Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

NCERT Solutions for Class 11 Chemistry Chapter- 5 Thermodynamics

ffImage

Thermodynamics Class 11 Chemistry Chapter 5 NCERT Solutions FREE PDF Download

Thermodynamics Class 11 NCERT solutions help to Uncover the principles governing energy transformations, from heat and work to the spontaneity of chemical reactions. Explore concepts like entropy and enthalpy, the balance of energy in the universe. With clear explanations and comprehensive solutions, Thermodynamics NCERT solutions equip you to conquer the complexities of thermodynamics. Access your FREE PDF download today and embark on a thrilling journey through the dynamic world of energy and its transformations.

toc-symbol
Table of Content
1. Thermodynamics Class 11 Chemistry Chapter 5 NCERT Solutions FREE PDF Download
2. Quick Insights of Thermodynamics NCERT Solutions
3. Access Class 11 Chemistry Chapter 5- Thermodynamics NCERT Solutions
4. Class 11 Chemistry Chapter 5 Quick Overview of Topics
5. Some Important Concepts and Formulas 
6. Important Points for Class 11 NCERT Solutions Chapter 5 Thermodynamics
7. Benefits of Referring to Vedantu’s NCERT Solutions for Class 11 Chemistry Chapter 5
8. Related Study Material Links for Chemistry Class 11 Chapter 5 NCERT Solutions
9. NCERT Solutions for Class 11 Chemistry - Chapter-wise Links
10. NCERT Solutions Class 11 Chemistry - Related Links
FAQs


Check out the Revised class 11 chemistry syllabus along with the Thermodynamics Class 11 Chemistry Chapter 5 NCERT Solutions and get started with Vedantu to embark on a journey of academic excellence!


Quick Insights of Thermodynamics NCERT Solutions

  • NCERT Solutions for Class 11 Chemistry Chapter 5 will give you insights into the Concepts of Systems and types of systems, surroundings, work, heat, energy, extensive and intensive properties, and state functions. 

  • Thermodynamics class 11 NCERT solutions section will give you crisp learnings on the First law of thermodynamics -internal energy and enthalpy, heat capacity and specific heat, measurement of U and H, Hess's law of constant heat summation, enthalpy of bond dissociation, combustion, formation, atomisation, sublimation, phase transition, ionisation, solution, and dilution.

  • The understanding related to topics like the Second Law of Thermodynamics Introduction of entropy as a state function, Gibb's energy change for spontaneous and nonspontaneous processes, criteria for equilibrium, Third Law of thermodynamics.

  • Using these class 11 chemistry chapter 5 NCERT solutions can help students analyse their level of preparation and understanding of concepts.

  • Class 11 Chemistry Chapter 5 NCERT solutions topics are included according to the revised academic year 2024-25 syllabus.

  • Class 11 chemistry chapter Thermodynamics NCERT solutions provide resources such as class notes, important concepts, and formulas exemplar solutions.

Competitive Exams after 12th Science
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow

Access Class 11 Chemistry Chapter 5- Thermodynamics NCERT Solutions

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. 

Ans: A thermodynamic state function is a quantity whose value is independent of the path. Functions like p, V, T etc. depend only on the state of a system and not on the path. 

Hence, alternative (ii) is correct. 


2. For the Process to Occur Under Adiabatic Conditions, the Correct Condition Is: 

(i) $\text{ }\!\!\Delta\!\!\text{ T=0}$ 

(ii) $\text{ }\!\!\Delta\!\!\text{ p=0}$ 

(iii) $\text{q=0}$ 

(iv) $\text{w=0}$ 

Ans: A system is said to be under adiabatic conditions if there is zero exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, $q=0$. Therefore, alternative (iii) is correct. 


3. The Enthalpies of All Elements in Their Standard States Are: 

(i) unity 

(ii) zero 

(iii) < 0 

(iv) different for each element 

Ans: The enthalpy of all elements in their standard state is zero. Therefore, alternative (ii) is correct.


4. $\text{ }\!\!\Delta\!\!\text{ }{{\text{U}}^{\text{ }\!\!\theta\!\!\text{ }}}$ of combustion of methane is $\text{-X}\,\text{kJ mo}{{\text{l}}^{\text{-1}}}$. The value of $\text{ }\!\!\Delta\!\!\text{ }{{\text{H}}^{\theta }}$is 

(i) = $\text{ }\!\!\Delta\!\!\text{ }{{\text{U}}^{\text{ }\!\!\theta\!\!\text{ }}}$

(ii) > $\text{ }\!\!\Delta\!\!\text{ }{{\text{U}}^{\text{ }\!\!\theta\!\!\text{ }}}$ 

(iii) < $\text{ }\!\!\Delta\!\!\text{ }{{\text{U}}^{\text{ }\!\!\theta\!\!\text{ }}}$ 

(iv) = 0 

Ans: Since 

\[\Delta {{H}^{\theta }}\text{ }=\text{ }\Delta {{U}^{\theta }}\text{ }+\text{ }\Delta {{n}_{g}}RT\] and\[\Delta {{U}^{\theta }}=-X\,kJ\text{ }mo{{l}^{-1}}\], 

\[\Delta {{H}^{\theta }}\text{ }=\text{ }(-X)\text{ }+\text{ }\Delta {{n}_{g}}RT\]

\[\Rightarrow \Delta {{H}^{\theta }}<\Delta {{U}^{\theta }}\]

Therefore, alternative (iii) is correct.


5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 $\text{kJ mo}{{\text{l}}^{\text{-1}}}$ ,-393.5 $\text{kJ mo}{{\text{l}}^{\text{-1}}}$, and -285.8 $\text{kJ mo}{{\text{l}}^{\text{-1}}}$ respectively. Enthalpy of formation of $\text{C}{{\text{H}}_{\text{4}}}\text{(g)}$ will be 

(i) -74.8 $\text{kJ mo}{{\text{l}}^{\text{-1}}}$

(ii) -52.27 $\text{kJ mo}{{\text{l}}^{\text{-1}}}$  

(iii) +74.8 $\text{kJ mo}{{\text{l}}^{\text{-1}}}$

(iv) +52.26 $\text{kJ mo}{{\text{l}}^{\text{-1}}}$  . 

Ans: According to the question, 

(i) $C{{H}_{4}}(g)+2{{O}_{2}}(g)\to C{{O}_{2}}(g)+2{{H}_{2}}O(l);{{\Delta }_{c}}{{H}^{\Theta }}=-890.3\,kJ\,mo{{l}^{-1}}$ 

(ii) $C(s)+2{{O}_{2}}(g)\to C{{O}_{2}}(g);{{\Delta }_{c}}{{H}^{\Theta }}=-393.5\,kJ\,mo{{l}^{-1}}$ 

(iii) $2{{H}_{2}}(g)+{{O}_{2}}(g)\to 2{{H}_{2}}O(l);{{\Delta }_{c}}{{H}^{\Theta }}=-285.8\,kJ\,mo{{l}^{-1}}$ 

Thus, the desired equation is the one that represents the formation of $C{{H}_{4}}(g)$that is as follows:

\[C(s)+2{{H}_{2}}(g)\to C{{H}_{4}}(g);{{\Delta }_{f}}{{H}_{C{{H}_{4}}}}={{\Delta }_{c}}{{H}_{c}}+2{{\Delta }_{c}}{{H}_{{{H}_{2}}}}-{{\Delta }_{c}}{{H}_{C{{O}_{2}}}}\]

Substituting the values in the above formula :

Enthalpy of formation$C{{H}_{4}}(g)$=$(-393.5)+2\times (-285.8)-(-890.3)=-74.8kJmo{{l}^{-1}}$ 

Therefore, alternative (i) is correct.


6. A Reaction, $\text{A + B }\to \text{C + D + q}$ 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 

(iv) possible at any temperature 

Ans: For a reaction to be spontaneous, $\Delta G$ should be negative $\Delta G\text{ }=\text{ }\Delta H\text{ }-\text{ }T\Delta S$ 

According to the question, for the given reaction, 

$\Delta S\text{ }=$ positive 

$\Delta H=$ negative (since heat is evolved) 

That results in $\Delta G=$ negative

Therefore, the reaction is spontaneous at any temperature. 

Hence, alternative (iv) is correct. 


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? 

Ans: According to the first law of thermodynamics, 

\[\Delta U=\text{ }q\text{ }+\text{ }W....(i)\] 

Where, 

$\Delta U$ = change in internal energy for a process 

q = heat 

W = work 

Given, 

q = + 701 J (Since heat is absorbed) 

W = -394 J (Since work is done by the system)

Substituting the values in expression (i), we get 

\[\Delta U=\text{ }701\text{ }J\text{ }+\text{ }\left( -394\text{ }J \right)\] 

\[\Delta U=\text{ }307\text{ }J\] 

Hence, the change in internal energy for the given process is 307 J.


8.The reaction of cyanamide, $\mathbf{N}{{\mathbf{H}}_{2}}\mathbf{CN}\left( \mathbf{s} \right)$ with dioxygen was carried out in a bomb calorimeter and $\text{ }\!\!\Delta\!\!\text{ U}$ was found to be -742.7 $\mathbf{KJ}\text{ }\mathbf{mo}{{\mathbf{l}}^{-1}}$ at 298 K. Calculate the enthalpy change for the reaction at 298 K. 

\[\text{N}{{\text{H}}_{\text{4}}}\text{C}{{\text{N}}_{\text{(g)}}}\text{+}\frac{\text{3}}{\text{2}}{{\text{O}}_{\text{2}}}_{\text{(g)}}\to {{\text{N}}_{\text{2(g)}}}\text{+C}{{\text{O}}_{\text{2(g)}}}\text{+}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{(l)}}}\] 

Ans: Enthalpy change for a reaction $\left( \Delta H \right)$ is given by the expression, 

\[\Delta H\text{ }=\text{ }\Delta U\text{ }+\text{ }\Delta {{n}_{g}}RT\] 

Where, 

$\Delta U$ = change in internal energy 

$\Delta {{n}_{g}}$ = change in number of moles 

For the given reaction, 

$\Delta {{n}_{g}}=\sum{{{n}_{g}}}$ (products) - $\sum{{{n}_{g}}}$(reactants)

$\Delta {{n}_{g}}=(2-1.5)$moles

$\Delta {{n}_{g}}=+0.5$moles

And, $\Delta U$= -742.7 $kJ\text{ }mo{{l}^{-1}}$ 

T = 298 K 

R = $8.314\times {{10}^{-3}}\,kJ\,mo{{l}^{-1}}{{K}^{-1}}$

Substituting the values in the expression of $\Delta H\text{ }$

\[\Delta H\text{ }=\text{ }\left( -742.7\text{ }kJ\text{ }mo{{l}^{-1}} \right)\text{ }+\text{ }\left( +0.5\text{ }mol \right)\text{ }\left( 298\text{ }K \right)8.314\times {{10}^{-3}}kJmo{{l}^{-1}}{{K}^{-1}}\] 

$\Delta H\text{ }$ = -742.7 + 1.2 

\[\Delta H\text{ }=\text{ }-741.5kJ\text{ }mo{{l}^{-1}}\] 


9. Calculate the number of kJ of heat necessary to raise the temperature of 60 g of aluminium from $\mathbf{35}{}^\circ C$ to $\text{55 }\!\!{}^\circ\!\!\text{ C}$. Molar heat capacity of Al is $\text{24J mo}{{\text{l}}^{\text{-1}}}{{\text{K}}^{\text{-1}}}$. 

Ans: From the expression of heat (q), 

\[q\text{ }=\text{ }m.\text{ }c.\text{ }\Delta T\] 

Where, 

c = molar heat capacity 

m = mass of substance 

$\Delta T$ = change in temperature 

Given, 

m = 60 g

c = $24J\text{ }mo{{l}^{-1}}{{K}^{-1}}$

\[Delta T=\left( 55-35 \right){}^\circ C\]

\[\Delta T=\left( 328-308 \right)K=20K\]

Substituting the values in the expression of heat:

\[q=\left( \frac{60}{27}mol \right)\left( 24\,Jmo{{l}^{-1}}{{K}^{-1}} \right)\left( 20K \right)\] 

q = 1066.7 J

q = 1.07 kJ


10. Calculate the enthalpy change on freezing of 1.0 mol of water at $\text{10}\text{.0 }\!\!{}^\circ\!\!\text{ C}$  to ice at $\text{-10}\text{.0 }\!\!{}^\circ\!\!\text{ C}$, ${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{fus}}}\text{H = 6}\text{.03 KJ mo}{{\text{l}}^{\text{-1}}}$  at $\text{0 }\!\!{}^\circ\!\!\text{ C}$.

\[\text{ }\!\!~\!\!\text{ }{{\text{C}}_{\text{p}}}\text{ }\left[ {{\text{H}}_{\text{2}}}\text{O}\left( \text{l} \right)\text{ } \right]\text{ = 75}\text{.3 J mo}{{\text{l}}^{\text{-1}}}\text{ }{{\text{K}}^{\text{-1}}}\]  

\[\text{ }\!\!~\!\!\text{ }{{\text{C}}_{\text{p}}}\text{ }\left[ {{\text{H}}_{\text{2}}}\text{O}\left( \text{s} \right)\text{ } \right]\text{ = 36}\text{.8 J mo}{{\text{l}}^{\text{-1}}}\text{ }{{\text{K}}^{\text{-1}}}\] .

Ans: Total enthalpy change involved in the transformation is the sum of the following changes: 

  1. Energy change involved in the transformation of 1 mol of water at $10.0{}^\circ C$  to 1mol of water at$0{}^\circ C$.

  2. Energy change involved in the transformation of 1 mol of water at $0{}^\circ C$to 1 mol of ice at $0{}^\circ C$. 

  3. Energy change involved in the transformation of 1 mol of ice at $0{}^\circ C$ to 1 mol of ice at $10{}^\circ C$. 

Total $\Delta H=~{{C}_{p}}\text{ }\left[ {{H}_{2}}O\left( l \right) \right]\text{ }\Delta T+\Delta {{H}_{freezing}}+~{{C}_{p}}\text{ }\left[ {{H}_{2}}O\left( s \right) \right]\text{ }\Delta T$ 

\[\Delta H=\left( 75.3\text{ }Jmo{{l}^{-1}}{{K}^{-1}} \right)\left( 0-10 \right)K+\left( -6.03\times {{10}^{3}}Jmo{{l}^{-1}} \right)+\left( 36.8\text{ }Jmo{{l}^{-1}}{{K}^{-1}} \right)\left( -10-0 \right)K\]

\[\Delta H=\text{ }-753\text{ }J\text{ }mo{{l}^{-1}}\text{ }-\text{ }6030\text{ }J\text{ }mo{{l}^{-1}}\text{ }-\text{ }368\text{ }J\text{ }mo{{l}^{-1}}\] 

\[\Delta H=\text{ }-7151\text{ }J\text{ }mo{{l}^{-1}}\] 

\[\Delta H=\text{ }-7.151\text{ }kJ\text{ }mo{{l}^{-1}}\] 

Hence, the enthalpy change involved in the transformation is $\text{ }-7.151\text{ }kJ\text{ }mo{{l}^{-1}}$ 


11. Enthalpy of combustion of carbon to carbon dioxide is $-\mathbf{393}.\mathbf{5}\text{ }\mathbf{kJ}\text{ }\mathbf{mo}{{\mathbf{l}}^{-1}}$ Calculate the heat released upon formation of 35.2 g of $\text{C}{{\text{O}}_{\text{2}}}$ from carbon and dioxygen gas. 

Ans: Formation of $C{{O}_{2}}$ from carbon and dioxygen gas can be represented as 

\[{{C}_{(s)}}+{{O}_{2(g)}}\ to C{{O}_{2(g)}}; \Delta H=-393.5kJ,mo{{l}^{-1}}\] (1mole=44g) 

Heat released in the formation of 44 g of $C{{O}_{2}}$ = 393.5 $kJ mol{{l}^{-1}}$ 

Heat released in the formation of 35.2 g of 

$C{{O}_{2}}=(393.5kJ)\times \frac{(35.2g)}{(44g)}=314.8\,kJ$

So, heat released upon formation of 35.2 g of $\text{C}{{\text{O}}_{\text{2}}}$from carbon and dioxygen gas is 314.8 kJ.


12. Enthalpies of formation of CO (g),$\text{C}{{\text{O}}_{\text{2}}}\text{(g)}$, ${{\text{N}}_{\text{2}}}\text{O(g)}$ and ${{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{(g)}$are -110 ,-393, 81 kJ and 9.7 $\mathbf{kJ}\text{ }\mathbf{mo}{{\mathbf{l}}^{\text{-1}}}$ respectively. Find the value of${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}\text{H}$for the reaction: 

\[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4(g)}}}\text{+3C}{{\text{O}}_{\text{(g)}}}\to {{\text{N}}_{\text{2}}}{{\text{O}}_{\text{(g)}}}\text{+3C}{{\text{O}}_{\text{2(g)}}}\] 

Ans: ${{\Delta }_{r}}H$ for a reaction is defined as the difference between ${{\Delta }_{f}}H$value of products and ${{\Delta }_{f}}H$value of reactants. 

${{\Delta }_{r}}H$= $\sum{{{\Delta }_{f}}H}$ (product) - $\sum{{{\Delta }_{f}}H}$(reactant)

For the given reaction, 

\[{{N}_{2}}{{O}_{4(g)}}+3C{{O}_{(g)}}\to {{N}_{2}}{{O}_{(g)}}+3C{{O}_{2(g)}}\] 

${{\Delta }_{r}}H=\left[ \left\{ {{\Delta }_{f}}H(N{{O}_{2}})+3{{\Delta }_{f}}H(C{{O}_{2}}) \right\}-\left\{ {{\Delta }_{f}}H({{N}_{2}}O)+3{{\Delta }_{f}}H(CO) \right\} \right]$ 

Substituting the values of ${{\Delta }_{f}}H$for CO (g),$C{{O}_{2}}(g)$, ${{N}_{2}}O(g)$ and ${{N}_{2}}{{O}_{4}}(g)$from the question, we get: 

\[{{\Delta }_{r}}H=\left[ \left\{ 81kJmo{{l}^{-1}}+3(-393)kJmo{{l}^{-1}} \right\}-\left\{ 9.7kJmo{{l}^{-1}}+3(-110)kJmo{{l}^{-1}} \right\} \right]\] 

\[{{\Delta }_{r}}H=-777.7kJ\,mo{{l}^{-1}}\] 

Hence, the value of ${{\Delta }_{r}}H$ for the reaction is $-777.7kJ\,mo{{l}^{-1}}$

 

13. Given 

\[{{\text{N}}_{\text{2(g)}}}\text{+3}{{\text{H}}_{\text{2(g)}}}\to \text{2N}{{\text{H}}_{\text{3(g)}}}\text{;}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{=}\,\text{-92}\text{.4kJ}\,\text{mo}{{\text{l}}^{\text{-1}}}\] 

What is the standard enthalpy of formation of $\text{N}{{\text{H}}_{\text{3}}}$ gas?

Ans: Standard enthalpy of formation of a compound is the change in 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. 

Re-writing the given equation for 1 mole of $N{{H}_{3}}(g)$is as follows:

\[\frac{1}{2}{{N}_{2(g)}}+\frac{3}{2}{{H}_{2(g)}}\to 2N{{H}_{3(g)}}\] 

Therefore, standard enthalpy of formation of $N{{H}_{3}}(g)$

= ${}^{1}/{}_{2}{{\Delta }_{r}}{{H}^{\theta }}$ 

\[{}^{1}/{}_{2}\text{ }\left( -92.4\text{ }kJ\text{ }mo{{l}^{-1}} \right)\] 

 \[-46.2\text{ }kJ\text{ }mo{{l}^{-1}}\] 


14. Calculate the standard enthalpy of formation of $\text{C}{{\text{H}}_{\text{3}}}\text{OH(}\ell \text{)}$ from the following data: 

\[\text{C}{{\text{H}}_{\text{3}}}\text{O}{{\text{H}}_{\text{(l)}}}\text{+}\frac{\text{3}}{\text{2}}{{\text{O}}_{\text{2(g)}}}\to \text{C}{{\text{O}}_{\text{2(g)}}}\text{+2}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{(l)}}}\text{,}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{=-726 }\!\!~\!\!\text{ kJ }\!\!~\!\!\text{ mo}{{\text{l}}^{\text{-1}}}\] 

\[{{\text{C}}_{\text{(g)}}}\text{+}{{\text{O}}_{\text{2(g)}}}\to \text{C}{{\text{O}}_{\text{2(g)}}}\text{;}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{c}}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{=-393 }\!\!~\!\!\text{ kJ }\!\!~\!\!\text{ mo}{{\text{l}}^{\text{-1}}}\] 

\[{{\text{H}}_{\text{2(g)}}}\text{+}\frac{\text{1}}{\text{2}}{{\text{O}}_{\text{2(g)}}}\to {{\text{H}}_{\text{2}}}{{\text{O}}_{\text{(l)}}}\text{;}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{=-286kJ }\!\!~\!\!\text{ mo}{{\text{l}}^{\text{-1}}}\] 

Ans: The reaction that takes place during the formation of $C{{H}_{3}}OH(\ell )$can be written as: 

\[C(s)+2{{H}_{2}}O(g)+\frac{1}{2}{{O}_{2}}(g)\to C{{H}_{3}}OH(\ell )\,\,\,\,\,(1)\] 

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as: 

Equation (ii) + 2 $\times $ equation (iii) - equation (i) 

\[{{\Delta }_{f}}{{H}^{\theta }}\left[ C{{H}_{3}}OH(\ell ) \right]={{\Delta }_{c}}{{H}^{\theta }}+2{{\Delta }_{f}}{{H}^{\theta }}[{{H}_{2}}O(l)]-{{\Delta }_{r}}{{H}^{\theta }}\] = (-393$kJ\,mo{{l}^{-1}}$ ) + 2(-286 $kJ\,mo{{l}^{-1}}$) - (-726 $kJ\,mo{{l}^{-1}}$) 

= (-393 - 572 + 726) $kJ\,mo{{l}^{-1}}$

Therefore, ${{\Delta }_{f}}{{H}^{\theta }}\left[ C{{H}_{3}}OH(\ell ) \right]=-239~kJmo{{l}^{-1}}$ 

 

15. Calculate the Enthalpy Change for the Process 

$\text{CC}{{\text{l}}_{\text{4}}}\left( \text{g} \right)\text{ }\to \text{ C}\left( \text{g} \right)\text{ + 4Cl}\left( \text{g} \right)$

and calculate bond enthalpy of C-Cl in $\text{CC}{{\text{l}}_{\text{4}}}\left( \text{g} \right)$.

\[{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{vap }}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{(CC}{{\text{l}}_{\text{4}}}\text{)=30}\text{.5 }\!\!~\!\!\text{ kJmo}{{\text{l}}^{\text{-1}}}\]

\[{{\text{ }\!\!\Delta\!\!\text{ }}_{f}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{(CC}{{\text{l}}_{\text{4}}}\text{)=-135}\text{.5 }\!\!~\!\!\text{ kJmo}{{\text{l}}^{\text{-1}}}\] 

\[{{\text{ }\!\!\Delta\!\!\text{ }}_{a}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{(C)=715}\text{.0 }\!\!~\!\!\text{ kJmo}{{\text{l}}^{\text{-1}}}\] where, ${{\text{ }\!\!\Delta\!\!\text{ }}_{a}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}$ is enthalpy of atomization

${{\text{ }\!\!\Delta\!\!\text{ }}_{a}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{(C}{{\text{l}}_{2}}\text{)=242 }\!\!~\!\!\text{ kJmo}{{\text{l}}^{\text{-1}}}$ 

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

  1. $C{{C}_{4(l)}}\to CC{{l}_{4(g)}}{{\Delta }_{vap}}{{H}^{\theta }}=30.5kJmo{{l}^{-1}}$ 

  2. ${{C}_{(s)}}\to {{C}_{(g)}}{{\Delta }_{a}}{{H}^{\theta }}=715.0~kJ~mo{{l}^{-1}}$ 

  3. $C{{l}_{2(g)}}\to 2C{{l}_{(g)}}{{\Delta }_{a}}{{H}^{\theta }}=242~kJ~mo{{l}^{-1}}$ 

  4. ${{C}_{(g)}}+4C{{l}_{(g)}}\to CC{{l}_{4(g)}}{{\Delta }_{f}}H=-135.5~kJ~mo{{l}^{-1}}$ 

Enthalpy change for the given process $CC{{l}_{4(g)}}\to {{C}_{(g)}}+4C{{l}_{(g)}}$ can be calculated using the following algebraic calculations as: 

Equation (ii) + 2 $\times $ Equation (iii) - Equation (i) - Equation (iv) 

\[\Delta H={{\Delta }_{a}}{{H}^{\theta }}(C)+2{{\Delta }_{a}}{{H}^{\theta }}(C{{l}_{2}})-{{\Delta }_{vap\text{ }}}{{H}^{\theta }}-{{\Delta }_{f}}H\]

 \[\left( 715.0~kJ~mo{{l}^{-1}} \right)+2\left( 242~kJ~mo{{l}^{-1}} \right)-\left( 30.5~kJ~mo{{l}^{-1}} \right)-\left( -135.5~kJ~mo{{l}^{-1}} \right)\]

Therefore, $\Delta H=1304~kJ~mo{{l}^{-1}}$ 

Bond enthalpy of C-Cl bond in $CC{{l}_{4}}(g)$ 

\[=\frac{1304}{4}kJmo{{l}^{-1}}\]

\[=326~kJ~mo{{l}^{-1}}\]  


16. For an isolated system, $\text{ }\!\!\Delta\!\!\text{ U=0}$ , what will be $\text{ }\!\!\Delta\!\!\text{ S}$?

Ans: $\Delta S$will be positive i.e., greater than zero. 

Since for an isolated system, $\Delta U=0$, hence $\Delta S$will be positive and the reaction will be spontaneous. 


17. For the reaction at 298 K, $\text{2A+B}\to \text{C}$ 

$\text{ }\!\!\Delta\!\!\text{ H= 400 kJ mo}{{\text{l}}^{\text{-1}}}$ and $\text{ }\!\!\Delta\!\!\text{ S= 0}\text{.2 kJ }{{\text{K}}^{\text{-1}}}\text{ mo}{{\text{l}}^{\text{-1}}}$ At what temperature will the reaction become spontaneous considering $\text{ }\!\!\Delta\!\!\text{ H}$ and $\text{ }\!\!\Delta\!\!\text{ S}$ to be constant over the temperature range?

Ans: From the expression, 

\[\Delta G=\text{ }\Delta H-T\Delta S\] 

Assuming the reaction at equilibrium, $\Delta T$for the reaction would be: 

\[T=(\Delta H-\Delta G)\frac{1}{\Delta S}\] 

\[=\frac{\Delta H}{\Delta S}\] 

($\Delta G$ = 0 at equilibrium) 

\[=\frac{400~kJ~mo{{l}^{-1}}}{0.2~kJ~{{K}^{-1}}~mo{{l}^{-1}}}\] 

T = 2000 K 

For the reaction to be spontaneous, $\Delta G$must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K. 


18. For the reaction, $\text{2Cl(g)}\to \text{C}{{\text{l}}_{\text{2}}}\text{(g)}$ What are the signs of ∆H and ∆S ? 

Ans: $\Delta H$ and $\Delta S$ are negative. 

The given reaction represents the formation of chlorine molecules from chlorine atoms. Here, bond formation is occurring. Therefore, energy is being released. Hence, $\Delta H$is negative. 

Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, $\Delta S$is negative for the given reaction. 


19. For the reaction 

$\text{2A}\left( \text{g} \right)\text{ + B}\left( \text{g} \right)\text{ }\to \text{ 2D}\left( \text{g} \right)$ 

$\text{ }\!\!\Delta\!\!\text{ }{{\text{U}}^{\text{ }\!\!\theta\!\!\text{ }}}$  = -10.5 kJ and $\text{ }\!\!\Delta\!\!\text{ }{{\text{S}}^{\text{ }\!\!\theta\!\!\text{ }}}$ = -44.1$\text{J}{{\text{K}}^{\text{-1}}}$ . 

Calculate $\text{ }\!\!\Delta\!\!\text{ }{{\text{G}}^{\text{ }\!\!\theta\!\!\text{ }}}$ for the reaction, and predict whether there action may occur spontaneously. 

Ans: For the given reaction,

\[2A\left( g \right)\text{ }+\text{ }B\left( g \right)\text{ }\to \text{ }2D\left( g \right)\] 

$\Delta {{n}_{g}}=2-(3)$ = -1 mole

Substituting the value of $\Delta {{U}^{\theta }}$ in the expression of$\Delta H$:

\[\Delta {{H}^{\theta }}=\Delta {{U}^{\theta }}+\Delta {{n}_{g}}RT\] 

\[=(-10.5~kJ)-(-1)\left( 8.314\times {{10}^{-3}}~kJ~{{K}^{-1}}~mo{{l}^{-1}} \right)(298~K)\] 

\[ -10.5 kJ - 2.48 kJ\] 

\[\Delta {{H}^{\theta }}=-12.98kJ\] 

Substituting the values of $\Delta {{H}^{\theta }}$ and $\Delta {{S}^{\theta }}$ in the expression of $\Delta {{G}^{\theta }}$: 

\[\Delta {{G}^{\theta }}\text{ }=\text{ }\Delta {{H}^{\theta }}\text{ }-T\Delta {{S}^{\theta }}\] 

\[=-12.98 kJ - (298 K) (-44.1) J\text{ }{{K}^{-1}}\] 

\[ =-12.98 kJ + 13.14 kJ\]

$\Delta {{G}^{\theta }}$= + 0.16 kJ 

Since $\Delta {{G}^{\theta }}$ for the reaction is positive, the reaction will not occur spontaneously.


20. The equilibrium constant for a reaction is 10. What will be the value of $\text{ }\!\!\Delta\!\!\text{ }{{\text{G}}^{\text{ }\!\!\theta\!\!\text{ }}}$? R = 8.314 $text{ }\!\!~\!\!\text{ J}{{\text{K}}^{\text{-1}}}\text{ mo}{{\text{l}}^{\text{-1}}}$ , T = 300 K. 

Ans: From the expression, 

\[\Delta {{G}^{\theta }}\text{ }=\text{ }-2.303\text{ }RTlogKeq\] 

$\Delta {{G}^{\theta }}$ for the reaction, 

\[= (2.303) (8.314 ~J{{K}^{-1}}\text{ }mo{{l}^{-1}}) (300 K) log10\] 

\[= -5744.14 ~Jmo{{l}^{-1}}\] 

\[= -5.744 k~Jmo{{l}^{-1}}\]


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

\[\frac{\text{1}}{\text{2}}\text{NO(g)+}\frac{\text{1}}{\text{2}}{{\text{O}}_{\text{2}}}\text{(g)}\to \text{N}{{\text{O}}_{\text{2}}}\text{(g):}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{=90kJmo}{{\text{l}}^{\text{-1}}}\] 

\[\text{N}{{\text{O}}_{\text{(g)}}}\text{+}\frac{\text{1}}{\text{2}}{{\text{O}}_{\text{2(g)}}}\to {{\text{O}}_{\text{2(g)}}}\text{:}{{\text{ }\!\!\Delta\!\!\text{ }}_{\text{r}}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}\text{=-74 }\!\!~\!\!\text{ kJ }\!\!~\!\!\text{ mo}{{\text{l}}^{\text{-1}}}\] 

Ans: The positive value of ${{\Delta }_{r}}H$ indicates that heat is absorbed during the formation of NO(g). This means that NO(g) has higher energy than the reactants (${{N}_{2}}$  and${{O}_{2}}$). Hence, NO(g) is unstable. The negative value of ${{\Delta }_{r}}H$indicates that heat is evolved during the formation of $N{{O}_{2}}(g)$ from NO(g) and ${{O}_{2}}$(g). The product, $N{{O}_{2}}(g)$is stabilized with minimum energy. 

Hence, unstable NO(g) changes to unstable $N{{O}_{2}}(g)$. 


22. Calculate the entropy change in surroundings when 1.00 mol of ${{\text{H}}_{\text{2}}}\text{O(l)}$ is formed under standard conditions. ${{\text{ }\!\!\Delta\!\!\text{ }}_{\text{f}}}{{\text{H}}^{\text{ }\!\!\theta\!\!\text{ }}}$ = -286 $\text{kJ mo}{{\text{l}}^{\text{-1}}}$ . 

Ans: It is given that 286 $kJ\text{ }mo{{l}^{-1}}$ of heat is evolved on the formation of 1 mol of ${{H}_{2}}O(l)$. Thus, an equal amount of heat will be absorbed by the surroundings.

\[{{q}_{surr}} = +286 kJ kJ\text{ }mo{{l}^{-1}}\]

Entropy change $\left( \Delta {{S}_{surr}} \right)$ for the surroundings \[=\frac{{{q}_{surr}}}{7}\] 

\[=\frac{286kJ\,mo{{l}^{-1}}}{298K}\] 

Therefore, $\left( \Delta {{S}_{surr}} \right)$= $959.73\,J\,mo{{l}^{-1}}{{K}^{-1}}$.


Class 11 Chemistry Chapter 5 Quick Overview of Topics

Class 11 Chemistry Chapter 5 NCERT Solutions -Quick Overview of Detailed Structure of Topics and Subtopics Covered.


Topic

Subtopics

Introduction to Thermodynamics

- Definition and scope

- System and surroundings

- Types of systems

- State variables and state functions

First Law of Thermodynamics

- Concept of internal energy

- Heat and work

- Sign conventions

- Enthalpy

- Heat capacity and specific heat

- Calorimetry

- Application of first law

Second Law of Thermodynamics

- Spontaneous processes

- Entropy

- Entropy changes in chemical reactions

- Gibbs free energy

- Gibbs free energy change and spontaneity

- Criteria for equilibrium

- Free energy and work

- Thermodynamic functions


Some Important Concepts and Formulas 

Class 11 NCERT solutions help the students to go through the formulas and concepts easily. Here find the Important formulas and concepts of Chapter 5- Chemical Thermodynamics to crack your exams.


  1. First Law of Thermodynamics:

    • $\Delta U = q + w$


  1. Enthalpy

    • $\Delta H = H_{\text{final}}-H_{\text{initial}}$

    • Change in enthalpy represents the heat absorbed or released at constant pressure.


  1. Heat Transfer:

    • $q = mc\Delta T$


  1. Gibbs Free Energy 

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


  1. Entropy

    • $\Delta S = \dfrac{q_{\text{rev}}}{T}$

    • Change in entropy represents the dispersal of energy in a system during a reversible process.


Important Points for Class 11 NCERT Solutions Chapter 5 Thermodynamics

Let us discuss some of the important points related to Thermodynamics here.


  • Thermodynamic Terms: The thermodynamics system is explained in this topic, which refers to the part of the universe where observations are made, while the rest of the universe is the environment. It also describes the many types of systems.

  • The Internal Energy as a State Function and its Applications: This topic teaches students how to compute energy changes in chemical systems as work and heat contributions.

  • Measurement of ∆U AND ∆H: Calorimetry: Students learn to detect energy changes associated with chemical or physical processes through an experimental technique in this module.

  • Enthalpy Change, ∆rH of a Reaction: Students will study how to convert reactants into products in a chemical reaction in this topic. The reaction enthalpy is the enthalpy change that occurs as a result of a reaction.

  • Enthalpies for Different Types of Reactions: Standard Enthalpy of Combustion, Enthalpy of Atomization, Bond Enthalpy, and Lattice Enthalpy, Enthalpy of Solution and Enthalpy of Dilution, and their computations are discussed in this topic.

  • Spontaneity: Spontaneity is defined as the ability to proceed in either a forward or backward direction (in the case of reactions) without the assistance of an external force.

  • Gibbs Energy Change and Equilibrium: This topic deals with the relationship between the equilibrium constant and free energy.


Benefits of Referring to Vedantu’s NCERT Solutions for Class 11 Chemistry Chapter 5

The Vedantu’s NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics provided here in PDFs offer various benefits, including:


  • Comprehensive Coverage: Detailed solutions for topics like laws of thermodynamics, enthalpy, entropy, and Gibbs free energy.

  • Expert Guidance: Curated by experienced educators for accurate and insightful answers.

  • Clarity and Precision: Clear, concise explanations using precise scientific terminology.

  • Exam Preparation: Aligned with the latest CBSE syllabus, including practice questions and sample papers.  Focused preparation with targeted questions and answers.

  • Accessibility: Free PDF download for easy offline access.

  • Core Concept Mastery: Understand fundamental principles and complex concepts clearly.

  • Application-Based Learning: Connect theoretical knowledge with practical scenarios.

  • Boost Analytical Skills: Improve problem-solving techniques for thermodynamic calculations.


Related Study Material Links for Chemistry Class 11 Chapter 5 NCERT Solutions

Students can access extra study materials on Thermodynamics. These resources are available for download and offer additional support for your studies.



Conclusion

Chemistry, the branch of science that deals with atoms, molecules, thermodynamics, state of matter, elements, etc., is a complex subject. Students are required to understand the concepts and theories behind each reaction.


Thermodynamics Class 11 includes subdivisions like open, closed, and isolated systems, internal energy as a state function, isothermal and free expansion of gas, enthalpy, etc. These topics have to be read in a detailed manner to understand the law of Thermodynamics better.


NCERT Solutions for Class 11 Chemistry - Chapter-wise Links


NCERT Solutions Class 11 Chemistry - Related Links

FAQs on NCERT Solutions for Class 11 Chemistry Chapter- 5 Thermodynamics

1. What are the Three Types of System in Class 11 Chapter 5 Thermodynamics?

Thermodynamics includes analysis of the quantity of matter or space which is called a system. This is covered by boundary and the area beyond the system is called the surrounding or universe. A boundary of the course is movable and fixable. The exchange of mass and energy can occur between system and surrounding. There are ideally three types of system:

  • Open System: This system in which a transfer of mass and energy can take place across the boundary.

  • Closed System: This system is the one where a transfer of energy can take place across its boundary and surrounding. There is no transfer of mass as it is fixed in a closed system.

  • Isolated System: It is a system where neither the transfer of energy or mass takes place across its boundary with surroundings.

2. What is the Thermodynamic Process?

A thermodynamic process involves the movement of heat energy within a system. There are four types of thermodynamic cycle, namely isobaric, isothermal, adiabatic and isochoric. It means:

  • Isobaric - This process is where the pressure of a system. Here the pressure is related to the amount of force applied by molecules to the wall of a container.

  • Isochoric - This process is one where the volume of the system remains constant. Its volume is the amount of space a material takes up, making it equal to a solid form of gas in a container. Here the molecules move fast, and the pressure increases but the container size is unchanging.

  • Adiabatic - In this system, the heat is not exchanged with the surroundings making it trapped.

  • Isothermal - This process is the one where the temperature remains constant.

3. What is Enthalpy or Heat Reaction according to Chapter 5 Class 11 Thermodynamics?

Enthalpy is termed as the heat of the reaction due to the chemical reaction that occurs when the pressure is constant. The thermodynamics measurement is used for calculating the amount of energy per mole in a reaction. Enthalpy is derived from volume, pressure and internal energy which is also a state function.


Heat reaction is used to calculate the heat of a reaction in a chemical process. A change in enthalpy can be used to measure the flow of heat in a calorimeter. This is used to evaluate a throttling process, also known as Joule Thomson Expansion. Moreover, enthalpy is used to calculate the minimum power used in a compressor.

4. How is studying Thermodynamics with the help of NCERT Solutions for Chapter 5 of Class 11 Chemistry helpful for exam preparation?

NCERT Solutions are the most important resource for preparing for exams as they are based on the NCERT pattern. Thermodynamics is also among the most asked topic in exams. And solving the important questions given in NCERT for the chapter becomes increasingly crucial. Thus, NCERT Solutions by Vedantu for Chapter 5 of Class 11 Chemistry, which is Thermodynamics, becomes very helpful for students as it covers all the critical sections of the chapter.

5. What is Thermodynamics according to NCERT Solutions for Chapter 5 of Class 11 Chemistry?

Thermodynamics is the scientific method of studying the link between different forms of energy and heat. It is one of the most crucial branches of Chemistry and important for students as they are expected to learn the concepts involved in reactions. The Thermodynamics class 11 explains all the concepts of thermodynamics very well with step by step explanation of the chemical reactions involved and showcasing the connection between the required form of energy and heat. The notes are available on Vedantu’s official website (vedantu.com) free of cost.

6. What is Thermodynamic equilibrium according to Chapter 5 of Class 11 Chemistry?

According to the Thermodynamics concepts, a system is said to be in Thermodynamic equilibrium if all Thermodynamic processes remain constant with passing time. When the Thermodynamic system is in chemical equilibrium (no change in chemical composition), mechanical equilibrium (no change in pressure) and thermal equilibrium (no change in temperature) and all other parameters stop varying with time, it is said to be in Thermodynamic equilibrium.

7. What is the Thermodynamics of a system according to Chapter 5 of Class 11 Chemistry?

As per thermodynamics, 3 types of systems exist:

  • Open system in which exchange of both matter and energy with surroundings is possible. For eg: Open cup of tea.

  • A closed system in which only energy exchange is possible but not an exchange of matter. For eg: Covered cup of tea

  • Isolated system in which exchange of neither energy nor matter Is possible in any way with its surroundings. For eg: Tea in an insulated thermos flask.

8. (a) What is a spontaneous process according to Chapter 5 of Class 11 Chemistry? Mention the conditions for a reaction to be spontaneous at constant temperature and pressure.

(b) Discuss the effect of temperature on the spontaneity of an exothermic reaction according to Chapter 5 of Class 11 Chemistry.

  • A process is considered to be spontaneous if it occurs on its own or under certain circumstances.

At constant temperature and pressure, G gives a set of conditions for spontaneity.

  • B will be positive and the process will be non-spontaneous if the temperature is so high that TS > H in magnitude.

If the temperature is lowered to the point where TS = H, G will be negative, and the process will occur spontaneously.

9. How is thermodynamics used in everyday life?

Thermodynamics finds application in everyday life through understanding energy transfers in cooking, heating, and cooling processes, elucidated in Class 11 Thermodynamics NCERT solutions.

10. What are the limitations of the first law of thermodynamics?

The first law of thermodynamics faces limitations in neglecting the direction of energy flow and inability to predict spontaneity, as detailed in class 11 chemistry chapter Thermodynamics NCERT solutions.

11. What is the basic principle of thermodynamics?

The basic principle of thermodynamics lies in the conservation of energy and the spontaneity of processes, outlined in Class 11 Chemistry Chapter 5 NCERT solutions.

12. What are three 3 applications of thermodynamics according the Class 11 Chapter 2?

Thermodynamics finds applications in power generation, refrigeration, and chemical engineering processes, showcasing its versatility and significance in various fields, highlighted in Thermodynamics class 11 chemistry NCERT solutions.