NCERT Solutions for Class 11 Chemistry Chapter 6

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

NCERT Solutions Class 11 Chemistry Thermodynamics – Free PDF Download

Chemistry, the branch of science which 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 for understanding the law of Thermodynamics better.

Many students fail to identify the critical sections of the chapter. Therefore, Thermodynamics Class 11 NCERT solutions are designed to cover topics important in exams. All the concepts have been elaborated in simple language for easing out the ideas.

Moreover, Class 11 Chemistry Thermodynamics solutions are available in PDF format for easy download. Students can avail this solution at their convenience without hassle.

NCERT Class 11 Chemistry Thermodynamics PDF

Thermodynamics Class 11 solution has twenty questions that require students to provide reasons for the equations. Take a glance:

1. Class 11 Chemistry Ch 6 NCERT Solutions

Solution - First question in this solution asks students to choose the correct option. Here the answer will be a thermodynamic state of function is whose value is independent of path.

2. Class 11 Chemistry Chapter 6

Solution - This second question asks young students to choose the condition for the process in a diabetic condition. Here the options are ∆T is 0, ∆p is 0, q is 0 and w is 0.

3. Ch 6 Chemistry Class 11

Solution - Third question in Thermodynamics Class 11 wants a student to choose the enthalpies of an element. Options in this question are unity, zero, less than zero, and different from each element.

4. Thermodynamics Class 11 Chemistry NCERT

Solution - Here ∆Uᶱ of combustion methane is – x kJ mol-1, students have to determine the value of ∆Hᶱ.

5. NCERT Solution of Thermodynamics Chemistry Class 11

Solution - This question requires students to find the enthalpy of combustion of methane, graphite and hydrogen at 298k, -890.3 kJ mol-1, -393.5 kJ mol-1 and -285.8 kJ mol-1. Here enthalpy of CH₄G is needed to be determined.

6. Thermodynamics Class 11 NCERT

Solution - In this question, the reaction to the equation A + B → C + D + q with positive entropy is needed to be determined.

7. Thermodynamics Class 11th

Solution - Question seven of the Thermodynamics Class 11  NCERT solutions give the variables that 701 J of heat is absorbed by a system and a system does 394 J  of work, is done by a system. Here students have to find the change in internal energy of the process.

8. Class 11th Chemistry Chapter 6 NCERT Solutions

Solution - Eighth question of this solution asks students to calculate the enthalpy change for a reaction. Here the reaction of cyanamide NH₂CN(S)  with dioxygen carried out in a bomb calorimeter ∆U was found -742.7 kJ mol 1 at 298k.

9. NCERT Solutions Class 11 Chemistry Thermodynamics PDF

Solution - Here a student is asked to calculate the number of K J of heat necessary to raise aluminium's temperature at 60gm from 35°C to 55°C. AI's Molar heat capacity mentioned is 24 j mol-1 K ̵ 1.

10. Thermodynamics Class 11 NCERT Solutions

Solution - In this question, young learners have to determine the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. Here H is 6.03 kJ mol-1 at 0°C.

Apart from these eight questions, the solution contains twelve questions requiring students to calculate heat released upon the formation of CO₂, value of ∆rH, standard enthalpy of formation of NH₃ and CH₃OH. They are also required to calculate bond enthalpy of C-CL in CCl₄ (g), the temperature at which the reaction becomes spontaneous, considering ∆H and ∆S to be constant over a temperature range.

Other questions like for a reaction 2 CL(g) →Cl₂(g) what are the signs of ∆H and ∆S, the value of ∆Gᶱ where R is 8.314 kJ mol-1 and T is 300 k needed to be calculated.

Why Study from CBSE Class 11 Chemistry Thermodynamics NCERT Solutions?

Thermodynamics Class 11  includes questions and answers structured in a way to ease the learning process. A student can follow the answer pattern to secure an impressive grade in CBSE Examination.

Following are the Benefits of Following NCERT Solution:

• Step by step explanation of the chemical reaction involved in the given equation.

• Answers are written in  a concise manner and to the point to help student fetch good marks.

• These solutions are prepared under the supervision of experts who offer quality reading material.

• Explanations are offered with images and tables for detailed information.

NCERT is known to impart quality education and learning to the students with their publications and online solutions. A student irrespective of a class needs a quality study guide for revision and to score satisfactory marks.

Solving exercises from CBSE NCERT Solutions can be helpful in this regard. They can also take help of online portals to procure Thermodynamics Class 11 solutions for free and in PDF format. CoolGyan is one such site that offers quality services for class 11 students at an affordable price.

So, study well and choose your reference material wisely.