NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics

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.

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:

• $\Delta U = q + w$

2. Enthalpy

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

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

3. Heat Transfer:

• $q = mc\Delta T$

4. Gibbs Free Energy 

• $\Delta G = \Delta H - T\Delta S$

5. 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