NCERT Solutions for Class 11 Chemistry: Chapter 6 (Thermodynamics)

NCERT Solutions for Class 11 Chemistry: Chapter 6 (Thermodynamics) can be found on this page. Here, students can access detailed, explanative solutions to all the intext and exercise questions listed in chapter 6 of the NCERT class 11 chemistry textbook. 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 Chemistry – Class 11, 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 which 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 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 Class 11 Chemistry Chapter 6 Chemical Thermodynamics

  1. Thermodynamic Terms
    • The System and the Surroundings
    • Types of Thermodynamic Systems
    • State of the System
    • Internal Energy as a State Function
  2. Applications
    • Work
    • Enthalpy, H
  3. Measurement Of ΔU And ΔH: Calorimetry
  4. Enthalpy Change and Reaction Enthalpy
  5. Enthalpies For Different Types Of Reactions
  6. Spontaneity
  7. Gibbs Energy Change And Equilibrium

NCERT Solutions for Class 11 Chemistry Chapter 6

Q-1: 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


(2) A quantity which is independent of path.


Functions like pressure, volume and temperature depends on the state of the system only and not on the path.

Q-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


3: q = 0


For an adiabatic process heat transfer is zero, i.e. q = 0.

Q-3: The enthalpies of all elements in their standard states are:

(1) Zero

(2) < 0

(3) Different for every element

(4) Unity


(1) Zero

Q-4: ∆U
of combustion of methane is – X kJ mol–1. The value of ∆H0 is

(a) > ΔUΘ\Delta U^{\Theta }

(b) = ΔUΘ\Delta U^{\Theta }

(c) = 0

(d) < ΔUΘ\Delta U^{\Theta }


(d) < ΔUΘ\Delta U^{\Theta }


ΔHΘ=ΔUΘ+ΔngRT\Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RT ; ΔUΘ\Delta U^{\Theta } = – Y kJmol1kJ mol^{-1},

ΔHΘ=(Y)+ΔngRTΔHΘ<ΔUΘ\Delta H^{\Theta } = ( – Y) + \Delta n_{g}RT \Rightarrow \Delta H^{\Theta } < \Delta U^{\Theta }

Q-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 CH4 (g) will be

(a) -52.27kJmol1mol^{-1}

(b) 52kJmol1mol^{-1}

(c) +74.8kJmol1mol^{-1}

(d) -74.8kJmol1mol^{-1}


(d) -74.8kJmol1mol^{-1}

à CH4(g)  + 2O2(g) \rightarrow CO2(g) + 2H2O(g)

ΔH=890.3kJmol1\Delta H = -890.3kJmol^{-1}

à C(s) + O2(g)  \rightarrow CO2(g)

ΔH=393.5kJmol1\Delta H = -393.5kJmol^{-1}

à 2H2(g) + O2(g) \rightarrow 2H2O(g)

ΔH=285.8kJmol1\Delta H = -285.8kJmol^{-1}

à C(s) + 2H2(g) \rightarrow CH4(g)

ΔfHCH4\Delta _{f}H_{CH_{4}}  = ΔcHc\Delta _{c}H_{c} + 2ΔfHH22\Delta _{f}H_{H_{2}}ΔfHCO2\Delta _{f}H_{CO_{2}}

= [ -393.5 +2(-285.8) – (-890.3)] kJmol1mol^{-1}

= -74.8kJmol1mol^{-1}

Q-6: A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be

(a) will be possible at low temperature only

(b) will be possible at high temperature only

(c) will be possible at any temperature

(d) won’t be possible at any temperature


(c) will be possible at any temperature

ΔG\Delta G should be –ve, for spontaneous reaction to occur

ΔG\Delta G = ΔH\Delta H – TΔS\Delta S

As per given in question,

ΔH\Delta H is –ve ( as heat is evolved)

ΔS\Delta S is +ve

Therefore, ΔG\Delta G is negative

So, the reaction will be possible at any temperature.

Q-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?


As per Thermodynamics 1st law,

ΔU\Delta U = q + W(i);

ΔU\Delta U internal energy = heat

W = work done

W = -594 J (work done by system)

q = +801 J (+ve as heat is absorbed)


ΔU\Delta U = 801 + (-594)

ΔU\Delta U = 207 J

Q-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 mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.

NH2CN(g) +3/2O2(g)à N2(g) + CO2(g) + H2O(l)


ΔH\Delta H is given by,

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_{g}RT………………(1)

Δng\Delta n_{g} = change in number of moles

ΔU\Delta U = change in internal energy


Δng=ng(product)ng(reactant)\Delta n_{g} = \sum n_{g}(product) – \sum n_{g}(reactant)

  = (2 – 2.5) moles

 Δng\Delta n_{g} = -0.5 moles


T =298K

ΔU\Delta U = -753.7 kJmol1kJmol^{-1}

R  = 8.314×103kJmol1K18.314\times 10^{-3}kJmol^{-1}K^{-1}

Now, from (1)

ΔH=(753.7kJmol1)+(0.5mol)(298K)(8.314×103kJmol1K1)\Delta H = (-753.7 kJmol^{-1}) + (-0.5mol)(298K)( 8.314\times 10^{-3}kJmol^{-1}K^{-1})

= -753.7 – 1.2

ΔH\Delta H = -754.9 kJmol1kJmol^{-1}

Q-9: 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.


Expression of heat(q),

q=mCPΔTq = mCP\Delta T;………………….(a)

ΔT\Delta T = Change in temperature

c = molar heat capacity

m = mass of substance

From (a)

q=(5027mol)(24mol1K1)(20K)q = ( \frac{50}{27}mol )(24mol^{-1}K^{-1})(20K)

q = 888.88 J q

Q-10: Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. ∆fusH = 6.03 kJ mol–1 at 0°C.

Cp[H2O(l)]=75.3J  mol1K1C_{p}[H_{2}O_{(l)}] = 75.3 J\;mol^{-1}K^{-1}

Cp[H2O(s)]=36.8J  mol1K1C_{p}[H_{2}O_{(s)}] = 36.8 J\;mol^{-1}K^{-1}


ΔHtotal\Delta H_{total} = sum of the changes given below:

(a) Energy change that occurs during transformation of 1 mole of water from 10C  to  0C10^{\circ}C\;to\;0^{\circ}C.

(b) Energy change that occurs during transformation of 1 mole of water at 0C0^{\circ}C  to  1 mole of ice at 0C0^{\circ}C.

(c) Energy change that occurs during transformation of 1 mole of ice from 0C  to  (10)C0^{\circ}C\;to\;(-10)^{\circ}C.

ΔHtotal=Cp[H2OCl]ΔT+ΔHfreezingCp[H2Os]ΔT\Delta H_{total} = C_{p}[H_{2}OCl]\Delta T + \Delta H_{freezing} C_{p}[H_{2}O_{s}]\Delta T

= (75.3 Jmol1K1J mol^{-1}K^{-1})(0 – 10)K + (-6.03*1000 Jmol1J mol^{-1}(-10-0)K

= -753 Jmol1J mol^{-1} – 6030Jmol1J mol^{-1} – 368Jmol1J mol^{-1}

= -7151 Jmol1J mol^{-1}

= -7.151kJmol1kJ mol^{-1}

Thus, the required change in enthalpy for given transformation is -7.151kJmol1kJ mol^{-1}.

Q-11: 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 ∆rH for the reaction:

N2O4(g) + 3CO(g)à  N2O(g) + 3 CO2(g)


ΔrH\Delta _{r}H for any reaction is defined as the fifference between ΔfH\Delta _{f}H value of products and ΔfH\Delta _{f}H value of reactants.”

ΔrH=ΔfH(products)ΔfH(reactants)\Delta _{r}H = \sum \Delta _{f}H (products) – \sum \Delta _{f}H (reactants)

Now, for

N2O4(g) + 3CO(g) à  N2O(g) + 3 CO2(g)

ΔrH=[(ΔfH(N2O)+(3ΔfH(CO2))(ΔfH(N2O4)+3ΔfH(CO))]\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))]

Now, substituting the given values in the above equation, we get:

ΔrH\Delta _{r}H = [{81kJmol1kJ mol^{-1} + 3(-393) kJmol1kJ mol^{-1}} – {9.7kJmol1kJ mol^{-1} + 3(-110) kJmol1kJ mol^{-1}}] ΔrH\Delta _{r}H = -777.7 kJmol1kJ mol^{-1}

Q-12 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.


Formation of carbon dioxide from di-oxygen and carbon gas is given as:

C(s) + O2(g) à CO2(g); ΔfH\Delta _{f}H = -393.5 kJmol1kJ mol^{-1}

1 mole CO2 = 44g

Heat released during formation of 44g CO2 = -393.5 kJmol1kJ mol^{-1}

Therefore, heat released during formation of 37.2g of CO2  can be calculated as

= 393.5kJmol144g×37.2g\frac{-393.5kJmol^{-1}}{44g}\times 37.2g

= -332.69 kJmol1kJ mol^{-1}

Q-13: Given N2 (g) + 3H2 (g) → 2NH3 (g) ; ∆rH0= –92.4 kJ mol–1
What is the standard enthalpy of formation of NH3 gas?


“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)N2(g) + (1.5)H2(g) à 2NH3(g)

Therefore, Standard Enthalpy for formation of ammonia gas

= (0.5) ΔrHΘ\Delta _{r}H^{\Theta }

= (0.5)(-92.4kJmol1kJ mol^{-1})

= -46.2kJmol1kJ mol^{-1}

Q-14: Calculate the standard enthalpy of formation of CH3OH(l) from the following data:

CH3OH(l) + (3/2)O2(g)à CO2(g) + 2H2O(l); ΔrHΘ\Delta _{r}H^{\Theta } = -726 kJmol1kJ mol^{-1}

C(g) + O2(g)à CO2(g); ΔcHΘ\Delta _{c}H_{\Theta } = -393 kJmol1kJ mol^{-1}

H2(g) + (1/2)O2(g)à H2O(l); ΔfHΘ\Delta _{f}H^{\Theta } = -286 kJmol1kJ mol^{-1}


C(s) + 2H2O(g) + (1/2)O2(g) à CH3OH(l) …………………………(i)

CH3OH(l) can be obtained as follows,

ΔfHΘ\Delta _{f}H_{\Theta } [CH3OH(l)] = ΔcHΘ\Delta _{c}H_{\Theta }

2ΔfHΘ\Delta _{f}H_{\Theta }ΔrHΘ\Delta _{r}H_{\Theta }

= (-393 kJmol1kJ mol^{-1}) +2(-286kJmol1kJ mol^{-1}) – (-726kJmol1kJ mol^{-1})

= (-393 – 572 + 726) kJmol1kJ mol^{-1}

= -239kJmol1kJ mol^{-1}

Thus, ΔfHΘ\Delta _{f}H_{\Theta } [CH3OH(l)] = -239kJmol1kJ mol^{-1}

Q-15: Calculate the enthalpy change for the process

CCl4(g)à C(g) + 4Cl(g) and determine the value of bond enthalpy for C-Cl in CCl4(g).

ΔvapHΘ\Delta _{vap}H^{\Theta } (CCl4) = 30.5 kJmol1kJ mol^{-1}.

ΔfHΘ\Delta _{f}H^{\Theta } (CCl4) = -135.5 kJmol1kJ mol^{-1}.

ΔaHΘ\Delta _{a}H^{\Theta } (C) = 715 kJmol1kJ mol^{-1},

ΔaHΘ\Delta _{a}H^{\Theta }  is a enthalpy of atomisation

ΔaHΘ\Delta _{a}H^{\Theta } (Cl2) = 242 kJmol1kJ mol^{-1}.



“ The chemical equations implying to the given values of enthalpies” are:

(1) CCl4(l) à CCl4(g) ; ΔvapHΘ\Delta _{vap}H^{\Theta } = 30.5 kJmol1kJ mol^{-1}

(2) C(s) à C(g) ΔaHΘ\Delta _{a}H^{\Theta } = 715 kJmol1kJ mol^{-1}

(3) Cl2(g) à 2Cl(g) ; ΔaHΘ\Delta _{a}H^{\Theta } = 242 kJmol1kJ mol^{-1}

(4) C(g) + 4Cl(g) à CCl4(g); ΔfHΘ\Delta _{f}H^{\Theta } = -135.5 kJmol1kJ mol^{-1} ΔH\Delta H for the process CCl4(g) à C(g) + 4Cl(g) can be measured as:

ΔH=ΔaHΘ(C)+2ΔaHΘ(Cl2)ΔvapHΘΔfH\Delta H = \Delta _{a}H^{\Theta }(C) + 2\Delta _{a}H^{\Theta }(Cl_{2}) – \Delta _{vap}H^{\Theta } – \Delta _{f}H

= (715kJmol1kJ mol^{-1}) + 2(kJmol1kJ mol^{-1}) – (30.5kJmol1kJ mol^{-1}) – (-135.5kJmol1kJ mol^{-1})

Therefore, H=1304kJmol1H = 1304kJmol^{-1}

The value of bond enthalpy for C-Cl in CCl4(g)

= 13044kJmol1\frac{1304}{4}kJmol^{-1}

= 326 kJmol1kJ mol^{-1}

Q-16: For an isolated system, ∆U = 0, what will be ∆S ?


ΔU\Delta U is positive ; ΔU\Delta U > 0.

As, ΔU\Delta U = 0 thenΔS\Delta S will be +ve, as a result reaction will be spontaneous.


Following reaction takes place at 298K,

2X + Y à Z

ΔH\Delta H = 400 kJmol1kJ mol^{-1}

ΔH\Delta H = 0.2kJmol1K1kJ mol^{-1}K^{-1}

Find the temperature at which the reaction become spontaneous considering ΔS\Delta S and ΔH\Delta H to be constant over the entire temperature range?



ΔG=ΔHTΔS\Delta G = \Delta H – T\Delta S

Let, the given reaction is at equilibrium, then ΔT\Delta T will be:

T = (ΔHΔG)1ΔS(\Delta H – \Delta G)\frac{1}{\Delta S} ΔHΔS\frac{\Delta H}{\Delta S}; (ΔG\Delta G = 0 at equilibrium)

= 400kJmol1kJ mol^{-1}/0.2kJmol1K1kJ mol^{-1}K^{-1}

Therefore, T = 2000K

Thus, for the spontaneous, ΔG\Delta G must be –ve and T > 2000K.

Q-18: 2Cl(g)à Cl2(g)

In above reaction what can be the sign for ΔS\Delta S and ΔH\Delta H?


ΔS\Delta S and ΔH\Delta H 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,  ΔH\Delta H is negative.

Also, 2 moles of Chlorine atoms is having more randomness than 1 mole of chlorine molecule. So, the spontaneity is decreased. Thus, ΔS\Delta S is negative.

Q-19: 2X(g) + Y(g)à 2D(g)

ΔUΘ\Delta U^{\Theta } = -10.5 kJ and ΔSΘ\Delta S^{\Theta } = -44.1JK1JK^{-1}

Determine ΔGΘ\Delta G^{\Theta } for the given reaction, and predict that whether given reaction can occur spontaneously or not.


2X(g) + Y(g) à 2D(g)

Δng\Delta n_{g} = 2 – 3

= -1 mole

Putting value of ΔUΘ\Delta U^{\Theta } in expression of ΔH\Delta H:

ΔHΘ=ΔUΘ+ΔngRT\Delta H^{\Theta } = \Delta U^{\Theta } + \Delta n_{g}RT

= (-10.5KJ) – (-1)( 8.314×103kJK1mol18.314\times 10^{-3}kJK^{-1}mol^{-1})(298K)

= -10.5kJ -2.48kJ

ΔHΘ\Delta H^{\Theta } = -12.98kJ

Putting value of ΔSΘ\Delta S^{\Theta } and ΔHΘ\Delta H^{\Theta } in expression of ΔGΘ\Delta G^{\Theta }:

ΔGΘ=ΔHΘTΔSΘ\Delta G^{\Theta } = \Delta H^{\Theta } – T\Delta S^{\Theta }

= -12.98kJ –(298K)(-44.1JK1JK^{-1})

= -12.98kJ +13.14kJ

ΔGΘ\Delta G^{\Theta } = 0.16kJ

As, ΔGΘ\Delta G^{\Theta } is positive, the reaction won’t occur spontaneously.

Q-20: The equilibrium constant for a reaction is 10. What will be the value of ∆G0? R = 8.314 JK–1 mol–1, T = 300 K.



ΔGΘ\Delta G^{\Theta } = 2.303RTlogeq-2.303RT\log eq

= (2.303)( 8.314×103kJK1mol18.314\times 10^{-3}kJK^{-1}mol^{-1})(300K) log10\log 10

= -5744.14Jmol1Jmol^{-1}


Q-21: Comment on the thermodynamic stability of NO(g), given,

(1/2)N2(g) + (1/2)O2(g); ΔrHΘ=90kJmol1\Delta _{r}H^{\Theta } = 90kJmol^{-1}

NO(g) + (1/2)O2(g)à NO2(g) ; ΔrHΘ=74kJmol1\Delta _{r}H^{\Theta } = -74kJmol^{-1}


The +ve value of ΔrH\Delta _{r}H represents that during NO(g) formation from O2 and N2,  heat is absorbed. The obtained product, NO(g) is having more energy than reactants. Thus, NO(g) is unstable.

The -ve value of ΔrH\Delta _{r}H represents that during NO2(g) formation from O2(g) and NO(g),  heat is evolved. The obtained product, NO2(g) gets stabilized with minimum energy.

Thus, unstable NO(g) converts into unstable  NO2(g).

Q-22: Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formedunder standard conditions. ∆f H0= –286 kJ mol–1.


ΔrHΘ=286kJmol1\Delta _{r}H^{\Theta } = -286kJmol^{-1} is given so that amount of heat is evolved during the formation of 1 mole of H2O(l).

Thus, the same heat will be absorbed by surrounding. Qsurr = +286kJmol1kJmol^{-1}.

Now, ΔSsurr\Delta S_{surr} = Qsurr/7

= 286kJmol1298K\frac{286kJmol^{-1}}{298K}

Therefore, ΔSsurr=959.73Jmol1K1\Delta S_{surr} = 959.73Jmol^{-1}K^{-1}

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The NCERT solutions for class 11 chemistry (chapter 6) provided on this page contains all the intext and exercise questions listed in chapter 6 of the NCERT chemistry textbook for class 11 and, therefore, cover many important questions that could be asked in examinations.

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