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NCERT · Class 11 · Chemistry

Chapter 5: Thermodynamics

Practice each question, then open the answer or worked solution.

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Textbook exercises

22 questions
Q1Written answer

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. 

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Worked solution

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. 


Q2Written answer

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) $\text{w=0}$ 

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Worked solution

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. 


Q3Written answer

The Enthalpies of All Elements in Their Standard States Are: 

(i) unity 

(ii) zero 

(iii) < 0 

(iv) different for each element 

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Worked solution

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


Q4Written answer

$\text{ }\!\!\Delta\!\!\text{ }{{\text{U}}^{\text{ }\!\!\theta\!\!\text{ }}}$ 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 

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Worked solution

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.


Q5Written answer

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 $\text{kJ mo}{{\text{l}}^{\text{-1}}}$  . 

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Worked solution

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.


Q6Written answer

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 

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Worked solution

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. 


Q7Written answer

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? 

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Worked solution

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.


Q8Written answer

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. 

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

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Worked solution

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 

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


Q9Written answer

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

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Worked solution

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


Q10Written answer

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 $\text{0 }\!\!{}^\circ\!\!\text{ C}$.

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

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

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Worked solution

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 $\text{ }-7.151\text{ }kJ\text{ }mo{{l}^{-1}}$