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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
Q11Written answer

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. 

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Worked solution

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.


Q12Written answer

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: 

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

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Worked solution

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 $-777.7kJ\,mo{{l}^{-1}}$

 

Q13Written answer

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?

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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 $N{{H}_{3}}(g)$

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

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

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


Q14Written answer

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

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

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

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

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

 

Q15Written answer

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

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

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Worked solution

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 $CC{{l}_{4}}(g)$ 

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

\[=326~kJ~mo{{l}^{-1}}\]  


Q16Written answer

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

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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. 


Q17Written answer

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?

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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. 


Q18Written answer

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

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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. 


Q19Written answer

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. 

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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.


Q20Written answer

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. 

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Ans: From the expression, 

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

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

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

\[= -5744.14 ~Jmo{{l}^{-1}}\] 

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