NCERT Solutions for Class 11 Chemistry chapter 6 Thermodynamics PDF – eSaral
Hey, are you a class 11 student and looking for ways to download NCERT Solutions for Class 11 Chemistry chapter 6 Thermodynamics PDF? If yes. Then read this post till the end.

In this article, we have listed NCERT Solutions for Class 11 Chemistry chapter 6 Thermodynamics in PDF that are prepared by Kota’s top IITian’s Faculties by keeping Simplicity in mind.

If you want to learn and understand class 11 Chemistry chapter 6 “Thermodynamics” in an easy way then you can use these solutions PDF.

NCERT Solutions helps students to Practice important concepts of subjects easily. Class 11 Chemistry solutions provide detailed explanations of all the NCERT questions that students can use to clear their doubts instantly.

If you want to score high in your class 11 Chemistry Exam then it is very important for you to have a good knowledge of all the important topics, so to learn and practice those topics you can use eSaral NCERT Solutions.

So, without wasting more time Let’s start.

### Download The PDF of NCERT Solutions for Class 11 Chemistry Chapter 6 “Thermodynamics”

Question 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

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

Question 2. For the process to occur under adiabatic conditions, the correct condition is:

(i) $\Delta \mathrm{T}=0$

(ii) $\Delta \mathrm{p}=0$

(ii) $\mathrm{q}=0$

(v) $w=0$

Solution. A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, $\mathrm{q}=\mathrm{O}$.

Question 3. The enthalpies of all elements in their standard states are:

(i) unity

(ii) zero

(iii) $<0$

(iv) different for each element

Solution: The enthalpy of all elements in their standard state is taken to be zero.

Question 4. $\Delta U^{\ominus}$ of combustion of methane is $-X \mathrm{~kJ} \mathrm{~mol}^{-1}$. The value of $\Delta \mathrm{H}^{\ominus}$ is

$(\mathrm{i})=\Delta \mathrm{U}^{\ominus}$

(ii) $>\Delta \mathrm{U} \ominus$

(iii) $<\Delta \mathrm{U}^{\ominus}$

(iv) $=0$

Solution: given $\Delta \mathrm{U}^{\theta}=-\mathrm{X} \mathrm{kJ} \mathrm{mol}^{-1}$,

Since $\Delta \mathrm{H}^{\theta}=\Delta \mathrm{U}^{\theta}+\Delta \mathrm{ngRT}$ and

$\Delta \mathrm{H}^{\theta}=(-\mathrm{X})+\Delta \mathrm{ngRT}$

$\Rightarrow \Delta \mathrm{H}^{\theta}<\Delta \mathrm{U}^{\theta}$

Therefore, alternative (iii) is correct.

Question 5. The enthalpy of combustion of methane, graphite and dihydrogen at $298 \mathrm{~K}$ are, $-890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$, and $-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. Enthalpy of formation of $\mathrm{CH}_{4}(\mathrm{~g})$ will be

(i) $-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(ii) $-52.27 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iii) $+74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iv) $+52.26 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution: According to the question,

(i) $\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$

$\Delta \mathrm{H}=-890.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(ii) $\mathrm{C}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})$

$\Delta \mathrm{H}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iii) $2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$

$\Delta \mathrm{H}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Thus, the desired equation is the one that represents the formation of $\mathrm{CH}_{4}(\mathrm{~g})$ i.e.,

$\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{4}(\mathrm{~g})$

$\Delta_{\mathrm{f}} \mathrm{H}_{\mathrm{CH}_{4}}=\Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{c}}+2 \Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{H}_{2}}-\Delta_{\mathrm{c}} \mathrm{H}_{\mathrm{CO}_{2}}$

$=[-393.5+2(-285.8)-(-890.3)] \mathrm{kJ} \mathrm{mol}^{-1}$

v$=-74.8 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$

$\therefore$ Enthalpy of formation of $\mathrm{CH}_{4}(\mathrm{~g})=-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Question 6. A reaction, $\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}+\mathrm{q}$ 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

(v) possible at any temperature

Solution: Given $: \Delta S=$ positive

\Delta \mathrm{H}=\text { negative (since heat is evolved) }

For a reaction to be spontaneous at constant pressure and temperature, $\Delta$ G should be negative.

$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$

$\frac{\Delta H-\Delta G}{\Delta S}=T$

According to the question, for the given reaction,

$\Rightarrow \Delta \mathrm{G}=$ negative

Therefore, the reaction is spontaneous at any temperature. Hence, alternative (iv) is correct.

Question 7. In a process, $701 \mathrm{~J}$ of heat is absorbed by a system and $394 \mathrm{~J}$ of work is done by the system. What is the change in internal energy for the process?

Solution: Given: heat is absorbed by a system $=701 \mathrm{~J}$

work is done by the system $=394 \mathrm{~J}$

According to the first law of thermodynamics,

Formula: $\Delta \mathrm{U}=$ change in internal energy for a process $(\Delta \mathrm{U})=\mathrm{q}+\mathrm{W}$(i)

$\mathrm{q}=+701 \mathrm{~J}$ (Since heat is absorbed than sign of $\mathrm{q}$ is positive )

$\mathrm{W}=-394 \mathrm{~J}$ (Since work is done by the system so sign of $\mathrm{w}$ is negative $)$

Substituting the values in a formula (i), we get

$\Delta \mathrm{U}=701 \mathrm{~J}+(-394 \mathrm{~J})$

$\Delta \mathrm{U}=307 \mathrm{~J}$

Hence, the change in internal energy for the given process is $307 \mathrm{~J}$.

Question 8. The reaction of cyanamide, $\mathrm{NH}_{2} \mathrm{CN}(g)$, with dioxygen was carried out in a bomb calorimeter, and $\Delta \mathrm{U}$ was found to be $-742.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $298 \mathrm{~K}$. Calculate the enthalpy change for the reaction at $298 \mathrm{~K}$.

$\mathrm{NH}_{2} \mathrm{CN}(\mathrm{g})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

Solution: Formula : Enthalpy change for a reaction $(\Delta \mathrm{H})=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$

Where,

$\Delta \mathrm{U}=$ change in internal energy $=-742.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta \mathrm{n}_{\mathrm{g}}=$ change in the number of moles $=($ moles of product $-$ moles of reactant $)$

For the given reaction,

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

$=(2-2.5)$ moles

$\Delta \mathrm{n}_{\mathrm{g}}=-0.5 \mathrm{moles}$

And,

We know $: \Delta \mathrm{U}=-742.7 \mathrm{~kJ} \mathrm{~mol}^{-1}=-742.7 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}$

$\mathrm{T}=298 \mathrm{~K}$

$\mathrm{R}=8.314 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}$

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

$\Delta \mathrm{H}=\left(-742.7 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\right)$

$+(-0.5 \mathrm{~mol})(298 \mathrm{~K})\left(8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$

$=-742700-1238.786$

$=743938.78 \mathrm{~J}$

$\Delta \mathrm{H}=-743.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Question 9. Calculate the number of $\mathrm{kJ}$ of heat necessary to raise the temperature of $60.0 \mathrm{~g}$ of aluminium from $35^{\circ} \mathrm{C}$ to $55^{\circ} \mathrm{C}$. Molar heat capacity of $\mathrm{Al}$ is $24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$.

Solution: molar mass of aluminium $=27$

From the expression of heat $(q)=m . c \cdot \Delta T$ Where,

$\mathrm{c}=$ molar heat capacity $=24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

$\mathrm{m}=$ mass of substance $=60.0 \mathrm{~g}$

$\Delta \mathrm{T}=$ change in temperature $=(55-35)=20$

Moles of Al= weight/molecular weight $=60 / 27$

Substituting the values in the expression of q:

$\mathrm{q}=\left(\frac{60}{27} \mathrm{~mol}\right)\left(24 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(20 \mathrm{~K})$

$q=1066.7 \mathrm{~J}$

$q=1.07 \mathrm{~kJ}$

Question 10. Calculate the enthalpy change on freezing of $1.0 \mathrm{~mol}$ of water at $10.0^{\circ} \mathrm{C}$ to ice at $-10.0^{\circ} \mathrm{C} \cdot \Delta_{\mathrm{fus}} \mathrm{H}=6.03 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $0^{\circ} \mathrm{C} .$

$\mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right]=75.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

$\mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\right]=36.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Solution: Given : mol of water $=1.0$

$\Delta_{\text {fus }} H=6.03 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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

(a) Energy change involved in the transformation of $1 \mathrm{~mol}$ of water at $10^{\circ} \mathrm{C}$ to $1 \mathrm{~mol}$ of water at $0^{\circ} \mathrm{C}$.

$\Delta \mathrm{H}=\mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right] \Delta \mathrm{T}$

(b) Energy change involved in the transformation of $1 \mathrm{~mol}$ of water at $0^{\circ}$ to $1 \mathrm{~mol}$ of ice at $0^{\circ} \mathrm{C}$

$\Delta \mathrm{H}=\mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\right] \Delta \mathrm{T}$

(c) Energy change involved in the transformation of $1 \mathrm{~mol}$ of ice at $0^{\circ} \mathrm{C}$ to $1 \mathrm{~mol}$ of ice at $-10^{\circ} \mathrm{C}$

$\Delta \mathrm{H}=\mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\right] \Delta \mathrm{T}$

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

$=\left(75.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(0-10) \mathrm{K}+\left(-6.03 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\right)$

$+\left(36.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(-10-0) \mathrm{K}$

$=-753 \mathrm{~J} \mathrm{~mol}^{-1}-6030 \mathrm{~J} \mathrm{~mol}^{-1}-368 \mathrm{~J} \mathrm{~mol}^{-1}$

$=-7151 \mathrm{Jmol}^{-1}$

$=-7.151 \mathrm{k}] \mathrm{mol}^{-1}$

Hence, the enthalpy change involved in the transformation is $-7.151 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

Question 11. Enthalpy of combustion of carbon to $\mathrm{CO}_{2}$ is $-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1} .$ Calculate the heat released upon formation of $35.2 \mathrm{~g}$ of $\mathrm{CO}_{2}$ from carbon and dioxygen gas.

Solution: Given: Enthalpy of combustion of carbon to $\mathrm{CO}_{2}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Weight of $\mathrm{CO}_{2}=35.2 \mathrm{~g}$

Formation of $\mathrm{CO}_{2}$ from carbon and dioxygen gas can be represented as:

$\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{f}} \mathrm{H}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\left(1\right.$ mole of $\mathrm{CO}_{2}=44 \mathrm{~g}$ of $\left.\mathrm{CO}_{2}\right)$

Heat released on formation of $44 \mathrm{~g} \mathrm{CO}_{2}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\therefore$ The heat released on formation of $35.2 \mathrm{~g} \mathrm{CO}_{2}$

$=\frac{-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}}{44 \mathrm{~g}} \times 35.2 \mathrm{~g}$

$=-314.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Question 12. Enthalpies of formation of $\mathrm{CO}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{~g}), \mathrm{N}_{2} \mathrm{O}(\mathrm{g})$ and $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})$ are $-110,-393,81$ and $9.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. Find the value of $\Delta_{\mathrm{r}} \mathrm{H}$ for the reaction:

$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+3 \mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{CO}_{2}(\mathrm{~g})$

Solution: Given: Enthalpies of formation of $\mathrm{CO}(\mathrm{g})=-110 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Enthalpies of formation of $\mathrm{CO}_{2}(\mathrm{~g})=-393 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Enthalpies of formation of $\mathrm{N}_{2} \mathrm{O}(\mathrm{g})=81 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$

Enthalpies of formation of $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})=9.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta \mathrm{rH}$ for a reaction is defined as the difference between $\Delta \mathrm{fH}$ value of products and $\Delta \mathrm{fH}$ value of reactants.

$\Delta_{\mathrm{r}} \mathrm{H}=\sum \Delta_{\mathrm{f}} \mathrm{H}$ (products) $-\sum \Delta_{\mathrm{f}} \mathrm{H}$ (reactants)

For the given reaction,

$\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})+3 \mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+3 \mathrm{CO}_{2}(\mathrm{~g})$

$\Delta_{\mathrm{r}} \mathrm{H}=\left[\left\{\Delta_{\mathrm{f}} \mathrm{H}\left(\mathrm{N}_{2} \mathrm{O}\right)+3 \Delta_{\mathrm{f}} \mathrm{H}\left(\mathrm{CO}_{2}\right)\right\}\right.$

$\left.-\left\{\Delta_{\mathrm{f}} \mathrm{H}\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)+3 \Delta_{\mathrm{f}} \mathrm{H}(\mathrm{CO})\right\}\right]$

Substituting the values of $\Delta_{\mathrm{f}} \mathrm{H}$ for $\mathrm{N}_{2} \mathrm{O}, \mathrm{CO}_{2}, \mathrm{~N}_{2} \mathrm{O}_{4}$ and CO from given data, we get:

$\Delta_{\mathrm{r}} \mathrm{H}=\left[\left\{81 \mathrm{~kJ} \mathrm{~mol}^{-1}+3(-393) \mathrm{kJ} \mathrm{mol}^{-1}\right\}\right.$

$\left.-\left\{9.7 \mathrm{~kJ} \mathrm{~mol}^{-1}+3(-110) \mathrm{k} \mathrm{mol}^{-1}\right\}\right]$

$\Delta_{\mathrm{r}} \mathrm{H}=-777.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Hence, the value of $\Delta_{\mathrm{r}} \mathrm{H}$ for the reaction is $-777.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

Question 13. Given

$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta_{\mathrm{r}} \mathrm{H}^{\ominus}$

$=-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$

What is the standard enthalpy of formation of $\mathrm{NH}_{3}$ gas?

Solution: Given $: \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) ; \Delta_{\mathrm{r}} \mathrm{H}^{\ominus}$

$=-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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 $\mathrm{NH}_{3}(\mathrm{~g})$,

$\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{NH}_{3}(\mathrm{~g})$

$\therefore$ Standard enthalpy of formation of $\mathrm{NH}_{3}(\mathrm{~g})=\frac{1}{2} \Delta_{\mathrm{r}} \mathrm{H}^{\ominus}$$\therefore$ Standard enthalpy of formation of $\mathrm{NH}_{3}(\mathrm{~g})=\frac{1}{2} \Delta_{\mathrm{r}} \mathrm{H}^{\ominus}$

$=\frac{1}{2}\left(-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

$=-46.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Question 14. Calculate the standard enthalpy of formation of $\mathrm{CH}_{3} \mathrm{OH}(1)$ from the following data:

$\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})$

$+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{\mathrm{r}} \mathrm{H}^{\ominus}=-726 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$

$\mathrm{C}$ (graphite) $+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\ominus}$

$=-393 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{\mathrm{f}} \mathrm{H}^{\ominus}$

$=-286 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$

Solution: The reaction that takes place during the formation of $\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})$ can be written as:

$\mathrm{C}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})$…(i)

C(graphite) $+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}) ; \Delta_{\mathrm{c}} \mathrm{H}^{\ominus}$

$=-393 \mathrm{~kJ} \mathrm{~mol}^{-1}(\mathrm{ii})$

$\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; \Delta_{\mathrm{f}} \mathrm{H}^{\ominus}$

$=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$. (iii)

$\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{~g}) \Delta_{\mathrm{r}} \mathrm{H}^{\ominus}$

$=726 \mathrm{~kJ} \mathrm{~mol}^{-1}(\mathrm{iv})$

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

Equation(ii) $+2 \times$ equation (iii) $+$ equation (iv)

$=\left(-393 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)+2\left(-286 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

$+\left(726 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

$=(-393-572+726) \mathrm{kJ} \mathrm{mol}^{-1}$

$\therefore \Delta_{\mathrm{f}} \mathrm{H}^{\ominus}\left[\mathrm{CH}_{3} \mathrm{OH}(1)\right]=-239 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$

Question 15. Calculate the enthalpy change for the process

$\mathrm{CCl}_{4}(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{Cl}(\mathrm{g})$

and calculate the bond enthalpy of $\mathrm{C}-\mathrm{Cl}$ in $\mathrm{CCl}_{4}(\mathrm{~g})$.

$\Delta_{\text {vap }} \mathrm{H}^{\ominus}\left(\mathrm{CCl}_{4}\right)=30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{\mathrm{f}} \mathrm{H}^{\ominus}\left(\mathrm{CCl}_{4}\right)=-135.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{\mathrm{a}} \mathrm{H}^{\ominus}(\mathrm{C})=715.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$, where $\Delta_{\mathrm{a}} \mathrm{H}^{\ominus}$ is enthalpy of atomization

$\Delta_{\mathrm{a}} \mathrm{H}^{\ominus}\left(\mathrm{Cl}_{2}\right)=242 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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

(i) $\mathrm{CCl}_{4}(\mathrm{l}) \rightarrow \mathrm{CCl}_{4}(\mathrm{~g})$

$\Delta_{\text {vap }} \mathrm{H}^{\ominus}\left(\mathrm{CCl}_{4}\right)=30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(ii) $\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{C}(\mathrm{g})$

$\Delta_{\mathrm{a}} \mathrm{H}^{\ominus}(\mathrm{C})=715.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iii) $\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Cl}(\mathrm{g})$

$\Delta_{\mathrm{a}} \mathrm{H}^{\ominus}\left(\mathrm{Cl}_{2}\right)=242 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iv) $\mathrm{C}(\mathrm{g}) \rightarrow 4 \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{CCl}_{4}(\mathrm{~g})$

$\Delta_{\mathrm{f}} \mathrm{H}^{\ominus}\left(\mathrm{CCl}_{4}\right)=-135.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The enthalpy change for the given process $\mathrm{CCl}_{4}(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{Cl}(\mathrm{g})$, can be calculated using the following algebraic calculations as:

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

$\Delta \mathrm{H}=\Delta_{\mathrm{a}} \mathrm{H}^{\ominus}(\mathrm{C})+2 \Delta_{\mathrm{a}} \mathrm{H}^{\ominus}\left(\mathrm{Cl}_{2}\right)$

$-\Delta_{\text {vap }} \mathrm{H}^{\ominus}\left(\mathrm{CCl}_{4}\right)-\Delta_{\mathrm{f}} \mathrm{H}^{\ominus}\left(\mathrm{CCl}_{4}\right)$

$=\left(715.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)+2\left(242 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

$-\left(30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)-\left(-135.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

$\therefore \Delta \mathrm{H}=1304 \mathrm{~kJ} \mathrm{~mol}^{-1}$

in $\mathrm{CCl}_{4}(\mathrm{~g}) 4 \mathrm{C}-\mathrm{Cl}$ is present

so Bond enthalpy of $\mathrm{C}-\mathrm{Cl}$ bond in $\mathrm{CCl}_{4}(\mathrm{~g})=\frac{1304}{4} \mathrm{~kJ} \mathrm{~mol}^{-1}$

$=326 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Question 16. For an isolated system, $\Delta U=0$, what will be $\Delta S$ ?

Solution Since $\Delta U=0$

i.e. The Changes in internal energy $(\Delta \mathrm{U})$ for an isolated system is zero So; it does not exchange any energy in any form with the surroundings.

But, The entropy tends to increase in case, $\Delta \mathrm{S}$ will be positive, i.e. greater than zero

of spontaneous reaction.

Question 17. For the reaction at $298 \mathrm{~K}$,

$2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}$

$\Delta \mathrm{H}=400 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta \mathrm{S}=0.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

At what temperature will the reaction become spontaneous, considering $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ to be constant over the temperature range?

Solution: From the expression,

$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$

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

$\mathrm{T}=\frac{(\Delta \mathrm{H}-\Delta \mathrm{G})}{\Delta \mathrm{S}}$

$=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}} \quad(\Delta \mathrm{G}=0$ at equilibrium)

$=\frac{400 \mathrm{~kJ} \mathrm{~mol}^{-1}}{0.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}$

$\mathrm{T}=2000 \mathrm{~K}$

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

Question 18. For the reaction,

$2 \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})$, what are the signs of $\Delta \mathrm{H}$ and $\Delta \mathrm{S}$ ?

Solution: The given reaction $2 \mathrm{Cl}(\mathrm{g}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g}) ;$ the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, $\Delta \mathrm{H}$ 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.

Question 19. For the reaction

$2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightarrow 2 \mathrm{D}(\mathrm{g})$

$\Delta \mathrm{U}^{\ominus}=-10.5 \mathrm{~kJ}$ and $\Delta \mathrm{S}^{\ominus}=-44.1 \mathrm{JK}^{-1}$

Calculate $\Delta \mathrm{G}^{\ominus}$ for the reaction, and predict whether the reaction may occur spontaneously.

Solution: For the given reaction,

$2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightarrow 2 \mathrm{D}(\mathrm{g})$

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

$=-1$ mole

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

$\Delta \mathrm{H}^{\ominus}=\Delta \mathrm{U}^{\ominus}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$

$\Delta \mathrm{H}^{\ominus}=\Delta \mathrm{U}^{\ominus}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT}$

$=\left(-10.5 \times 10^{3} \mathrm{~J}\right)+(-1)\left(8.314 \mathrm{Jmol}^{-1}\right)(298 \mathrm{~K})$

$=-10500 \mathrm{~J}-2477.572 \mathrm{~J}$

$=12977.572 \mathrm{~J}$

$\Delta \mathrm{H}^{\ominus}=-12.98 \mathrm{~kJ}$

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

$\Delta \mathrm{G}^{\ominus}=\Delta \mathrm{H}^{\ominus}-\mathrm{T} \Delta \mathrm{S}^{\ominus}$

$=-12977.572 \mathrm{~J}-(298 \mathrm{~K})\left(-44.1 \mathrm{JK}^{-1}\right)$

$=-12977.572 \mathrm{~J}+1314.18 \mathrm{~J}$

$\Delta \mathrm{G}^{\ominus}=164.228 \mathrm{~J}$

$\Delta \mathrm{G}^{\ominus}=0.164228 \mathrm{k} \mathrm{J}$

Since $\Delta \mathrm{G}^{\ominus}$ for the reaction is positive so the reaction is non- spontaneously.

Question 20. The equilibrium constant for a reaction is $10 .$ What will be the value of $\Delta \mathrm{G}^{\ominus}$ ?

$\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~T}=300 \mathrm{~K}$

Solution: $\Delta \mathrm{G}=\Delta \mathrm{G}^{\ominus}+2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{eq}}$

At equilibrium $\Delta \mathrm{G}=0$ and $\mathrm{Q}=\mathrm{K}_{\mathrm{eq}}$

Hance $\Delta \mathrm{G}^{\ominus}=-2.303 \mathrm{RT} \log \mathrm{K}_{e q}$

$\Delta \mathrm{G}^{\ominus}$ for the reaction,

$=(2.303)\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)(300 \mathrm{~K}) \log 10$

$=-5744.14 \mathrm{~J} \mathrm{~mol}^{-1}$

$=-5.744 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Question 21. Comment on the thermodynamic stability of $\mathrm{NO}(\mathrm{g})$, given

$\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})$

$\rightarrow \mathrm{NO}(\mathrm{g}) ; \Delta_{\mathrm{r}} \mathrm{H}^{\ominus}=90 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{NO}_{2}(\mathrm{~g}): \Delta_{\mathrm{r}} \mathrm{H}^{\ominus}$

$=-74 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution: The positive value of $\Delta_{\mathrm{r}} \mathrm{H}$ indicates that heat is absorbed during the formation of $\mathrm{NO}(\mathrm{g})$. This means that $\mathrm{NO}(\mathrm{g})$ has higher energy than the reactants $\left(\mathrm{N}_{2}\right.$ and $\left.\mathrm{O}_{2}\right) .$ Hence, $\mathrm{NO}(\mathrm{g})$ is unstable. The negative value of $\Delta_{\mathrm{r}} \mathrm{H}$ indicates that heat is evolved during the

formation of $\mathrm{NO}_{2}(\mathrm{~g})$ from $\mathrm{NO}(\mathrm{g})$ and $\mathrm{O}_{2}(\mathrm{~g})$. The product, $\mathrm{NO}_{2}(\mathrm{~g})$ is stabilized with minimum energy.

Hence, unstable $\mathrm{NO}(\mathrm{g})$ changes to unstable $\mathrm{NO}_{2}(\mathrm{~g})$.

Question 22. Calculate the entropy change in surroundings when $1.00 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ is formed under standard conditions. $\Delta_{\mathrm{f}} \mathrm{H}^{\ominus}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$.

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

Entropy change of the surroundings $\left(\Delta \mathrm{S}_{\text {surr }}\right)=\frac{\text { qsur }}{\mathrm{T}(\mathrm{K})}$

$=\frac{286 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}}{298 \mathrm{~K}}$

$\therefore \Delta \mathrm{S}_{\text {surr }}=959.73 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$

Class 11 Chemistry Notes.

Class 11 Chemistry Book Chapterwise.

Class 11 Chemistry Exemplar Chapterwise.

If you have any Confusion related to NCERT Solutions for Class 11 Chemistry chapter 6 Thermodynamics PDF then feel free to ask in the comments section down below.

To watch Free Learning Videos on Class 11 Chemistry by Kota’s top IITian’s Faculties Install the eSaral App