NCERT Solutions For Class 11 Chemistry Chapter 6
NCERT Solutions for Class 11 Chemistry chapter 6 Thermodynamics PDF – eSaral
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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 $\left(\Delta \mathrm{S}_{\text {surr }}\right)=\frac{\text { qsur }}{\mathrm{T}(\mathrm{K})}$

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

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

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