NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

Edited By Shivani Poonia | Updated on May 08, 2025 11:38 AM IST

Thermodynamics is an important branch of Chemistry. It deals with energy, its transformation, and the laws governing the entire process of transformation of energy. Have you ever wondered why a hot cup of tea cools down after some time or why ice melts, so the only reason behind this kind of phenomenon is thermodynamics. NCERT Class 11 Chapter 5 plays a very important role in Chemistry as it is the basic chapter that builds the foundations of advanced chapters and it also helps to explain why changes occur around us.

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This Story also Contains
  1. Download PDF Of NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics
  2. NCERT Solutions for Class 11 Chemistry Thermodynamics (Exercise Questions)
  3. Class 11 Chemistry NCERT Chapter 7: Higher Order Thinking Skills (HOTS)
  4. Approach to Solve Questions of Chapter 5 Thermodynamics
  5. NCERT Syllabus Class 11 Chemistry Thermodynamics-Topics
  6. What Extra Should Students Study Beyond NCERT for JEE/NEET?
  7. NCERT Solutions for Class 11 Chemistry
  8. Importance of Learning Thermodynamics for JEE and NEET Exams
  9. NCERT solutions for class 11 subject-wise
  10. Benefits of NCERT Solutions for Class 11 Chemistry Thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics
NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

In Thermodynamics, there are 22 questions in the exercise. The solutions of thermodynamics are prepared in a very comprehensive manner. These NCERT solutions of class 11 can help you prepare for the class 11 final examination and 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.

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Download PDF Of NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

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NCERT Solutions for Class 11 Chemistry Thermodynamics (Exercise Questions)

Question 5.1 Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of the path
(iii) used to determine pressure-volume work
(iv) whose value depends on temperature only.

Answer :

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

So, the correct option is (ii)

Question 5.2 For the process to occur under adiabatic conditions, the correct condition is:

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

Answer :

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

Question 5.3 The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element

Answer :

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

Question 5.4 $\Delta U ^\ominus$ of combustion of methane is $- X KJ mol ^{-1}$ . The value of $\Delta H ^\ominus$ is
(i) $= \Delta U ^\ominus$
(ii) $> \Delta U ^\ominus$
(iii) $< \Delta U ^\ominus$
(iv) = 0

Answer :

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).

Question 5.5 The enthalpy of combustion of methane, graphite, and dihydrogen at 298 K are $-890.KJ mol ^{-1} , -393.5 KJ mol ^{-1} \: \:$ 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 $kJ mol^{-1}$

Answer :

$CH_4+2O_{2}\rightarrow CO_{2}+2H_{2}O\ \Delta H$ ................. -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)

Question 5.6 A reaction, A + B $\rightarrow$ 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
(v) possible at any temperature

Answer :

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

According to the 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)

Question 5.7 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?

Answer :

The first law of Thermodynamics states 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 the equation of the first law we get,

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

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

Question 5.8 The reaction of cyanamide, $NH_2 CN$ (s), with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be $-742.7 KJ mol ^{-1}$ at 298 K. Calculate the enthalpy change for the reaction at 298 K.

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

Answer :

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,
$\Delta H = (-742.7)+0.5(298)(8.314\times 10^{-3})$
$=-741.5 kJmol^{-1}$

Question 5.9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminum from $35 ^\circ C$ to $55 ^\circ C$. Molar heat capacity of Al is $24 J mol ^{-1} K ^{-1}$

Answer :

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

Question 5.10 Calculate the enthalpy change on freezing of 1.0 mol of water at $10.0 ^\circ C$ to ice at $- 10.0 ^\circ C$.

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

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

Answer :

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

Question 5.11 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.

Answer :

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 =
$\frac{-393.5}{44}\times 35.2 = -314.8 kJ/mol$

Question 5.12 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 $\Delta _ r H$ for the reaction:

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

Answer :

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

Question 5.13 Given $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) ;\Delta rH = -92.4 kJ mol ^{-1}$ .What is the standard enthalpy of the formation of $NH_3$ gas?

Answer :

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

Question 5.14 Calculate the standard enthalpy of the formation of CH3OH(l) from the following data:

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

Answer :

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

Answer :

We have the following chemical reaction 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

Question 5.16 For an isolated system, $\Delta U = 0$, what will be $\Delta S$?

Answer :

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

Question 5.17 For the reaction at 298 K,
2A + B $\rightarrow$ C
$\Delta H = 400 kJ mol^ {-1 }\: \: and \: \: \Delta S = 0.2 kJ K ^{-1} mol^{-1}$
At what temperature will the reaction become spontaneous considering $\Delta H$ and $\Delta S$ to be constant over the temperature range?

Answer :

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 negative. So, for the given reaction T should be greater than 2000 K

Question 5.18 For the reaction, $2 Cl (g) \rightarrow Cl_2 ( g)$ , what are the signs of $\Delta H \: \: and \: \: \Delta S$?

Answer :

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 one mole of chlorine. So, spontaneity is decreased. Hence $\Delta S$ is negative.

Question 5.19 For the reaction
$2 A (g) + B ( g ) \rightarrow 2 D ( g)$
$\Delta U ^ \ominus = -10.5 KJ \: \: and \: \: \Delta S ^\ominus = - 44.1 JK^{-1}$
Calculate $\Delta G ^ \ominus$ for the reaction, and predict whether the reaction may occur spontaneously.

Answer :

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.

Question 5.20 The equilibrium constant for a reaction is 10. What will be the value of

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

Answer :

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 5.21 Comment on the thermodynamic stability of NO(g), given

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

Answer :

The formation of $NO$ is unstable because $\Delta _rH$ is positive which 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 established with minimum energy. Hence unstable $NO$ changes to stable $NO_{2}$.

Question 5.22 Calculate the entropy change in surroundings when 1.00 mol of $H_2 O (l)$ is formed under standard conditions. $\Delta _f H ^ \ominus = -286 KJ mol ^ { -1}$

Answer :

-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

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 its surroundings. After completing NCERT solutions for Thermodynamics you will be able to explain terms like system, surroundings, and boundary; In this chapter, we will also discuss the first law of thermodynamics and second laws of thermodynamics and express them mathematically.

Class 11 Chemistry NCERT Chapter 7: Higher Order Thinking Skills (HOTS)

Question:

At standard conditions, if the change in the enthalpy for the following reaction is $-109 \mathrm{kJmol}^{-1}$.

$$
\mathrm{H}_{2(g)}+B r_{2(g)} \rightarrow 2 \mathrm{HBr}_{(g)}
$$
Given that bond energy of $\mathrm{H}_2$ and $\mathrm{Br}_2$ is $435 \mathrm{kJmol}^{-1}$ and $192 \mathrm{kJmol}^{-1}$ respectively, What is the bond energy (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of HBr ?

1) 736

2) 518

3) 259

4) 368

Solution:

$$
\begin{aligned}
& \quad \mathrm{H}_2 \rightarrow 2 \mathrm{H} \quad \mathrm{~B} \cdot E_{H_2} \\
& \mathrm{Br}_2 \rightarrow 2 \mathrm{Br} \quad \mathrm{~B} \cdot E_{B r_2} \\
& 2 \mathrm{H}+2 \mathrm{Br} \rightarrow 2 \mathrm{HBr} \quad-2 \mathrm{~B} \cdot E_{H B r} \\
& \mathrm{H}_2+\mathrm{Br}_2 \rightarrow 2 \mathrm{HBr}
\end{aligned}
$$


From the Bond energy concept, we have

$$
\begin{aligned}
& B E_{H_2}+B E_{B r_2}-2 B E_{H B r}=\Delta H_{H B r} \\
\rightarrow & 435+192-2 x=-109 \\
\rightarrow & 2 x=736 \\
\rightarrow & x=368 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$

Hence, the answer is the option (4).

Approach to Solve Questions of Chapter 5 Thermodynamics

Thermodynamics comes under physical chemistry. This chapter is about learning the concepts and then applying them to solve the numerical. The approach to solving numerical should be simple yet time-saving. For this, you first need to understand the key concepts as a system, surroundings, internal energy, work, heat, enthalpy, etc. For solving any questions first give it a thorough reading and note down all the information given in the question. Begin by identifying what is given in the question and what needs to be found. Carefully note the units and convert them if necessary to ensure consistency. Use the appropriate formula based on the concept like $\Delta U=q+w$ (change in internal energy $=$ heat + work done) or $\Delta H=\Delta U+P \Delta V$ (enthalpy change).

Let us understand it by taking an example. Suppose the question says "A system absorbs 300 J of heat and does 100 J of work. Calculate the change in internal energy."
Solution- Here, heat absorbed $q=+300 J$ (positive as it is absorbed) and work done by the system $w=$ $-100 J$ (negative as energy is leaving the system).
Now use the formula:

$$
\Delta U=q+w=300+(-100)=200 J
$$


So, the change in internal energy is 200 J.

Practice helps in mastering thermodynamics numerical. Start with simpler problems and gradually move to ones involving concepts like enthalpy change, work in isothermal or adiabatic processes, and heat capacity.

NCERT Syllabus Class 11 Chemistry Thermodynamics-Topics

5.1 Thermodynamic Terms
5.2 Applications
5.3 Measurement of $\Delta U$ and $\Delta H$: Calorimetry
5.4 Enthalpy Change, $\Delta H$ of a Reaction – Reaction Enthalpy
5.5 Enthalpies for Different Types of Reactions
5.6 Spontaneity
5.7 Gibbs Energy Change and Equilibrium

What Extra Should Students Study Beyond NCERT for JEE/NEET?

ConceptJEENCERT
Thermodynamics
Thermodynamics: Properties Of System
Path, State Function, Types Of Process
Reversible, Irreversible, Polytropic Process
Thermodynamic Equilibrium
Zeroth Law Of Thermodynamics
Heat And Work
Internal Energy
Isothermal Reversible And Isothermal Irreversible
Graphical Representation Of Work Done In Thermodynamics
Heat Capacity
Relation Between Cp And Cv
Thermochemistry And Enthalpy For Chemical Reaction
Standard Enthalpy And Enthalpy Of Formation
Enthalpy Of Combustion
Enthalpy Of Dissociation, Atomisation And Phase Change
Lattice Enthalpy, Hydration Enthalpy And Enthalpy Of Solution
Enthalpy Of Neutralisation
Ionization And Electron Gain Enthalpy
Resonance Enthalpy
Kirchoff’s Equation
Born Habers Cycle
Hess’s Law
Bomb Calorimeter
Calculation Of Changes In S For Different Process
Spontaneity Criteria Through Entropy
Spontaneity Criteria Through Enthalpy (H) And Entropy (S)
Gibbs Energy And Change In Gibbs Energy
Spontaneity Criteria With Gibbs Energy (G)
Gibbs Energy At Equilibrium
NEET/JEE Offline Coaching
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NCERT Solutions for Class 11 Chemistry

Importance of Learning Thermodynamics for JEE and NEET Exams

The Thermodynamics chapter in Class 11 NCERT Chemistry offers a bunch to students preparing for JEE and NEET:

  • It builds a strong conceptual foundation for physical chemistry. The key principles like internal energy, enthalpy, entropy, Gibbs free energy, etc. are quite important from an exam point of view.
  • The most beneficial part is the numerical problems. The numerical will make you understand the concept deeply and will also help to improve speed and accuracy in handling formula-based and conceptual questions.
  • Many JEE and NEET questions are directly based on NCERT content. A lot of definitions and laws like the First Law of Thermodynamics, Hess's Law, spontaneity criteria, etc. are often asked in the exams.
  • Thermodynamic concepts are common to both Chemistry and Physics so it will give dual preparation benefits.

NCERT solutions for class 11 subject-wise

Benefits of NCERT Solutions for Class 11 Chemistry Thermodynamics

  • The comprehensive answers given in the CBSE NCERT solutions for class 11 chemistry Thermodynamics will help understand the chapter easily.
  • The revision will be quite easier because 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 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:

Frequently Asked Questions (FAQs)

1. What is the relationship between enthalpy and heat capacity?

The enthalpy and heat capacity is related by the formula $C_p=\left(\frac{d H}{d T}\right)_{p^{\prime}}$ where $C_p$ is the heat capacity at constant pressure. It means that heat capacity is the rate of change of enthalpy with respect to temperature at constant pressure.

2. Explain the concept of entropy in Class 11 Thermodynamics.

A system's degree of disorder or unpredictability is measured by its entropy.  It helps determine the direction of spontaneous processes. The greater is the entropy of a system, higher will be the randomness.  

3. What is the Gibbs free energy equation, and how is it used?

Gibbs Free Energy Equation is $G=H-T S$; it's used to predict spontaneity. If $\Delta G<0$ then the process is spontaneous. The value of G, depends on the enthaply and entropy of the system.

4. How do we calculate work done in isothermal and adiabatic processes?

Work in Isothermal Process is $W=n R T \ln \frac{V_f}{V_i}$; in Adiabatic Process, $W=\frac{P_i V_i-P_f V_f}{\gamma-1}$, where no heat exchange occurs.

5. What is the significance of the Hess’s Law of Constant Heat Summation?

According to Hess's Law, the overall enthalpy change of a reaction remains constant regardless of the number of steps involved. It allows us to compute unknown enthalpy changes using known reactions.

6. What is the first law of thermodynamics in NCERT Class 11 Chemistry?

The first law of thermodynamics states that energy can neither be created nor be destroyed it can only transfer from one form to another form.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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