NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

# NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

Edited By Sumit Saini | Updated on Aug 18, 2022 04:14 PM IST

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics- Thermodynamics is the branch of science which deals with the transfer of energy or changes in energy in physical or chemical processes. This is an important chapter for class 11th students because some concepts of this chapter will also be applied in the next chapter which is equilibrium and also some in the 12th class NCERT book chapter like chemical kinetics etc.

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In NCERT solutions for class 11 chemistry chapter 6 Thermodynamics, there are 22 questions in the exercise. The NCERT solutions for class 11 chemistry chapter 6 thermodynamics are prepared in a very comprehensive manner. These NCERT solutions of class 11 can help you in the preparation of class 11 final examination as well as in the competitive exams like JEE Mains, NEET, etc. If you are looking for an answer from any other chapter even from any other class then go with NCERT Solutions , there you will get all the answers of NCERT easily.

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics- Exercise Questions

A state function refers to the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of path.
Like- $p, V,T$ depends on the state of the system, not on the path.

So, the correct option is (ii)

(i) $\Delta T = 0$
(ii) $\Delta p = 0$
(iii) q = 0
(iv) w = 0

In adiabatic condition, the heat exchange between the system and the surrounding through its boundary is not allowed.
So, the correct option is (iii) $q$ = 0

The enthalpies of all elements in their standard states are Zero.
So, the correct option is (ii)

Equation of combustion of methane-

$CH_4(g)+\:2O_2(g)\rightarrow CO_2(g)\:+\:2H_2O(l)$

Since, $\Delta H^o=\Delta U^o+\Delta n_gRT$ and $\Delta U^o=-X\:KJ\:mol^{-1}$

from the equation $\Delta n_g=(n_p-n_R)=1-3=-2$

$\Delta H^o=-X-2RT$

$\Delta H^o<\Delta U^o$

so the correct option is (iii).

(i) –74.8 $kJ mol^{-1}$

(ii) –52.27 $kJ mol^{-1}$

(iii) +74.8 $kJ mol^{-1}$

(iv) +52.26 $kJ mol^{-1}$

$CH_4+2O_{2}\rightarrow CO_{2}+2H_{2}O\ \Delta H$ ................. -890.3 kJ/mol
$C+2O_{2}\rightarrow CO_{2}$ $\Delta H$ ..................-393.5 kJ/mol
$2H_{2}+O_{2}\rightarrow 2H_{2}O$ $\Delta H$ ..................-258.8 kJ/mol

So, the required equation is to get the formation of $CH_4 (g)$ by combining these three equations-

Thus, $C+2H_{2}\rightarrow CH_{4}$

$\Delta _fH_{CH_{4}} = \Delta _cH_c+2\Delta _cH_{H_2}$
= [-393.5 + 2(-285.8) + 890.3]
= -74.8 kJ/mol

therefore, the enthalpy of formation of methane is -74.8 kJ/mol. So, the correct option is (i)

For the reaction to be feasible the $\Delta G$ should be negative
$\Delta G = \Delta H - T\Delta S$

According to question,
$\Delta S$ = positive unit
$\Delta H$ = negative (as heat is evolved in the reaction)
Overall the $\Delta G$ is negative.

Therefore the reaction is possible at any temperature. So, the correct option is (iv)

The first law of Thermodynamics state that,
$\Delta U = q+W$

where, $\Delta U$ = change in internal energy for the process
q = heat and W = work

Given that,
q = +701 J (heat is absorbed)
W = - 394 (work is done by the system)

Substituting the value in equation of first law we get,

$\Delta U = 701+(-394) =307J$

So, the change in internal energy for the process is 307 J

$NH_2 CN ( g ) + \frac{3}{2} O_2 (g) \rightarrow N_2 + CO_2 ( g ) H_2 O ( l)$

Given information,
$\Delta U = -742.7kJmol^{-1}$
T = 298 K
R = 8.314 $\times 10^{-3}$

$\Delta n_g =$ $n_g$ (products) $- n_g$ (reactants)
= (2 - 1.5)
= 0.5 moles

The enthalpy change for the reaction is expressed as;

$\Delta H = \Delta U+n_gRT$
where, $\Delta U$ = change in internal energy and
$\Delta n_g =$ change in no. of moles

By putting the values we get,
$\Delta H = (-742.7)+0.5(298)(8.314\times 10^{-3})$
$=-741.5 kJmol^{-1}$

We know that
$q = mc\Delta T$
m = mass of the substance
c = heat capacity
$\Delta T$ = change in temperature

By putting all these values we get,
$q = (\frac{60}{27}mol)(24 J/mol/K)(20K)$
= 1066.7 J
= 1.07 kJ

$\Delta _{ fus }H = 6.03 KJ mol ^{-1}$ at $0 \degree C$ .

$Cp [H_2O(l)] = 75.3 J mol^{-1} K^{-1}\\\\.\: \: \: \: Cp [H_2O(s)] = 36.8 J mol^ { -1}K^{-1}$

Total enthalpy change is equal to the summation of all the energy required at three different stages-
(i) from $10^o C$ to $0^o C$ of 1 mol of water (water $\rightarrow$ water )
(ii) from $0^o C$ to $0^o C$ of 1 mol of ice (water $\rightarrow$ ice)
(iii) from $0^o C$ to $-10^oC$ of 1 mole of ice (ice $\rightarrow$ ice)

So, the total enthalpy change
$\Delta H = C_p[H_2O(l)](\Delta T)+\Delta H_(freezing)+C_p[H_2O(s)]\Delta T$

$\\ = (75.3)(-10)+(-6.03\times 10^{3})+(36.8)(-10)\\ =-753-6030 - 368\\ =-7151 Jmol^{-1}\\ = 7.151kJ/mol$
Hence the total enthalpy change in the transformation process is 7.151 kJ/mol

Formation of carbon dioxide from carbon and dioxygen reaction is-

$C+O_{2}\rightarrow CO_{2}$ ............. $\Delta H_f$ = -393.5 kJ/mol

1 mole of $CO_2$ = 44 g
So, the heat released in the formation of 44g of $CO_2$ = -393.5 kJ/mol
therefore, In 35.5 g of $CO_2$ the amount of heat released =
$\frac{-393.5}{44}\times 35.2 = -314.8 kJ/mol$

$N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)$

Given,
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 $KJ mol ^{-1}$
$N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)$
We know that the $\Delta _rH=\sum \Delta _fH$ (product) $-\sum \Delta _fH$ (reactants)

For the above reaction,

$\Delta _rH =[{\Delta _fH(N_{2}O)+3\Delta _fH(CO_{2})}]-[\Delta _fH(N_{2}O_{4})+3\Delta _fH(CO)]$
Substituting the given values we get,
$\\= [81+3(-393)]-[9.7+3(-110)]\\ =-777.7 kJ/mol$
Thus the value of $\Delta _ r H$ of the reaction is -777.7 kJ/mol

the standard enthalpy of formation of any compound is the required change in enthalpy of formation of 1 mole of a substance in its standard form from its constituent elements.

Thus we can re-write the reaction as;
$\frac{1}{2}N_{2}+\frac{3}{2}H_{2}\rightarrow NH_{3}$

Therefore, standard enthalpy of formation of ammonia is = $\frac{1}{2}\Delta _rH^\Theta$
= 1/2 (-92.4)
= -46.2 kJ/mol

$CH_3OH (l) + 3/2 O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \Delta _r H = -726 kJ mol^{-1}\\\\ .\: \: \: \: C(graphite) + O_2(g) \rightarrow CO_2(g) ;\Delta_ cH = -393 kJ mol ^ {-1}\\\\ .\: \: \: H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l) ; \Delta_ f H = - 286 kJ mol^{-1}.$

for the formation of $CH_3OH$ the reaction is ,

$C+2H_{2}O+\frac{1}{2}O_{2}\rightarrow CH_3OH$
This can be obtained by the following expressions-
required eq = eq (i) + 2 (eq iii) - eq (i)

$\Delta _fH[CH_{3}OH] = \Delta _cH^\Theta +2\Delta _fH[H_{2}O]-\Delta _rH^\Theta$
$=(-393) + 2(-286)-(-726)$
$=-239 kJmol^{-1}$

Question 6.15 Calculate the enthalpy change for the process $CCl_4(g)\rightarrow C(g) + 4 Cl(g)$ and calculate bond enthalpy of $C-Cl$ in $CCl_4(g)$
$\Delta _{vap} H ^{\ominus }(CCl_4) = 30.5 kJ mol^{-1}.\\\\ .\: \: \: \: \: \Delta _f H ^\ominus (CCl_4) = -135.5 kJ mol^{-1}.\\\\.\: \: \: \: \Delta _ a H ^ \ominus (C) = 715.0 kJ mol ^{-1} , where \: \: \Delta_ a H ^\ominus \: \: is\: \: enthalpy \: \: of \: \: atomisation\\\\.\: \: \: \: \Delta _ a H ^\ominus (Cl_2) = 242 kJ mol ^{-1}$

We have the following chemical reactions equations-

$(1)\ CCl_{4}(l)\rightarrow CCl_{4}(g)$ ................ $\Delta _{vap}H^o = 30.5kJ/mol$
$(2)\ C(s)\rightarrow C(g)$ ........................ $\Delta _{a}H^o =715.0kJ/mol$
$(3)\ Cl2(g)\rightarrow 2Cl(g)$ .................. $\Delta _{a}H^o =242kJ/mol$
$(4)\ C(g)+4Cl\rightarrow CCl_4(g)$ ............ $\Delta _{f}H =-135.5kJ/mol$
The enthalpy change for the process $CCl_4(g)\rightarrow C(g) + 4 Cl(g)$ by the above reaction is calculated as;

$\Delta H =\Delta _aH^o(C)+2(\Delta _aH^o(Cl_{2}))-\Delta _{vap}H^o-\Delta _fH$
$= [(715)+2(242)-(30.5)-(-135.5)] kJ/mol$
$=1304 kJmol^{-1}$

And the bond enthalpy of C-Cl bond in $CCl_4$ (g) = 1304/4 = 326 kJ/mol

Since $\Delta U = 0$
So, the $\Delta S>0$ (positive) therefore the reaction will be feasible.

From the equation,

$\Delta G = \Delta H-T\Delta S$
Suppose the reaction is at equilibrium, So the change in temperature is given as;

$T=\frac{\Delta H -\Delta G}{\Delta S}$
$= \frac{400}{0.2} =2000K$ ( $\Delta G$ at equilibrium is zero)
To reaction should be spontaneous, $\Delta G$ should be neagtive. So, that for the given reaction T should be greater than 2000 K

The given reaction represents the formation of a chlorine molecule from its atom. Bond formation taking place, therefore the energy is released during this. So, $\Delta H$ is negative.

Two moles of an atom have more randomness than the one mole of chlorine. So, spontaneity is decreased. Hence $\Delta S$ is negative.

Given reaction is $2 A (g) + B ( g ) \rightarrow 2 D ( g)$

We know that, $\Delta G^o = \Delta H^o-T\Delta S^o$
and $\Delta H^o = \Delta U^o+\Delta n_{g}RT$
here $\Delta n_{g}= 2 - (3) = -1$

and $\Delta U ^ \ominus = -10.5 KJ \: \: and \: \: \Delta S ^\ominus = - 44.1 JK^{-1}$

substituting the given values in equations-
$\Delta G = (-10.5+8.314\times 10^{-3}\times 298)-298(-44.1)$
$\\\Delta G = (-12.98 +13.14 )kJ\\ \Delta G= +0.16 kJ$

Hence the reaction will not occur spontaneously because the $\Delta G ^ \ominus$ value is positive for this reaction.

$\Delta G ^ \ominus ? R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K$

Given values are,
$R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K$
and equilibrium constant = 10

It is known that,
$\Delta G^o = -2.303\ RT \log K_{eq}$
$\\=-2.303(8.314)(300)\log 10\\ =-5744.14 Jmol^{-1}\\ =-5.744 kJmol^{-1}$

Hence the value of $\Delta G^o$ is -5.744 kJ/mol

Question

$1/2 N_2(g) + 1/2 O_2(g)\rightarrow NO(g) ; \Delta _ rH ^ \ominus = 90 kJ mol ^{-1}\\\\ .\: \: \: NO(g) + 1/2 O_2(g) \rightarrow NO_ 2(g) : \Delta _ r H^ \ominus = -74 kJ mol ^{-1}$

The formation of $NO$ is unstable because $\Delta _rH$ is positive it means heat is absorbed during the reaction. So, $NO$ (g) has higher energy than its reactants $N_{2}$ and $O_{2}$ .
On the other hand, $NO_{2}$ is stable because $\Delta _rH$ is negative means heat is released during the formation of $NO_{2}$ . It is stablised with minimum energy. Hence unstable $NO$ changes to stable $NO_{2}$ .

-286 kJ/mol heat is evolved when 1 mol of the water molecule is formed. It means the same amount of energy is absorbed by the surrounding also.

$q_{surr}= +286$ kJ/mol
We know that,
Entropy changes for the surrounding is = q(surr)/ temp.
= 286000/ 298

= 959.73 J/mol/K

### More About Thermodynamics Class 11 NCERT Chemistry Chapter

Thermodynamics enables us to study energy changes quantitatively and to make useful predictions. For these, we divide the universe into the system and the surroundings. In thermodynamics, the system is the part of the universe where observations are made and the remaining universe constitutes surroundings. After completing NCERT solutions for class 11 chemistry chapter 6 Thermodynamics you will be able to explain the terms like system, surroundings, and boundary; explain three types of system open, close and isolated systems; explain work, heat, and internal energy, etc. In this chapter, we will also discuss the first law of thermodynamics and second laws of thermodynamics and express them mathematically, internal energy, enthalpy, entropy, Gibbs energy. You will find NCERT solutions for class 11 chemistry chapter 6 Thermodynamics by scrolling down.

Important points of NCERT solutions for class 11 chemistry chapter 6 Thermodynamics-

1. System and Surroundings - in thermodynamics, the system is the part of the universe where observations are made and the remaining universe constitutes surroundings. The universe = The system+ The Surroundings .

2. The First Law of Thermodynamics- according to it under any circumstances, energy cannot be created and nor be destroyed, it can be converted from one form to another form with the interaction of work, heat and internal energy. Mathematically, this is represented as: $\Delta U=q+w$ , where w is the work done on or by the system, q is the heat exchanged between the system and its surroundings and ΔU is the total change in internal energy of the system .

3. The second law of thermodynamics- it states that entropy of the entire universe(total entropy of both the system and its surroundings), as an isolated system, will always increase over time.

NCERT Syllabus Class 11 Chemistry Chapter 6 Thermodynamics-Topics

6.1 Thermodynamic Terms

6.2 Applications

6.3 Measurement of âU and âH: Calorimetry

6.4 Enthalpy Change, âH of a Reaction – Reaction Enthalpy

6.5 Enthalpies for Different Types of Reactions

6.6 Spontaneity

6.7 Gibbs Energy Change and Equilibrium

NCERT Solutions for Class 11 Chemistry

 Chapter 1 Chapter-2 Chapter-3 Chapter-4 Chapter-5 States of Matter Chapter-6 6 Thermodynamics Chapter-7 Chapter-8 Chapter-9 Chapter-10 Chapter-11 Chapter-12 Chapter-13 Chapter-14

NCERT solutions for class 11 subject wise

 lass 11 Maths lass 11 Chemistry lass 11 Physics

Benefits of NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

• The comprehensive answers given in the CBSE NCERT solutions for class 11 chemistry chapter 6 Thermodynamics will be helpful in understanding chapter easily.
• Revision will be quiet easier because with the detailed solutions will help you to remember the concepts and get very good marks in your class.
• Homework problems won't bother you anymore, all you need to do is check the detailed NCERT solutions for class 11 chemistry Chapter 6 Thermodynamics and you are good to go.
• If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.

Also Check NCERT Books and NCERT Syllabus here:

1. Question: Whether the unit thermodynamics is helpful in higher studies?

Yes, it is helpful to a large extent

2. What is the official website of NCERT

official website of NCERT: http://www.ncert.nic.in/

3. What is the weightage of the chapter thermodynamics for CBSE board exam

Weightage of thermodynamics for CBSE board exam is of 8 marks

4. What is the weightage of NCERT Chapter Thermodynamics for NEET exam

Weightage of Thermodynamics for NEET exam is 3%

5. Where can I find complete solutions of NCERT class 11 Chemistry
6. What is the weightage of NCERT Chapter Thermodynamics for JEE mains

8 marks

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