex] is –ve ( as heat is evolved) [latex]\Delta S[/latex] is +veTherefore, [latex]\Delta G[/latex] is negative
So, the reaction will be possible at any temperature.
Q7: 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?
Ans:
As per Thermodynamics 1^{st} law,
[latex]\Delta U[/latex] = q + W(i); [latex]\Delta U[/latex] internal energy = heatW = work done
W = 594 J (work done by system)
q = +801 J (+ve as heat is absorbed)
Now,
[latex]\Delta U[/latex] = 801 + (594) [latex]\Delta U[/latex] = 207 J
Q8: The reaction of cyanamide, NH_{2}CN_{(s)}, with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol^{–1} at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH_{2}CN_{(g)} + 3/2 O_{2(g)} → N_{2(g)} + CO_{2(g)} + H_{2}O_{(l)}
Ans:
[latex]\Delta H[/latex] is given by, [latex]\Delta H = \Delta U + \Delta n_{g}RT[/latex]………………(1) [latex]\Delta n_{g}[/latex] = change in number of moles [latex]\Delta U[/latex] = change in internal energyHere,
[latex]\Delta n_{g} = \sum n_{g}(product) – \sum n_{g}(reactant)[/latex]_{ }= (2 – 1.5) moles
_{ }[latex]\Delta n_{g}[/latex] = 0.5 moles
Here,
T =298K
[latex]\Delta U[/latex] = 742.7 [latex]kJmol^{1}[/latex]R = [latex]8.314\times 10^{3}kJmol^{1}K^{1}[/latex]
Now, from (1)
[latex]\Delta H = (742.7 kJmol^{1}) + (0.5mol)(298K)( 8.314\times 10^{3}kJmol^{1}K^{1})[/latex]= 742.7 + 1.2
[latex]\Delta H[/latex] = 741.5 [latex]kJmol^{1}[/latex]
Q9: Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^{–1} K^{–1}.
Ans:
Expression of heat(q),
[latex]q = mCP\Delta T[/latex];………………….(a) [latex]\Delta T[/latex] = Change in temperaturec = molar heat capacity
m = mass of substance
From (a)
[latex]q = ( \frac{60}{27}mol )(24mol^{1}K^{1})(20K)[/latex]q = 1066.67 J = 1.067 KJ
Q10: Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C. ∆_{fus}H = 6.03 kJ mol^{–1} at 0°C.
[latex]C_{p}[H_{2}O_{(l)}] = 75.3 J\;mol^{1}K^{1}[/latex]
[latex]C_{p}[H_{2}O_{(s)}] = 36.8 J\;mol^{1}K^{1}[/latex]
Ans:
[latex]\Delta H_{total}[/latex] = sum of the changes given below:(a) Energy change that occurs during transformation of 1 mole of water from [latex]10^{\circ}C\;to\;0^{\circ}C[/latex].
(b) Energy change that occurs during transformation of 1 mole of water at [latex]0^{\circ}C[/latex] to 1 mole of ice at [latex]0^{\circ}C[/latex].
(c) Energy change that occurs during transformation of 1 mole of ice from [latex]0^{\circ}C\;to\;(10)^{\circ}C[/latex].
[latex]\Delta H_{total} = C_{p}[H_{2}OCl]\Delta T + \Delta H_{freezing} C_{p}[H_{2}O_{s}]\Delta T[/latex]= (75.3 [latex]J mol^{1}K^{1}[/latex])(0 – 10)K + (6.03*1000 [latex]J mol^{1}[/latex](100)K
= 753 [latex]J mol^{1}[/latex] – 6030[latex]J mol^{1}[/latex] – 368[latex]J mol^{1}[/latex]
= 7151 [latex]J mol^{1}[/latex]
= 7.151[latex]kJ mol^{1}[/latex]
Thus, the required change in enthalpy for given transformation is 7.151[latex]kJ mol^{1}[/latex].
Q11 Enthalpy of combustion of carbon to CO_{2} is –393.5 kJ mol^{–1}. Calculate the heat released upon formation of 35.2 g of CO_{2} from carbon and dioxygen gas.
Ans:
Formation of carbon dioxide from dioxygen and carbon gas is given as:
C_{(s)} + O_{2(g)} → CO_{2(g)}; [latex]\Delta _{f}H[/latex] = 393.5 [latex]kJ mol^{1}[/latex]
1 mole CO_{2} = 44g
Heat released during formation of 44g CO_{2} = 393.5 [latex]kJ mol^{1}[/latex]
Therefore, heat released during formation of 35.2g of CO_{2} can be calculated as
= [latex]\frac{393.5kJmol^{1}}{44g}\times 35.2g[/latex]
= 314.8 [latex]kJ mol^{1}[/latex]
Q12: Enthalpies of formation of CO _{(g)}, CO_{2 (g)}, N_{2}O _{(g)} and N_{2}O_{4}_{(g)} are –110, – 393, 81 and 9.7 kJ mol^{–1} respectively. Find the value of ∆_{r}H for the reaction:
N_{2}O_{4(g)} + 3CO_{(g)}→ _{ }N_{2}O_{(g)} + 3 CO_{2(g)}
Ans:
“[latex]\Delta _{r}H[/latex] for any reaction is defined as the fifference between [latex]\Delta _{f}H[/latex] value of products and [latex]\Delta _{f}H[/latex] value of reactants.”
[latex]\Delta _{r}H = \sum \Delta _{f}H (products) – \sum \Delta _{f}H (reactants)[/latex]Now, for
N_{2}O_{4(g)} + 3CO_{(g)} à _{ }N_{2}O_{(g)} + 3 CO_{2(g)}
[latex]\Delta _{r}H = [(\Delta _{f}H(N_{2}O) + (3\Delta _{f}H(CO_{2})) – (\Delta _{f}H(N_{2}O_{4}) + 3\Delta _{f}H(CO))][/latex]Now, substituting the given values in the above equation, we get:
[latex]\Delta _{r}H[/latex] = [{81[latex]kJ mol^{1}[/latex] + 3(393) [latex]kJ mol^{1}[/latex]} – {9.7[latex]kJ mol^{1}[/latex] + 3(110) [latex]kJ mol^{1}[/latex]}] [latex]\Delta _{r}H[/latex] = 777.7 [latex]kJ mol^{1}[/latex]
Q13: Given N_{2} (g) + 3H_{2} (g) → 2NH_{3} (g) ; ∆_{r}H^{0}= –92.4 kJ mol^{–1}
What is the standard enthalpy of formation of NH_{3} gas?
Ans:
“Standard enthalpy of formation of a compound is the enthalpy that takes place during the formation of 1 mole of a substance in its standard form, from its constituent elements in their standard state.”
Dividing the chemical equation given in the question by 2, we get
(0.5)N_{2(g)} + (1.5)H_{2(g)} → 2NH_{3(g)}
Therefore, Standard Enthalpy for formation of ammonia gas
= (0.5) [latex]\Delta _{r}H^{\Theta }[/latex]
= (0.5)(92.4[latex]kJ mol^{1}[/latex])
= 46.2[latex]kJ mol^{1}[/latex]
Q14: Calculate the standard enthalpy of formation of CH_{3}OH_{(l)} from the following data:
CH_{3}OH_{(l)} + 3/2 O_{2(g)}→ CO_{2(g)} + 2H_{2}O_{(l)}; [latex]\Delta _{r}H^{\Theta }[/latex] = 726 [latex]kJ mol^{1}[/latex]
C_{(g)} + O_{2(g)}→ CO_{2(g)}; [latex]\Delta _{c}H_{\Theta }[/latex] = 393 [latex]kJ mol^{1}[/latex]
H_{2(g)} + 1/2 O_{2(g)}→ H_{2}O_{(l)}; [latex]\Delta _{f}H^{\Theta }[/latex] = 286 [latex]kJ mol^{1}[/latex]
Ans:
C_{(s)} + 2H_{2}O_{(g)} + (1/2)O_{2(g)} → CH_{3}OH_{(l)} …………………………(i)
CH_{3}OH_{(l)} can be obtained as follows,
[latex]\Delta _{f}H_{\Theta }[/latex] [CH_{3}OH_{(l)}] = [latex]\Delta _{c}H_{\Theta }[/latex]2[latex]\Delta _{f}H_{\Theta }[/latex] – [latex]\Delta _{r}H_{\Theta }[/latex]
= (393 [latex]kJ mol^{1}[/latex]) +2(286[latex]kJ mol^{1}[/latex]) – (726[latex]kJ mol^{1}[/latex])
= (393 – 572 + 726) [latex]kJ mol^{1}[/latex]
= 239[latex]kJ mol^{1}[/latex]
Thus, [latex]\Delta _{f}H_{\Theta }[/latex] [CH_{3}OH_{(l)}] = 239[latex]kJ mol^{1}[/latex]
Q15: Calculate the enthalpy change for the process
CCl_{4(g)}→ C_{(g)} + 4Cl_{(g)} and determine the value of bond enthalpy for CCl in CCl_{4(g)}.
[latex]\Delta _{vap}H^{\Theta }[/latex] (CCl_{4}) = 30.5 [latex]kJ mol^{1}[/latex].
[latex]\Delta _{f}H^{\Theta }[/latex] (CCl_{4}) = 135.5 [latex]kJ mol^{1}[/latex].
[latex]\Delta _{a}H^{\Theta }[/latex] (C) = 715 [latex]kJ mol^{1}[/latex],
[latex]\Delta _{a}H^{\Theta }[/latex] is a enthalpy of atomisation
[latex]\Delta _{a}H^{\Theta }[/latex] (Cl_{2}) = 242 [latex]kJ mol^{1}[/latex].
Ans:
“ The chemical equations implying to the given values of enthalpies” are:
(1) CCl_{4(l)} à CCl_{4(g)} ; [latex]\Delta _{vap}H^{\Theta }[/latex] = 30.5 [latex]kJ mol^{1}[/latex]
(2) C_{(s)} à C_{(g)} [latex]\Delta _{a}H^{\Theta }[/latex] = 715 [latex]kJ mol^{1}[/latex]
(3) Cl_{2(g)} à 2Cl_{(g)} ; [latex]\Delta _{a}H^{\Theta }[/latex] = 242 [latex]kJ mol^{1}[/latex]
(4) C_{(g)} + 4Cl_{(g)} à CCl4(g); [latex]\Delta _{f}H^{\Theta }[/latex] = 135.5 [latex]kJ mol^{1}[/latex] [latex]\Delta H[/latex] for the process CCl_{4(g)} à C_{(g)} + 4Cl_{(g) }can be measured as:
[latex]\Delta H = \Delta _{a}H^{\Theta }(C) + 2\Delta _{a}H^{\Theta }(Cl_{2}) – \Delta _{vap}H^{\Theta } – \Delta _{f}H[/latex]= (715[latex]kJ mol^{1}[/latex]) + 2([latex]kJ mol^{1}[/latex]) – (30.5[latex]kJ mol^{1}[/latex]) – (135.5[latex]kJ mol^{1}[/latex])
Therefore, [latex]H = 1304kJmol^{1}[/latex]
The value of bond enthalpy for CCl in CCl_{4(g)}
= [latex]\frac{1304}{4}kJmol^{1}[/latex]
= 326 [latex]kJ mol^{1}[/latex]
Q16: For an isolated system, ∆U = 0, what will be ∆S?
Ans:
[latex]\Delta U[/latex] is positive ; [latex]\Delta U[/latex] > 0.As, [latex]\Delta U[/latex] = 0 then[latex]\Delta S[/latex] will be +ve, as a result reaction will be spontaneous.
Q17: For the reaction at 298K,
2A + B → C
[latex]\Delta H[/latex] = 400 [latex]kJ mol^{1}[/latex]
[latex]\Delta H[/latex] = 0.2 [latex]kJ mol^{1}K^{1}[/latex]
At what temperature will the reaction become spontaneous considering [latex]\Delta S[/latex] and [latex]\Delta H[/latex] to be constant over the temperature range?
Ans:
Now,
[latex]\Delta G = \Delta H – T\Delta S[/latex]Let, the given reaction is at equilibrium, then [latex]\Delta T[/latex] will be:
T = [latex](\Delta H – \Delta G)\frac{1}{\Delta S}[/latex] [latex]\frac{\Delta H}{\Delta S}[/latex]; ([latex]\Delta G[/latex] = 0 at equilibrium)
= 400[latex]kJ mol^{1}[/latex]/0.2[latex]kJ mol^{1}K^{1}[/latex]
Therefore, T = 2000K
Thus, for the spontaneous, [latex]\Delta G[/latex] must be –ve and T > 2000K.
Q18: For the reaction
2Cl_{(g)}→ Cl_{2(g)}
What are the signs of [latex]\Delta S[/latex] and [latex]\Delta H[/latex]?
Ans:
[latex]\Delta S[/latex] and [latex]\Delta H[/latex] are having negative sign.The reaction given in the question represents the formation of Cl molecule from Cl atoms. As the formation of bond takes place in the given reaction. So, energy is released. So, [latex]\Delta H[/latex] is negative.
Also, 2 moles of Chlorine atoms are having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus, [latex]\Delta S[/latex] is negative.
Q19: For the reaction
2A_{(g)} + B_{(g)}→ 2D_{(g)}
[latex]\Delta U^{\Theta }[/latex] = 10.5 kJ and [latex]\Delta S^{\Theta }[/latex] = 44.1[latex]JK^{1}[/latex]
Calculate [latex]\Delta G^{\Theta }[/latex] for the reaction, and predict whether the reaction may occur spontaneously.
Ans:
2A_{(g)} + B_{(g)} → 2D_{(g)}
[latex]\Delta n_{g}[/latex] = 2 – 3= 1 mole
Putting value of [latex]\Delta U^{\Theta }[/latex] in expression of [latex]\Delta H[/latex]:
[latex]\Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RT[/latex]= (10.5KJ) – (1)( [latex]8.314\times 10^{3}kJK^{1}mol^{1}[/latex])(298K)
= 10.5kJ 2.48kJ
[latex]\Delta H^{\Theta }[/latex] = 12.98kJPutting value of [latex]\Delta S^{\Theta }[/latex] and [latex]\Delta H^{\Theta }[/latex] in expression of [latex]\Delta G^{\Theta }[/latex]:
[latex]\Delta G^{\Theta } = \Delta H^{\Theta } – T\Delta S^{\Theta }[/latex]= 12.98kJ –(298K)(44.1[latex]JK^{1}[/latex])
= 12.98kJ +13.14kJ
[latex]\Delta G^{\Theta }[/latex] = 0.16kJAs, [latex]\Delta G^{\Theta }[/latex] is positive, the reaction won’t occur spontaneously.
Q20: The equilibrium constant for a reaction is 10. What will be the value of ∆G_{0}? R = 8.314 JK^{–1} mol^{–1}, T = 300 K.
Ans:
Now,
[latex]\Delta G^{\Theta }[/latex] = [latex]2.303RT\log eq[/latex]= (2.303)( [latex]8.314\times kJK^{1}mol^{1}[/latex])(300K) [latex]\log 10[/latex]
= 5744.14[latex]Jmol^{1}[/latex]
= 5.744[latex]kJmol^{1}[/latex]
Q21: Comment on the thermodynamic stability of NO(g), given,
(1/2)N_{2(g)} + (1/2)O_{2(g) }→ NO_{(g)}; [latex]\Delta _{r}H^{\Theta } = 90kJmol^{1}[/latex]
NO_{(g)} + (1/2)O_{2(g)} → NO_{2(g)}; [latex]\Delta _{r}H^{\Theta } = 74kJmol^{1}[/latex]
Ans:
The +ve value of [latex]\Delta _{r}H[/latex] represents that during NO_{(g)} formation from O_{2} and N_{2}, heat is absorbed. The obtained product, NO_{(g)} is having more energy than reactants. Thus, NO_{(g)} is unstable.
The ve value of [latex]\Delta _{r}H[/latex] represents that during NO_{2(g)} formation from O_{2(g)} and NO_{(g)}, heat is evolved. The obtained product, NO_{2(g)} gets stabilized with minimum energy.
Thus, unstable NO_{(g) }converts into stable NO_{2(g).}
Q22: Calculate the entropy change in surroundings when 1.00 mol of H_{2}O(l) is formed under standard conditions. ∆_{f} H^{0}= –286 kJ mol^{–1}.
Ans:
[latex]\Delta _{r}H^{\Theta } = 286kJmol^{1}[/latex] is given so that amount of heat is evolved during the formation of 1 mole of H_{2}O_{(l)}.Thus, the same heat will be absorbed by surrounding Q_{surr} = +286[latex]kJmol^{1}[/latex].
Now, [latex]\Delta S_{surr}[/latex] = Q_{surr}/7
= [latex]\frac{286kJmol^{1}}{298K}[/latex]
Therefore, [latex]\Delta S_{surr} = 959.73Jmol^{1}K^{1}[/latex]
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NCERT Exemplar for class 11 chemistry Chapter 6 
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