Chemical Thermodynamics – NEET UG ChemistryPractice Questions & PYQs

The feasibility of a chemical process at constant temperature and pressure depends on the Gibbs free energy change, determined by the relation $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$. Before calculating this, the enthalpy change must be determined from the internal energy change using the formula $\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT$, where $\Delta n_g$ is the difference in stoichiometric coefficients between gaseous products and gaseous reactants. In this reaction, $\Delta n_g$ is calculated as $2 - (2 + 1)$, which equals $-1$. Using the provided values, the enthalpy change is $\Delta H^{\ominus} = -10 \text{ kJ mol}^{-1} + (-1)(8.31 \text{ J K}^{-1} \text{ mol}^{-1})(298 \text{ K}) \times 10^{-3} \text{ kJ J}^{-1}$, simplifying to $-10 - 2.47638$, or $-12.47638 \text{ kJ mol}^{-1}$.

The entropic component, $T\Delta S^{\ominus}$, is found by multiplying 298 K by $-44 \text{ J K}^{-1} \text{ mol}^{-1}$, yielding $-13112 \text{ J mol}^{-1}$, which is equivalent to $-13.112 \text{ kJ mol}^{-1}$. Substituting these values back into the Gibbs equation gives $\Delta G^{\ominus} = -12.47638 \text{ kJ mol}^{-1} - (-13.112 \text{ kJ mol}^{-1})$, resulting in $+0.63562 \text{ kJ mol}^{-1}$, which is approximately $+0.63568 \text{ kJ mol}^{-1}$. Since the resulting Gibbs free energy change is positive, the reaction is non-spontaneous at 298 K.

The work done by an ideal gas during a reversible isothermal process is calculated using the formula $w = -nRT \ln(V_f/V_i)$. Given $n = 1$ mol and $R = 2$ cal K⁻¹ mol⁻¹, we can determine the work for each isothermal process individually.

For process 1, which occurs at a constant temperature $T_1$ with the volume changing from $V_1$ to $V_2$, the work done is $w_1 = -1 \times 2 \times T_1 \times \ln(V_2/V_1) = -2T_1 \ln(V_2/V_1)$. Similarly, for process 3, which is isothermal at temperature $T_2$ with the volume changing from $V_3$ to $V_4$, the work done is $w_3 = -1 \times 2 \times T_2 \times \ln(V_4/V_3) = -2T_2 \ln(V_4/V_3)$. Summing the work done in these two processes results in the expression:

$w_1 + w_3 = -2T_1 \ln \frac{V_2}{V_1} - 2T_2 \ln \frac{V_4}{V_3}$

The First Law of Thermodynamics relates the change in internal energy of a system to the heat transferred and the work performed, expressed as $\Delta U = q + w$. According to standard conventions, heat absorbed by a system is positive, while work done by a system on its surroundings is negative because it represents energy extracted from the system.

Since 500 J is absorbed as heat, the heat term $q$ is 500 J. Because the system performs 200 J of work on the surroundings, the work term $w$ is -200 J. Substituting these values into the formula provides
$\Delta U = 500 - 200 = 300$
confirming the resulting change in internal energy is 300 J.

Free expansion describes an ideal gas expanding into a vacuum, a condition where the system does no work and exchanges zero heat with its environment. Because this process is adiabatic and involves no interaction with the surroundings, there is no heat transfer to cause any change in the entropy of the surroundings, making $\Delta S_{\text{surroundings}} = 0$.

For the system, the change in entropy is calculated using the formula $\Delta S_{\text{system}} = nR \ln \frac{V_2}{V_1}$. Substituting the given values of $n = 2$ moles, an initial volume of $10$ L, and a final volume of $100$ L results in $\Delta S_{\text{system}} = 2R \ln \frac{100}{10}$, which simplifies to $2R \ln(10)$. Using the standard value $\ln(10) \approx 2.303$, the calculation becomes $2 \times 2.303 R$, yielding $\Delta S_{\text{system}} = 4.606 R$.

When a reversible process like protein denaturation reaches equilibrium, the standard Gibbs free energy change relates to the equilibrium constant through the expression $\Delta G^\circ = -RT \ln K$. Because the concentrations of the native and denatured states are equal, the equilibrium constant is 1, which makes the natural logarithm of 1 equal to zero, thereby setting the standard Gibbs free energy change to zero. Under these conditions, the thermodynamic relationship $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ = 0$ simplifies to $\Delta H^\circ = T \Delta S^\circ$, which indicates that the entropy change is directly proportional to the enthalpy change divided by the absolute temperature.

To perform the calculation, convert the temperature from 60 °C to the absolute scale by adding 273, resulting in 333 K. Using the expression $\Delta S^\circ = \frac{\Delta H^\circ}{T}$, substitute the standard enthalpy change of 666 kJ mol^-1 and the absolute temperature of 333 K into the equation. Dividing the enthalpy value by the temperature yields a standard entropy change of 2.0 kJ K^-1 mol^-1.

The total enthalpy of formation of a compound remains constant regardless of the path taken, a principle known as Hess's Law. We can relate the formation of the solid compound $\text{BaSO}_4(s)$ to the formation enthalpies of its individual components to determine the standard heat of formation of $\text{Ba}^{2+}$. The enthalpy change for the formation of $\text{BaSO}_4(s)$ is -349 kcal/mol. This process reflects the formation of the $\text{SO}_4^{2-}(aq)$ ion, with a formation enthalpy of -216 kcal/mol, and the subsequent combination of the $\text{Ba}^{2+}$ and $\text{SO}_4^{2-}$ ions, which releases 4.5 kcal/mol of heat upon crystallization.

By expressing the standard heat of formation of $\text{Ba}^{2+}$ using the relationship between the solid formation enthalpy and the ionic components, we define the following equation:
$\Delta H_f (\text{Ba}^{2+}) = \Delta H_f (\text{BaSO}_4) - \Delta H_f (\text{SO}_4^{2-}) - \Delta H_{\text{crystallization}}$
We determine the unknown value by subtracting the formation energy of the sulfate ion and the energy released during crystallization from the total formation energy of the barium sulfate solid. Substituting the given values into the formula:
$\Delta H_f (\text{Ba}^{2+}) = -349 - (-216) - (-4.5)$
This calculation results in -128.5 kcal/mol for the standard heat of formation.

The reaction $\mathrm{C}(\mathrm{s}) + 2\mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{CH}_4(\mathrm{g})$ has a $\Delta H$ value of -74.8 kJ mol$^{-1}$, characterizing it as an exothermic process. In exothermic reactions, the chemical system releases energy to its surroundings, meaning the potential energy of the product, $\mathrm{CH}_4$, is lower than the potential energy of the reactants, $\mathrm{C}$ and $2\mathrm{H}_2$. Consequently, an accurate energy diagram must depict the energy level of the reactants positioned above that of the products to illustrate the net release of heat.

When an ideal gas expands into a vacuum, the external pressure $P_{ext}$ is zero, which makes the mechanical work done by the system, calculated as $W = -P_{ext} \Delta V$, equal to zero. Because the process is isothermal, the temperature remains constant, implying that the change in internal energy $\Delta U$ for an ideal gas is also zero. Based on the first law of thermodynamics, expressed as:
$\Delta U = q + W$
since both the change in internal energy and the work done are zero, it follows that the heat exchanged $q$ must also be zero.

The work done during the reversible isothermal expansion of an ideal gas is determined by the logarithmic relationship between the initial and final pressure states. Because the gas expands and performs work on the surroundings, the calculated value is negative according to standard thermodynamic conventions. Before calculating the work, the temperature must be converted from Celsius to Kelvin, which results in 298 K. The mathematical expression for the work done in this process is given by the equation:
$W = -2.303 n R T \log \left( \frac{P_1}{P_f} \right)$
Substituting the known values of $n = 1$ mole, $R = 2 \text{ cal } K^{-1} \text{ mol}^{-1}$, $T = 298 \text{ K}$, $P_1 = 20 \text{ atm}$, and $P_f = 10 \text{ atm}$ into this formula leads to the following expression:
$W = -2.303 \times 1 \times 2 \times 298 \times \log \left( \frac{20}{10} \right)$
This simplifies to an expression involving the logarithm of 2. Using the given approximation that $\log(2) = 0.3$, the calculation becomes $W = -2.303 \times 1 \times 2 \times 298 \times 0.3$, which results in a final value of -413.14 calories.

Thermodynamic processes are classified based on the specific physical parameter that remains constant throughout the change of state. An isothermal process operates at a fixed temperature, whereas an isochoric process is characterized by maintaining a constant volume. An isobaric process occurs under constant pressure conditions, and an adiabatic process is distinct because it involves no exchange of heat ($q=0$) between the system and its surroundings.

Entropy serves as a measure of a system's disorder, where processes that increase particle motion or the number of independent particles typically lead to a higher entropy state. Phase transitions like liquid evaporation clearly increase disorder, and chemical decompositions such as $2\,\mathrm{NaHCO}_{3(\mathrm{s})} \rightarrow \mathrm{Na}_2\mathrm{CO}_{3(\mathrm{s})} + \mathrm{CO}_{2(\mathrm{g})} + \mathrm{H}_2\mathrm{O}_{(\mathrm{g})}$ also increase entropy because they generate multiple gaseous products from a solid reactant. Likewise, the dissociation $\mathrm{Cl}_{2(\mathrm{g})} \rightarrow 2\,\mathrm{Cl}_{(\mathrm{g})}$ promotes randomness by increasing the mole count of gaseous species, whereas cooling a crystalline solid from $130\,\text{K}$ to $0\,\text{K}$ restricts motion and forces the system toward a more ordered arrangement, decreasing entropy.

The relationship between enthalpy change and internal energy change for reactions involving gaseous species is defined by the equation:
$\Delta H = \Delta U + \Delta n_g R T$
To determine the internal energy change, we must first calculate the change in the number of gaseous moles, $\Delta n_g$, which is the difference between the sum of the stoichiometric coefficients of the gaseous products and the gaseous reactants. For the given reaction, there are two moles of gaseous product and a total of four moles of gaseous reactant, resulting in $\Delta n_g = 2 - 4 = -2$.

Because the enthalpy change is provided as $15 \, \mathrm{kJ}$, it must be converted to $15000 \, \mathrm{J}$ to remain consistent with the units of the universal gas constant, $R = 8.314 \, \mathrm{J/K \cdot mol}$. Rearranging the fundamental thermodynamic relation to solve for the internal energy change yields:
$\Delta U = \Delta H - \Delta n_g R T$
Substituting the known values into this expression gives $\Delta U = 15000 - (-2 \times 8.314 \times 300)$. Evaluating this product leads to $15000 + 4988.4$, resulting in a final value of $19988.4 \, \mathrm{J}$.

Endothermic reactions are characterized by the absorption of heat from the surroundings, which indicates that the system undergoes a positive change in enthalpy. Because the heat transferred at constant pressure is defined as $q_p = \Delta_r H$, the uptake of energy necessitates that both $q_p$ and $\Delta_r H$ are positive values. This signifies that the system gains thermal energy throughout the process, confirming that both the enthalpy change and the heat term must be positive.

Enthalpy is fundamentally defined as the sum of a system's internal energy and its pressure-volume product, expressed as $H = U + PV$. In the context of chemical reactions involving gaseous components, the change in enthalpy accounts for the work done during expansion or compression, which is mathematically represented by the term $\Delta (PV)$. By applying the ideal gas law where $\Delta (PV)$ equals $\Delta n_g R T$, we arrive at the standard thermodynamic relation $\Delta H = \Delta U + \Delta n_g R T$.

Enthalpy change is an extensive property, meaning the energy associated with a chemical process depends directly on the stoichiometric amounts of the substances involved. The provided equation $2 \mathrm{H}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{H}_2\mathrm{O}(\mathrm{g})$ has a $\Delta_r H^{\circ} = -483.64$ kJ, representing the heat released when forming two moles of water. To find the enthalpy change for the decomposition of one mole of water, we must reverse the reaction and halve the coefficients, which requires changing the sign of the energy to positive and dividing the magnitude by two. Applying this logic, the enthalpy change is 241.82 kJ.

The hybridization of a carbon atom is determined by its number of electron domains, where three regions of electron density correspond to $sp^2$ hybridization and four regions correspond to $sp^3$ hybridization. Following IUPAC nomenclature priority, the carbonyl carbon within the ester functional group is assigned as $C_1$. This $C_1$ atom forms a double bond with oxygen and two single bonds with other atoms, resulting in three electron domains, which defines its state as $sp^2$. The subsequent carbon, $C_2$, is a methylene unit that forms four distinct single bonds, establishing four electron domains and assigning it the $sp^3$ hybridization.

In thermodynamics, the work performed during a gas process is represented by the geometric area underneath the curve on a $P-V$ diagram. By examining the extent of the shaded region bounded by each path and the horizontal volume axis, it becomes clear that the curve in option B encompasses the largest area among the choices provided. Because a larger area directly corresponds to a higher quantity of work, this specific path signifies the state where the system performs maximum work.

For an ideal gas, the internal energy is solely a function of temperature, which implies that any change in internal energy depends entirely on the change in temperature. In an isothermal process, the temperature remains constant throughout the entire expansion from 1 L to 10 L, meaning that $\Delta T = 0$. Since the relationship for internal energy change is defined by $\Delta U = n C_v \Delta T$, substituting a value of zero for the temperature change yields $\Delta U = 0$ J. Therefore, the initial temperature of 300 K does not lead to any change in the system's internal energy during this process.

The internal energy of an ideal gas depends strictly on its temperature, described by the relation $\Delta U = n C_v \Delta T$; consequently, since an isothermal process requires $\Delta T = 0$, the change in internal energy is necessarily $\Delta U = 0$. Furthermore, because any irreversible expansion is a spontaneous process, the second law of thermodynamics dictates that the total entropy of the universe must increase, implying that $\Delta S_{\text{total}} \neq 0$.

The relationship between molar heat capacities at constant pressure ($C_{P}$) and constant volume ($C_{V}$) for an ideal gas stems from the thermodynamic definition of enthalpy, defined as $\Delta H = \Delta U + \Delta(PV)$. For one mole of an ideal gas, the change in internal energy is $\Delta U = C_{V} \Delta T$, the change in enthalpy is $\Delta H = C_{P} \Delta T$, and the ideal gas law allows us to express the change in the product of pressure and volume as $\Delta(PV) = R \Delta T$. Substituting these values into the enthalpy expression yields:
$C_{P} \Delta T = C_{V} \Delta T + R \Delta T$
Dividing this entire equation by $\Delta T$ confirms the fundamental relationship $C_{P} - C_{V} = R$.