Thermodynamics

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

$ \begin{aligned} \Delta \mathrm{U}^{\circ} & =-10 \mathrm{~kJ} / \mathrm{mol} \\ \Delta \mathrm{~S}^{\circ} & =-44 \mathrm{~J} / \mathrm{K} \\ \Delta \mathrm{H}^{\circ} & =\Delta \mathrm{U}^{\circ}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\ \Delta \mathrm{H}^{\circ} & =-10-\frac{1 \times 298 \times(8.31)}{1000}=-10-2.48=-12.48 \mathrm{~kJ} / \mathrm{mol} \\ \Delta \mathrm{G}^{\circ} & =\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{~S}^{\circ} \\ & =-12.48-\frac{298 \times(-44)}{1000} \\ & =-12.48+13.112 \\ & =+0.632 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $

Since $\Delta \mathrm{G}^{\circ}$ comes out to be positive, so given process is non-spontaneous.

From first law of thermodynamics

$ \begin{aligned} \Delta U & =q+w \\ q & =+500 \mathrm{~J} \\ w & =-200 \mathrm{~J} \\ \Delta U & =500-200 \\ & =300 \mathrm{~J} \end{aligned} $

$ \begin{aligned} &\text { During free expansion }\\ &\begin{aligned} & \Delta S_{\text {system }}=n R \ln \frac{V_2}{V_1} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2.303 \times 2 \times R \times \log \frac{100}{10} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=4.606 R \\ & \Delta S_{\text {surroundings }}=0 \text { as } q=0 \end{aligned} \end{aligned} $

$ \begin{aligned} &\mathrm{w}_1 \rightarrow \text { isothermal reversible process }\\ &\begin{aligned} & w_1=-n R T_1 \ln \frac{V_2}{V_1} \\ & w_1=-1 \times R \times T_1 \ln \frac{V_2}{V_1} \\ & w_3=-n \times R \times T_2 \ln \frac{V_4}{V_3} \\ & w_3=-1 \times R \times T_2 \ln \frac{V_4}{V_3} \\ & w_1+w_3=-n R T_1 \ln \left(\frac{V_2}{V_1}\right)-n R T_2 \ln \left(\frac{V_4}{V_3}\right) \\ & w_1+w_3=-2 T_1 \ln \left(\frac{V_2}{V_1}\right)-2 T_2 \ln \left(\frac{V_4}{V_3}\right) \end{aligned} \end{aligned} $

$\Delta \mathrm{S}=\frac{\Delta \mathrm{H}}{\mathrm{T}}$

$ \begin{aligned} & =\frac{666}{333} \\ & =2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \end{aligned} $

Let's consider the following equations:

Formation of $\mathrm{SO}_4^{2-}$:

$ \mathrm{S} + 2 \mathrm{O}_2 + 2 \mathrm{e}^{-} \longrightarrow \mathrm{SO}_4^{2-} \quad \Delta \mathrm{H}_{\mathrm{f}} = -216 \mathrm{kcal/mol} $

Crystallization of $\mathrm{BaSO}_4(\mathrm{s})$:

$ \mathrm{Ba}^{2+}(\mathrm{g}) + \mathrm{SO}_4^{2-}(\mathrm{g}) \longrightarrow \mathrm{BaSO}_4(\mathrm{s}) \quad \Delta \mathrm{H}_{\text{crystallisation}} = -4.5 \mathrm{kcal/mol} $

Formation of $\mathrm{BaSO}_4(\mathrm{s})$:

$ \mathrm{Ba} + \mathrm{S} + 2 \mathrm{O}_2 \longrightarrow \mathrm{BaSO}_4(\mathrm{s}) \quad \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{BaSO}_4) = -349 \mathrm{kcal/mol} $

We need to find the standard heat of formation for $ \mathrm{Ba}^{2+} $.

To determine this, apply the following relationship derived from the equations:

$ \mathrm{Ba}(\mathrm{s}) \longrightarrow \mathrm{Ba}^{2+}(\mathrm{g}) + 2 \mathrm{e}^{-} $

Using the data provided:

Start from equation (3), then subtract equations (1) and (2):

$ -349 - (-4.5) - (-216) $

Simplifying the expression:

$ -349 + 4.5 + 216 $

$ = -349 + 220.5 $

$ = -128.5 \, \mathrm{kcal/mol} $

This calculation shows that the standard heat of formation for $ \mathrm{Ba}^{2+} $ is $-128.5 \, \mathrm{kcal/mol}$.

$\Delta \mathrm{H}=-74.8 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$, it is an exothermic reaction.

So, accurate representation is

Since it is isothermal, $\Delta \mathrm{T}=0$

$\Delta \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}=0$

Since expansion is taking place against vacuum

$\begin{aligned} & P_{\text {ext }}=0 \\ & W=-P_{\text {ext }} \Delta V=0 \end{aligned}$

From first law of thermodynamics,

$\begin{aligned} & \Delta U=q+W \\ & 0=q+0 \\ & q=0 \end{aligned}$

For endothermic reactions,

$ \begin{aligned} & \Delta H_r=\text { positive (Heat is absorbed) } \\ & \Delta H=E_{a(f)}-E_{a(b)} \end{aligned} $

To determine the internal energy change (ΔU) for the reaction at $300 \mathrm{~K}$, we will use the relationship between enthalpy change (ΔH) and internal energy change:

$\Delta H = \Delta U + \Delta nRT$

where $\Delta H$ is the enthalpy change, $\Delta U$ is the internal energy change, $\Delta n$ is the change in moles of gas, $R$ is the universal gas constant, and $T$ is the temperature.

Given:

• $\Delta H = +15 \mathrm{~kJ} = 15000 \mathrm{~J}$
• $T = 300 \mathrm{~K}$
• $R = 8.314 \, \mathrm{J \, mol^{-1} \, K^{-1}}$
• $\Delta n = \text{moles of products} - \text{moles of reactants}$

From the balanced chemical equation:

$\mathrm{A}_2(\mathrm{~g}) + 3 \mathrm{B}_2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}_3(\mathrm{~g})$

Moles of reactants = 1 (for $\mathrm{A}_2$) + 3 (for $\mathrm{B}_2$) = 4 moles

Moles of products = 2 (for $\mathrm{AB}_3$)

Therefore, $\Delta n$ = 2 (products) - 4 (reactants) = -2

Now, substituting these values into the equation:

$\Delta H = \Delta U + \Delta nRT$

$15000 = \Delta U + (-2)(8.314)(300)$

Simplify the equation:

$15000 = \Delta U - 4988.4$

Therefore:

$\Delta U = 15000 + 4988.4 = 19988.4 \mathrm{~J}$

Hence, the internal energy change is:

Option A: $19988.4 \mathrm{~J}$

The concept of entropy in thermodynamics refers to the degree of randomness or disorder in a system. An increase in entropy is generally associated with processes in which disorder increases. Let's analyze each option listed:

A. A liquid evaporates to vapour.

During the evaporation of a liquid to form a vapour, the molecules of the substance move from a relatively ordered state (liquid) to a more disordered state (vapour). In a vapour, the molecules have more freedom of motion and are less confined than in a liquid. This transition from liquid to vapour increases the randomness or disorder of the system, hence, the entropy increases.

B. Temperature of a crystalline solid lowered from $130 \mathrm{~K}$ to $0 \mathrm{~K}$.

Reducing the temperature of a crystalline solid generally decreases the entropy of the system. As the temperature decreases, the molecular motion within the solid becomes more restricted, leading to a decrease in randomness. At absolute zero ($0 \mathrm{~K}$), the entropy is at its lowest possible value (ideally zero for a perfect crystal), as the molecular motion is minimized to only quantum mechanical vibrations.

C. $2 \mathrm{NaHCO}_{3(s)} \rightarrow \mathrm{Na}_2 \mathrm{CO}_{3(s)}+\mathrm{CO}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{(g)}$

In this chemical reaction, solid sodium bicarbonate decomposes to form solid sodium carbonate, carbon dioxide gas, and water vapor. The formation of gases from a solid significantly increases the entropy of the system because gases have much higher randomness due to their free molecular motion compared to solids.

D. $\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{Cl}_{(g)}$

The dissociation of chlorine gas ($\mathrm{Cl}_2$) into atomic chlorine ($\mathrm{Cl}$) represents a transition from a diatomic molecule to two separate atoms. This increases the number of particles in the gas phase, leading to increased randomness and disorder, hence, increasing the entropy.

Conclusion: Based on our analysis, processes A, C, and D involve an increase in entropy, as they each lead to greater disorder or randomness in the system. Option B is incorrect as it wrongly includes lowering the temperature of a solid as increasing entropy. Therefore, the correct answer is:

Option C :

A, C and D

The question involves matching different thermodynamic processes with their corresponding description or characteristics. Let's analyze these processes:

• Isothermal process (A): This process occurs at a constant temperature. Hence, A should match with "Carried out at constant temperature" (II).
• Isochoric process (B): This process occurs at a constant volume. Therefore, B should be matched with "Carried out at constant volume" (III).
• Isobaric process (C): This process happens at a constant pressure, making C correspond to "Carried out at constant pressure" (IV).
• Adiabatic process (D): During this process, there is no heat exchange with the surroundings. Consequently, D should be paired with "No heat exchange" (I).

Reviewing the options:

• Option A: A-IV, B-III, C-II, D-I
• Option B: A-IV, B-II, C-III, D-I
• Option C: A-I, B-II, C-III, D-IV
• Option D: A-II, B-III, C-IV, D-I

The correct matches, as elaborated above, are A-II, B-III, C-IV, D-I. Therefore, the correct answer here is Option D.

The work done in a reversible isothermal process can be calculated using the formula:

$ W = -nRT \ln\left(\frac{V_f}{V_i}\right) $

Here:

• $ W $ is the work done by the gas.
• $ n $ is the number of moles of the gas.
• $ R $ is the universal gas constant.
• $ T $ is the temperature in Kelvin.
• $ V_i $ and $ V_f $ are the initial and final volumes of the gas, respectively.

However, since the volumes are not directly provided but the pressures are given, we use the ideal gas law, $ PV = nRT $, to relate pressures and volumes at the same temperature and amount of gas:

For an ideal gas undergoing a change at constant temperature, we can also write the work done in terms of the initial and final pressures:

$ W = -nRT \ln\left(\frac{P_i}{P_f}\right) $

where:

• $ P_i $ and $ P_f $ are the initial and final pressures, respectively.

Given:

• $ n = 1 $ (one mole of hydrogen)
• $ T = 25^{\circ} \mathrm{C} = 25 + 273.15 = 298.15 \, \mathrm{K} $
• $ R = 2.0 \, \mathrm{cal} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} $
• $ P_i = 20 \, \mathrm{atm} $
• $ P_f = 10 \, \mathrm{atm} $

Substituting all values into the formula:

$ W = -1 \cdot 2.0 \cdot 298.15 \ln\left(\frac{20}{10}\right) $

$ W = -2.0 \cdot 298.15 \cdot \ln(2) $

Now, solving this using the value of $ \ln(2) \approx 0.693 $:

$ W \approx -2.0 \cdot 298.15 \cdot 0.693 $

$ W \approx -413.14 \, \text{calories} $

Thus, the work done during the process is about $ -413.14 \, \text{calories} $, indicating that this amount of energy was done by the system (expansion work being done by the gas against external pressure), and hence is negative as it indicates work done by the system. Therefore, the correct option is:

Option B: $ -413.14 \, \text{calories} $

Decomposition for 1 mole of water

$\begin{aligned} & \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) ; \Delta \mathrm{H}=+\frac{483.64}{2} \\ & \Delta \mathrm{H}=+241.82 \mathrm{~kJ} \end{aligned}$

$\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$

$ \begin{aligned} {[A] } & =2 \mathrm{~mol} \mathrm{~L}^{-1} \\ {[B] } & =3 \mathrm{~mol} \mathrm{~L}^{-1} \\ {[C] } & =10 \mathrm{~mol} \mathrm{~L}^{-1} \\ {[D] } & =6 \mathrm{~mol} \mathrm{~L}^{-1} \\ \Delta \mathrm{G}^{0} & =-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{eq}} \\ & =-2.303 \mathrm{RT} \log \frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]} \\ & =-2.303 \times 2 \times 300 \times \log \frac{10 \times 6}{2 \times 3} \\ & =-2.303 \times 2 \times 300 \times \log 10 \\ & =-1381.8 \mathrm{~cal} \end{aligned} $

$\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n_g R T}$

$\Delta$U = nC_v$\Delta$T

For isothermal condition; $\Delta$T = 0

$\therefore$ $\Delta$U = 0

At constant temperature and amount

${P_1}{V_1} = {P_2}{V_2}$

${P_1}{V_1} = {{{P_1}} \over 3}{V_2}$ [$\therefore$ ${P_2} = {{{P_1}} \over 3}$]

${V_2} = 3{V_1}$

mole of ${O_2}(g) = {{3.2} \over {32}} = 0.1$ mole

Volume of ${O_2}(g) = (0.1 \times 22.4)$ L = 2.24 L

At STP (V₁)

${V_2} = 3{V_1} = 3 \times 2.24$

= 6.72 L

Work done under any thermodynamic process can be determined by area under the 'p-V' graph. As it can be observed maximum area is covered in option '2'.

1.

2.

3.

4.

To determine the heat released when 35.2 g of CO₂ is formed from carbon and oxygen gas, we will follow these steps:

1. Calculate the number of moles of CO₂:

The molar mass of CO₂ can be calculated as follows:

Carbon (C): 12.01 g/mol

Oxygen (O): 16.00 g/mol

So, the molar mass of CO₂ is:

$12.01 + (16.00 \times 2) = 12.01 + 32.00 = 44.01 \, \text{g/mol}$

The number of moles of CO₂ in 35.2 g is calculated using:

$\text{Number of moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{35.2 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.8 \, \text{moles}$

2. Calculate the heat released for 0.8 moles of CO₂:

The heat of combination given is -393.5 kJ/mol. This is the energy released when 1 mole of CO₂ is formed:

$\text{Heat released} = \text{Number of moles} \times \text{Heat of combination}$

$\text{Heat released} = 0.8 \, \text{moles} \times -393.5 \, \text{kJ/mol} = -314.8 \, \text{kJ}$

After rounding, the heat released is approximately -315 kJ.

Answer: Option D $-$ 315 kJ.