NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

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• NCERT Exemplar Solutions Class 11 Chemistry Thermodynamics
• NCERT Notes Class 11 Chemistry Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 5 (Exercise Questions with Answers)

Get detailed and accurate class 11 chemistry thermodynamics question answer covering all exercise questions with step by step answers. These solutions help you understand important concepts like enthalpy, internal energy, entropy, and Gibbs free energy.

Class 11 Chemistry NCERT Chapter 5: Higher Order Thinking Skills (HOTS)

Enhance your conceptual understanding with HOTS questions from thermodynamics ncert solutions. These questions help you think deeper and prepare better for CBSE exams and entrance tests like NEET and JEE.

Question 1. 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

Answer:

$
\begin{aligned}
& \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 correct answer is option (4).

Question 2. The correct statement amongst the following is :

(1) The term 'standard state' implies that the temperature is $0^{\circ} \mathrm{C}$

(2) The standard state of pure gas is the pure gas at a pressure of 1 bar and temperature 273 K

(3) $\boldsymbol{\Delta}_{\mathrm{f}} \mathrm{H}_{298}^{\boldsymbol{\theta}}$ is zero for $\mathrm{O}(\mathrm{g})$

(4) $\boldsymbol{\Delta}_{\mathrm{f}} \mathrm{H}_{298}^{\boldsymbol{\theta}}$ is zero for $\mathrm{O}_2(\mathrm{~g})$

Answer:

Option (1)

It asserts that the term "standard state" implies that the temperature is $0^{\circ} \mathrm{C}(273 \mathrm{~K})$.
In fact, "standard state" typically specifies a standard pressure (1 bar) and a chosen temperature-commonly 298 K for tabulated thermodynamic data-not necessarily $0^{\circ} \mathrm{C}$.

Thus, Option 1 is incorrect.

Option (2)

It claims that the standard state for a pure gas is defined at 1 bar and 273 K .
In modern thermochemistry, the standard state generally means the pure substance at 1 bar, but the standard temperature most commonly used is $298 \mathrm{~K}\left(25^{\circ} \mathrm{C}\right)$, not 273 K $\left(0^{\circ} \mathrm{C}\right)$.

Thus, Option 2 is incorrect.

Option (3)

It claims that $\Delta_f H_{298}^{\circ}$ is zero for $O(g)$ (atomic oxygen).
However, the standard state of oxygen is diatomic $O_2(g)$, not atomic oxygen. Since $O(g)$ is not the most stable form of oxygen under standard conditions, its formation enthalpy is not zero.

Hence, Option 3 is incorrect.

Option (4)

$\Delta_{\mathrm{f}} \mathrm{H}_{298}^0$ for $\mathrm{O}_2$ is zero and not for O.

The standard enthalpy of formation of any element in its most stable form is zero, regardless of temperature (though typically reported at 298 K ).

Thus, option (4) is correct

Hence, the correct answer is option (4).

Question 3. The hydration energies of $\mathrm{K}^{+}$and $\mathrm{Cl}^{-}$are -x and -y $\mathrm{kJ} / \mathrm{mol}$ respectively. If lattice energy of KCl is -z $\mathrm{kJ} / \mathrm{mol}$, then the heat of solution of KCl is :

(1) $+x-y-z$

(2) $x+y+z$

(3) $z-(x+y)$

(4) $-z-(x+y)$

Answer:

$\begin{aligned} & \Delta \mathrm{H}_{\text {solution }}=\left(-\Delta \mathrm{H}_{\text {L.E. }}\right)+\Delta \mathrm{H}_{\text {hydration }} \\ & =-(-z)+(-\mathrm{x}-\mathrm{y}) \\ & =\mathrm{z}-(\mathrm{x}+\mathrm{y})\end{aligned}$

Hence, the correct answer is option (3).

Question 4. Let us consider a reversible reaction at temperature, T .
In this reaction, both $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ were observed to have positive values. If the equilibrium temperature is Te , then the reaction becomes spontaneous at :

(1) $\mathrm{T}=\mathrm{Te}$

(2) $\mathrm{Te}>\mathrm{T}$

(3) $\mathrm{T}>\mathrm{Te}$

(4) $\mathrm{Te}=5 \mathrm{~T}$

Answer:

For reaction to be spontaneous according to $2^{\text {nd }}$ law:

$\begin{aligned}
& \Delta \mathrm{G}<0 \\
& \Rightarrow \Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{~S}<0 \\
& \Rightarrow \mathrm{~T}>\left(\frac{\Delta \mathrm{H}}{\Delta \mathrm{~S}}\right)=\mathrm{T}_{\mathrm{e}} \\
& \Rightarrow \mathrm{~T}>\mathrm{T}_{\mathrm{e}}
\end{aligned}$

Hence, the correct answer is option (3).

Question 5. 500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm . The final temperature and the change in internal energy respectively are :

Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

(1) 348 K and 300 J

(2) 378 K and 300 J

(3) 368 K and 500 J

(4) 378 K and 500 J

Answer:

Heat transferred:

\( q = 500 \) J

Moles of Argon: \( n = 0.5 \) mol

Initial temperature: \( T_i = 298 \) K

Argon is a monoatomic gas, so its molar heat capacity at constant volume:

\[ C_V = \frac{3}{2} R = \frac{3}{2} \times 8.314 = 12.471 \text{ J/mol·K} \]

Change in Internal Energy (\(\Delta U\))

Since no work is done at constant volume,

\[ \Delta U = q = 500 \text{ J} \]

Final Temperature (\(T_f\))

Using the formula:

\[ \Delta U = n C_V (T_f - T_i) \] \[ 500 = (0.5) \times (12.471) \times (T_f - 298) \] \[ T_f - 298 = \frac{500}{6.2355} \approx 80.2 \] \[ T_f = 298 + 80.2 = 378.2 \approx 378 \text{ K} \]

Final Temperature = 378 K

Change in Internal Energy = 500 J

Hence, the correct answer is option (4).

Question 6: Consider the given data :
(a) $\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}$

$\Delta \mathrm{H}=69.01 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(b) $\mathrm{HCl}(\mathrm{g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}$

$\Delta \mathrm{H}=-72.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Choose the correct statement :

(1) Dissolution of gas in water is an endothermic process

(2) The heat of solution depends on the amount of solvent.

(3) The heat of dilution for the $\mathrm{HCl}\left(\mathrm{HCl} \cdot 10 \mathrm{H}_2 \mathrm{O}\right.$ to $\mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}$ ) is $3.78 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

(4) The heat of formation of HCl solution is represented by both (a) and (b)

Answer:

From the given information
$\Delta \mathrm{H}$ is negative so it means dissolution of gas $\mathrm{HCl}(\mathrm{g})$ is exothermic.

$\mathrm{HCl}(\mathrm{~g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} \cdot 10 \mathrm{H}_2 \mathrm{O}$ ....(1)

$\begin{aligned}
& \Delta \mathrm{H}_1=-69.01 \frac{\mathrm{~kJ}}{\mathrm{~mol}} \\
& \mathrm{HCl}(\mathrm{~g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl} \cdot 40 \mathrm{H}_2 \mathrm{O}
\end{aligned}$ .......(2)

$\Delta \mathrm{H}_2=-72.79 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$

Hence heat of solution depends upon amount of solvent
By equation......... (2) - equation .....(1)

$\mathrm{HCl} .10 \mathrm{H}_2 \mathrm{O}+30 \mathrm{H}_2 \mathrm{O}(\ell) \rightarrow \mathrm{HCl} .40 \mathrm{H}_2 \mathrm{O}$

So Heat of dilution $=-72.79-(-69.01)$

$=-3.78 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$
Hence option (3) is incorrect.
For heat of formation reactant should be in elemental form hence option (4) is incorrect

Hence, the correct answer is option (2).

Question 7: 1 mole of $\mathrm{O}_2$ gas is adiabatically and irreversibly expanded in one step from 27 atm & 300 K to 3 atm . Calculate the final temperature of the system after the expansion -
(to the nearest integer).

(1) 224K

(2) 390K

(3) 412K

(4) 500K

Answer:

(1) diabatic psocess $(q=0)$ and Irreversible expansion.
(2) $\mathrm{O}_2$ is diatomic gas.
(3) Molar heat capacity $\left(C_v\right)=\frac{5}{2} R$

$
P_1=27 \mathrm{~atm} \quad T_1=300 \mathrm{~K}
$
$
P_{\text {ext }}=3 a t i n \quad P_2(\text { final pressure })=3 a t i n
$

(4) Apply FLOT:- $\Delta U=q+\omega$.

$
\begin{aligned}
& q=0 \\
& \therefore \quad \Delta U=w .
\end{aligned}
$
Change in Internal evergy $\Delta U=n C_V\left(T_2-T_1\right)$

$
\begin{aligned}
& \omega=-\operatorname{Pext}\left(V_2-V_1\right) \\
& n C_V\left(T_2-T_1\right)=-\operatorname{Pext}\left(V_2-V_1\right) \\
& n C_V\left(T_2-T_1\right)=-\operatorname{Pext}\left(\frac{n R T_2}{2}-\frac{n R T_1}{P_1}\right)
\end{aligned}
$

$
\begin{aligned}
& C_v=2.5 R \\
& 2.5\left(T_2-300\right)=-3\left(\frac{T_2}{3}-\frac{300}{27}\right)
\end{aligned}
$

simplify it.

$
T_g=223.8 \mathrm{~K} \simeq 224 \mathrm{~K}
$

Hence, the correct answer is option (1).

Question 8: Consider the following cases:
(A) $\mathrm{O}+e^{-} \longrightarrow \mathrm{O}^{-}$
(B) $\mathrm{O}^{-}+e^{-} \longrightarrow \mathrm{O}^{2-}$
(C) $\mathrm{F}+\mathrm{e}^{-} \longrightarrow \mathrm{F}^{-}$
(D) $\mathrm{He}+e^{-} \longrightarrow \mathrm{He}^{-}$
(E) $\mathrm{Na} \longrightarrow \mathrm{Na}^{+}+e^{-}$

In which of the above mentioned cases, $\triangle H$ for the process is negative

(1) A,B,C,D

(2) A,C,D

(3) A,C

(4) C, D, E

Answer:

Only A and C cases are exothermic, that's why $\triangle H$ is negative, and the rest of the cases are endothermic.

Hence, the correct answer is option (3).

Question 9: 500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm . The final temperature and the change in internal energy respectively are :

Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

(1) 348 K and 300 J

(2) 378 K and 300 J

(3) 368 K and 500 J

(4) 378 K and 500 J

Answer:

Heat transferred:

\( q = 500 \) J

Moles of Argon: \( n = 0.5 \) mol

Initial temperature: \( T_i = 298 \) K

Argon is a monoatomic gas, so its molar heat capacity at constant volume:

\[ C_V = \frac{3}{2} R = \frac{3}{2} \times 8.314 = 12.471 \text{ J/mol·K} \]

Change in Internal Energy (\(\Delta U\))

Since no work is done at constant volume,

\[ \Delta U = q = 500 \text{ J} \]

Final Temperature (\(T_f\))

Using the formula:

\[ \Delta U = n C_V (T_f - T_i) \] \[ 500 = (0.5) \times (12.471) \times (T_f - 298) \] \[ T_f - 298 = \frac{500}{6.2355} \approx 80.2 \] \[ T_f = 298 + 80.2 = 378.2 \approx 378 \text{ K} \]

Final Temperature = 378 K

Change in Internal Energy = 500 J

Hence, the correct answer is option (4).

Approach to Solve Questions of Class 11 Chemistry Chapter 5

The basic understanding to solve class 11 chemistry chapter 5 thermodynamics solutions is essential for building a strong foundation in physical chemistry. The approaches given below help students in developing the right techniques to solve questions.

1. Learning about the basic concepts like system, surrounding, internal energy, work, heat, enthalpy and Laws of Thermodynamics is very important for solving questions. These topics also help in understanding energy changes in chemical reactions and predicting the direction of physical and chemical change.

2). Before solving any question, first give it a thorough reading and note down all the information given in the question. Identify given data like pressure, volume, temperature, etc. and what we need to find.

3). Using the correct formula is very important for solving class 11 chemistry chapter 5 thermodynamics question answer

(i). Internal Energy Change

$\Delta U=U_{\text {final }}-U_{\text {initial }}$

(ii). First Law of Thermodynamics

$\Delta U=q+W$

(iii). Work Done in Expansion/Compression

$W=-P \Delta V$

(iv). Enthalpy and Enthalpy Change

$H=U+P V$

$\Delta H=\Delta U+P \Delta V$

(v). Standard Enthalpy Change of Reaction

$\Delta H_{\text {reaction }}^{\circ}=\sum \Delta H_{\text {products }}^{\circ}-\sum \Delta H_{\text {reactants }}^{\circ}$

(vi). Gibbs Free Energy

$\Delta G=\Delta H-T \Delta S$

4). Be careful while dealing with units and convert them if necessary to ensure consistency like temperature in Kelvin or volume in litres, etc.

5). Practice again and again, as it will help in mastering thermodynamics numerically.

• Start with simpler problems
• Then gradually move to ones involving concepts like enthalpy change, work in isothermal or adiabatic processes, and heat capacity.

6). Students must practice thermodynamics class 11 question answer from the NCERT books are asked directly in the boards and other competitive exams. Do previous year questions from NEET and JEE to get used to question patterns.

Topics and Subtopics Covered in the NCERT Textbook

All the topics and subtopics discussed in class 11 chemistry chapter 5 thermodynamics solutions are listed below. These solutions of NCERT provide a clear and stepwise explanation of each concept, helping students understand the laws of thermodynamics and their applications.

5.1 Thermodynamics Terms

5.1.1 The System and the Surroundings

5.1.2 Types of the System

5.1.3 The State of the System

5.1.4 The Internal Energy as a State Function

5.2 Applications

5.2.1 Work

5.2.2 Enthalpy

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 the NCERT for JEE/NEET?

Beyond the NCERT, students should focus on important concepts, formulas, different types of processes, and laws of thermodynamics according to the table given below. This additional practice with thermodynamics class 11 question answer will enhance their analytical thinking and prepare them for competitive examinations as well.

What Students Learn from NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

These chapter 5 thermodynamics solutions help students to develop a strong foundation in thermodynamics. These solutions explain stepwise explanations and solved examples, making it easier for students to understand concepts

• Using these solutions basic concepts related to energy, work, and heat can be understood more easily
• Laws of thermodynamics and their applications in physical and chemical processes are explained well in these thermodynamics ncert solutions.
• Here students will learn about different types of thermodynamic systems and processes
• Concepts like enthalpy, internal energy, entropy, and Gibbs free energy are explained well using these class 11 chemistry chapter 5 thermodynamics solutions.

Importance of Class 11 Chemistry Chapter 5 Thermodynamics Solutions

The thermodynamics class 11 question answer explains concepts that helps students understand energy changes in physical and chemical process

• These solutions explain key concepts like system, surroundings, work, heat, internal energy, enthalpy, and laws of thermodynamics.
• Using these solutions students can learn how energy is conserved and transferred during chemical reactions.
• The class 11 chemistry thermodynamics question answer explains principles of thermodynamics.

NCERT Solutions for Class 11 Chemistry

Besides the NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics, students can also access the complete set of Class 11 NCERT chapter wise solutions provided below for thorough preparation.

NCERT Solutions for Class 11 Subject-Wise

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