NCERT Solutions For Class 11 Chemistry Chapter 6

NCERT Solutions Class 11 Chemistry Chemical Thermodynamics

NCERT Solutions For Class 11 Chemistry Chapter 6 is provided here. This topic is an extremely important topic in chemistry and is important for class 11, Class 12 and for competitive exams like JEE and NEET. Students must have a good knowledge of the topic in order to excel in the examination. We at BYJU'S provide the NCERT Solutions For Class 11 Chemistry Chemical Thermodynamics PDF which students can download. Practicing all the questions will be very helpful for the students as many questions are framed from the topic.

NCERT Solutions for Class 11 Chemistry Chapter 6 includes the topic - Thermodynamic Terms: The system and the surroundings, Types of System: Open, Closed, Isolated. The State of the system. The Internal energy as a state Function: Work, Heat, The general case. Applications: Work, Isothermal and free expansion of an Ideal Gas. Enthalpy, Extensive and Intensive Properties, Heat capacity, The relation between Cp and Cv for an ideal gas. Measurement of ∆U and ∆H Calorimetry. Gibbs energy change and equilibrium, Enthalpy Change. Enthalpies for different types of reactions. Lattice enthalpy. Spontaneity. NCERT Solutions Class 11 Chemistry Chemical Thermodynamics PDF is provided here for better understanding and clarification of the chapter.

Thermodynamics is a branch of science. It deals with the relationship between heat and other forms of energy. A part of the universe where observations are made is called system. Surrounding is part of universe excluding system. Based on the exchange of energy and matter, system is classified in 3 types: Closed system, Open system, Isolated system. In 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 isolated system both energy and matter cannot be exchanged with the surrounding. This was a brief on Chemical Thermodynamics.

Q-1: A thermodynamic state function is _______.

1. A quantity which depends upon temperature only.

2. A quantity which determines pressure-volume work.

3. A quantity which is independent of path.

4. A quantity which determines heat changes.


(3) 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: Which of the following is a correct conditions for adiabatic condition to occur.

1. q = 0

2. w = 0

3. Δp=0

4. ΔT=0


1: q = 0


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

Q-3: The value of enthalpy for all elements in standard state is _____.

(1) Zero

(2) < 0

(3) Different for every element

(4) Unity


(1) Zero

Q-4: For combustion of methane ΔUΘ is – Y kJmol1. Then value of ΔHΘ is ____.

(a) > ΔUΘ

(b) = ΔUΘ

(c) = 0

(d) < ΔUΘ


(d) < ΔUΘ


ΔHΘ=ΔUΘ+ΔngRT ; ΔUΘ = – Y kJmol1,


Q-5: For, Methane, di-hydrogen and and graphite the enthalpy of combustion at 298K are given -890.3kJ mol1, -285.8kJmol1 and -393.5kJmol1 respectively. Find the enthalpy of formation of Methane gas?

(a) -52.27kJmol1

(b) 52kJmol1

(c) +74.8kJmol1

(d) -74.8kJmol1


(d) -74.8kJmol1

à CH4(g)  + 2O2(g) CO2(g) + 2H2O(g)


à C(s) + O2(g)   CO2(g)


à 2H2(g) + O2(g) 2H2O(g)


à C(s) + 2H2(g) CH4(g)

ΔfHCH4  = ΔcHc + 2ΔfHH2ΔfHCO2

= [ -393.5 +2(-285.8) – (-890.3)] kJmol1

= -74.8kJmol1

Q-6: A reaction, X + Y à U + V + q is having a +ve entropy change. Then the reaction ____.

(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 should be –ve, for spontaneous reaction to occur


As per given in question,

ΔH is –ve ( as heat is evolved)

ΔS is +ve

Therefore, ΔG is negative

So, the reaction will be possible at any temperature.

Q-7: In the process, system absorbs 801 J and work done by the system is 594 J. Find ΔU for the given process.



As per Thermodynamics 1st law,

ΔU = q + W(i);

ΔU internal energy = heat

W = work done

W = -594 J (work done by system)

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


ΔU = 801 + (-594)

ΔU = 207 J

Q-8: The reaction given below was done in bomb calorimeter, and at 298K we get, ΔU = -753.7 kJ mol1. Find ΔH at 298K.

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


ΔH is given by,


Δng = change in number of moles

ΔU = change in internal energy



  = (2 – 2.5) moles

 Δng = -0.5 moles


T =298K

ΔU = -753.7 kJmol1

R  = 8.314×103kJmol1K1

Now, from (1)


= -753.7 – 1.2

ΔH = -754.9 kJmol1

Q-9: Calculate the heat (in kJ) required for 50.0 g aluminium to raise the temperature from 45Cto65C. For aluminium molar haet capacity is 24 Jmol1K1


Expression of heat(q),


ΔT = Change in temperature

c = molar heat capacity

m = mass of substance

From (a)


q = 888.88 J q

Q-10: Calculate ΔH for transformation of 1 mole of water into ice from10Cto(10)C.. ΔfusH=6.03kJmol1at10C.




ΔHtotal = sum of the changes given below:

(a) Energy change that occurs during transformation of 1 mole of water from 10Cto0C.

(b) Energy change that occurs during transformation of 1 mole of water at 0C  to  1 mole of ice at 0C.

(c) Energy change that occurs during transformation of 1 mole of ice from 0Cto(10)C.


= (75.3 Jmol1K1)(0 – 10)K + (-6.03*1000 Jmol1(-10-0)K

= -753 Jmol1 – 6030Jmol1 – 368Jmol1

= -7151 Jmol1

= -7.151kJmol1

Thus, the required change in enthalpy for given transformation is -7.151kJmol1.

Q-11: Enthalpies of formation for CO2(g), CO(g), N2O4(g), N2O(g) are -393kJmol1,-110kJmol1, 9.7kJmol1 and 81kJmol1 respectively. Then, ΔrH = _____.

N2O4(g) + 3CO(g) à  N2O(g) + 3 CO2(g)


ΔrH for any reaction is defined as the fifference between ΔfH value of products and ΔfH value of reactants.”


Now, for

N2O4(g) + 3CO(g) à  N2O(g) + 3 CO2(g)


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

ΔrH = [{81kJmol1 + 3(-393) kJmol1} – {9.7kJmol1 + 3(-110) kJmol1}]

ΔrH = -777.7 kJmol1


Q-12 Enthalpy of combustion of C to CO2 is -393.5 kJmol1. Determine the heat released on the formation of 37.2g of CO2 from dioxygen and carbon.


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

C(s) + O2(g) à CO2(g); ΔfH = -393.5 kJmol1

1 mole CO2 = 44g

Heat released during formation of 44g CO2 = -393.5 kJmol1

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

= 393.5kJmol144g×37.2g

= -332.69 kJmol1


Q-13: N2(g) + 3H2(g) à 2NH3(g) ; ΔrHΘ = -92.4 kJmol1

Standard Enthalpy for formation of ammonia gas is _____.


“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

= (0.5)(-92.4kJmol1)

= -46.2kJmol1


Q-14: Determine Standard Enthalpy of formation for CH3OH(l) from the data given below:

CH3OH(l) + (3/2)O2(g) à CO2(g) + 2H2O(l); ΔrHΘ = -726 kJmol1

C(g) + O2(g) à CO2(g); ΔcHΘ = -393 kJmol1

H2(g) + (1/2)O2(g) à H2O(l); ΔfHΘ = -286 kJmol1



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

CH3OH(l) can be obtained as follows,

ΔfHΘ [CH3OH(l)] = ΔcHΘ


= (-393 kJmol1) +2(-286kJmol1) – (-726kJmol1)

= (-393 – 572 + 726) kJmol1

= -239kJmol1

Thus, ΔfHΘ [CH3OH(l)] = -239kJmol1


Q-15: Calculate ΔH for the following process

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

ΔvapHΘ (CCl4) = 30.5 kJmol1.

ΔfHΘ (CCl4) = -135.5 kJmol1.

ΔaHΘ (C) = 715 kJmol1,

ΔaHΘ  is a enthalpy of atomisation

ΔaHΘ (Cl2) = 242 kJmol1.



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

(1) CCl4(l) à CCl4(g) ; ΔvapHΘ = 30.5 kJmol1

(2) C(s) à C(g) ΔaHΘ = 715 kJmol1

(3) Cl2(g) à 2Cl(g) ; ΔaHΘ = 242 kJmol1

(4) C(g) + 4Cl(g) à CCl4(g); ΔfHΘ = -135.5 kJmol1

ΔH for the process CCl4(g) à C(g) + 4Cl(g) can be measured as:


= (715kJmol1) + 2(kJmol1) – (30.5kJmol1) – (-135.5kJmol1)

Therefore, H=1304kJmol1

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

= 13044kJmol1

= 326 kJmol1


Q-16: ΔU = 0 for isolated system, then what will be ΔU?


ΔU is positive ; ΔU > 0.

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


Following reaction takes place at 298K,

2X + Y à Z

ΔH = 400 kJmol1

ΔH = 0.2kJmol1K1

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




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


ΔHΔS; (ΔG = 0 at equilibrium)

= 400kJmol1/0.2kJmol1K1

Therefore, T = 2000K

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


Q-18: 2Cl(g) à Cl2(g)

In above reaction what can be the sign for ΔS and ΔH?


ΔS and ΔH 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 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 is negative.


Q-19: 2X(g) + Y(g) à 2D(g)

ΔUΘ = -10.5 kJ and ΔSΘ = -44.1JK1

Determine ΔGΘ for the given reaction, and predict that whether given reaction can occur spontaneously or not.


2X(g) + Y(g) à 2D(g)

Δng = 2 – 3

= -1 mole

Putting value of ΔUΘ in expression of ΔH:


= (-10.5KJ) – (-1)( 8.314×103kJK1mol1)(298K)

= -10.5kJ -2.48kJ

ΔHΘ = -12.98kJ

Putting value of ΔSΘ and ΔHΘ in expression of ΔGΘ:


= -12.98kJ –(298K)(-44.1JK1)

= -12.98kJ +13.14kJ

ΔGΘ = 0.16kJ

As, ΔGΘ is positive, the reaction won’t occur spontaneously.


Q-20: Find the value of ΔGΘ for the reaction, if equilibrium is given 10.given that T = 300K and R =  8.314×103kJK1mol1.



ΔGΘ = 2.303RTlogeq

= (2.303)( 8.314×103kJK1mol1)(300K) log10

= -5744.14Jmol1



Q-21: What can be said about the thermodynamic stability of NO(g), given

(1/2)N2(g) + (1/2)O2(g); ΔrHΘ=90kJmol1

NO(g) + (1/2)O2(g) à NO2(g) ; ΔrHΘ=74kJmol1


The +ve value of ΔrH 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 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: Determine ΔS in surrounding given that  mole of H2O(l) is formed I standard condition. ΔrHΘ=286kJmol1.


ΔrHΘ=286kJmol1 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 = +286kJmol1.

Now, ΔSsurr = Qsurr/7

= 286kJmol1298K

Therefore, ΔSsurr=959.73Jmol1K1

Continue learning the Laws of Thermodynamics and its applications with Byju’s video lectures.

Downloadable files of ebooks, notes, pdfs are available at BYJU’S for students to start their exam preparation without delay. Once you have registered with BYJU’S access to online course is very simple. To know more about the difference between system and surrounding, distinguish between closed, open and isolated systems, Internal energy, heat, work, First law of thermodynamics, Calculation of energy change, Correlation between ∆U and ∆H, Hess’s law, Extensive and intensive properties, ∆G etc sign up on BYJU’S the learning app. Apart from NCERT Solutions for Class 11 Chemistry Chapter 6, get NCERT solutions for other chemistry chapters - NCERT Solutions for Class 11 Chemistry as well. Students can download worksheets, assignments, NCERT Class 11 Chemistry pdf and other study materials for exam preparation and score better marks.

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