Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

The net work done by an ideal gas during a cyclic process on a pressure-volume diagram is equivalent to the area enclosed by the cycle. Because the transformation traces a triangular path on the coordinate plane, the work is calculated using the geometric relation $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$.

By observing the plot, we find the base spans the volume interval of $30 - 10$ and the height spans the pressure interval of $30 - 10$. Substituting these values into the area formula gives $W = \frac{1}{2} \times (30 - 10) \times (30 - 10) = 200$. Since the product of kilopascals and cubic decimeters results in standard units of energy, the total work done in the process is 200 J.

The enthalpy of neutralization remains constant for strong acids and bases because the reaction in an aqueous medium simplifies to the same fundamental process regardless of the specific chemicals involved. Since strong acids and bases completely dissociate, the net reaction is always the combination of one mole of $\mathrm{H}^+$ ions with one mole of $\mathrm{OH}^-$ ions to form one mole of water, which invariably releases -57 kJ/mol of energy. This underlying reaction mechanism explains why the assertion holds true, as the specific identity of the strong monobasic acid or monoacidic base does not alter the thermodynamic outcome of the neutralization.

The first law of thermodynamics states that the change in internal energy is equal to the sum of the heat transferred and the work done, represented by the equation
$\Delta U = Q + W$
For an ideal gas undergoing an isothermal process, the internal energy remains constant, meaning $\Delta U = 0$. Consequently, the expression simplifies to $Q = -W$, which indicates that the heat transferred to the system is exactly equal to the work done during the expansion.

Because the gas expands against a constant opposing pressure, the process is irreversible and the work done is given by
$W = -P_{\text{ext}} \Delta V$
The external pressure $P_{\text{ext}}$ is 80 kPa, which is $80 \times 10^3$ Pa. The change in volume is the difference between the final and initial states, $45 \text{ dm}^3 - 30 \text{ dm}^3 = 15 \text{ dm}^3$, which converts to $15 \times 10^{-3} \text{ m}^3$. Substituting these values into the work equation gives $W = -(80 \times 10^3) \times (15 \times 10^{-3}) = -1200$ J. Since $Q = -W$, the total heat transferred is 1200 J.

Hess's Law allows the enthalpy of hydration to be determined by recognizing that the heat of solution of the anhydrous salt is equivalent to the sum of the heat of hydration and the heat of solution of the hydrated salt. The relationship is expressed by the equation $-70\ \mathrm{kJ\ mol}^{-1} = \Delta H_{\text{hydration}} + 12\ \mathrm{kJ\ mol}^{-1}$. Subtracting $12\ \mathrm{kJ\ mol}^{-1}$ from both sides yields $\Delta H_{\text{hydration}} = -82\ \mathrm{kJ\ mol}^{-1}$. With the heat of hydration defined as $-x\ \mathrm{kJ}$, equating the terms leads to $-x = -82$, confirming that $x$ is 82.

In a reversible isothermal expansion of an ideal gas, the work done reflects the energy change associated with volume expansion at a constant temperature. The relationship between the system properties and the work performed is given by the formula:

$w = -2.303 nRT \log \left( \frac{V_2}{V_1} \right)$

Substituting the values 5 moles for $n$, 8.314 J K^-1 mol^-1 for $R$, 300 K for $T$, 10 L for $V_1$, and 100 L for $V_2$ into the expression leads to:

$w = -2.303 \times 5 \times 8.314 \times 300 \times \log \left( \frac{100}{10} \right)$

Because $\log(10)$ is 1, the calculation simplifies to -2.303 * 5 * 8.314 * 300. Multiplying these factors results in a total work value of -28720.713 J. By matching this result to the expression $-x \text{ J}$, the value of $x$ is 28721.

A chemical reaction proceeds spontaneously when the change in Gibbs free energy, $\Delta G$, is negative, a condition determined by the relationship $\Delta G = \Delta H - T\Delta S$. The reaction reaches equilibrium at the point where $\Delta G = 0$, which defines the threshold temperature through the expression $T = \frac{\Delta H}{\Delta S}$. Substituting the provided enthalpy of 400 kJ mol^-1 and entropy of 0.2 kJ mol^-1 K^-1 into this formula yields $T = \frac{400}{0.2} = 2000$ K. Because both enthalpy and entropy are positive, the entropy term outweighs the enthalpy term to make the reaction spontaneous only when the temperature rises above this value.

Hess's Law states that the total enthalpy change for a chemical process remains constant regardless of the path taken, allowing us to compute unknown heats of reaction by combining known equations. To construct the target reaction $3\,\mathrm{C}_{(s)} + \mathrm{Fe}_2\mathrm{O}_{3(s)} \rightarrow 2\,\mathrm{Fe}_{(s)} + 3\,\mathrm{CO}_{(g)}$, the provided thermochemical data must be manipulated algebraically.

Reversing the first equation places $\mathrm{Fe}_2\mathrm{O}_{3(s)}$ on the reactant side and changes the enthalpy sign to $+822$ kJ/mol. Multiplying the second equation by $3$ provides the required coefficients for $\mathrm{C}_{(s)}$ and $\mathrm{CO}_{(g)}$, resulting in an enthalpy change of $3 \times (-110) = -330$ kJ/mol. Combining these processed equations eliminates the oxygen molecules from both sides of the final expression, and the sum of the resulting enthalpy values, $822 + (-330)$, produces a total of $492$ kJ/mol.

The standard Gibbs free energy change is related to the equilibrium constant by the equation $\Delta G^{\circ} = -RT \ln K$. Substituting the given values of 300 K for temperature, 8.314 J K^-1 mol^-1 for the gas constant, and 10 for the equilibrium constant produces an energy magnitude of 5744.14 J mol^-1. Converting this result into kilojoules per mole gives 5.74414 kJ mol^-1. To express this value in terms of $10^{-1}$ kJ mol^-1, we shift the decimal to obtain 57.4414, identifying 57 as the required integer value.

Free expansion implies that a gas expands into a vacuum, meaning the external pressure is zero and consequently no work is performed, resulting in $w = 0$. Since the process is adiabatic, the system is thermally isolated and does not exchange heat with its surroundings, so $q = 0$. According to the first law of thermodynamics, $\Delta U = q + w$, and substituting these values yields a change in internal energy of $\Delta U = 0$. Because the internal energy of an ideal gas is exclusively a function of temperature, this lack of change in internal energy dictates that the temperature remains constant, meaning $\Delta T = 0$.

For any adiabatic expansion of an ideal gas, the heat exchange is zero, which means the change in internal energy equals the work done by the system, expressed as $\Delta U = w$, or $n C_V \Delta T = -P_{\text{ext}} \Delta V$. Because the volume doubles, the change in volume $\Delta V$ is exactly equal to the initial volume $V_1$. Using the ideal gas law $P_1 V_1 = n R T_1$, we can substitute $V_1$ with $\frac{n R T_1}{P_1}$ into the work expression, where the work done becomes $-P_{\text{ext}} (\frac{n R T_1}{P_1})$.

By substituting these expressions into the first law equation, we obtain $\frac{5}{2} n R (T_2 - T_1) = -P_{\text{ext}} (\frac{n R T_1}{P_1})$. We can cancel the moles $n$ and the gas constant $R$ from both sides, leaving $\frac{5}{2} (T_2 - T_1) = -P_{\text{ext}} (\frac{T_1}{P_1})$. Plugging in the initial values of $T_1 = 298$, $P_1 = 5$, and $P_{\text{ext}} = 1$, the expression becomes $\frac{5}{2} (T_2 - 298) = -1 (\frac{298}{5})$. Solving for $T_2$ gives $T_2 - 298 = -23.84$, resulting in a final temperature of 274 K when rounded to the nearest integer.

The connection between the heat of combustion at constant pressure, which represents the enthalpy change $\Delta H$, and the heat of combustion at constant volume, representing the internal energy change $\Delta U$, relies on the equation $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ represents the change in the number of gaseous moles. For the combustion of solid benzoic acid, the stoichiometry is defined by $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}(\mathrm{s}) + \frac{15}{2} \mathrm{O}_2(\mathrm{g}) \longrightarrow 7 \mathrm{CO}_2(\mathrm{g}) + 3 \mathrm{H}_2\mathrm{O}(\ell)$, which indicates that carbon dioxide and oxygen are the only gases involved in the reaction. Calculating the difference between the moles of gaseous products and gaseous reactants gives $\Delta n_g = 7 - 7.5 = -0.5$. Converting the temperature 27 °C to 300 K and substituting these values into the thermodynamic relationship results in $\Delta H = -321.30 + (-0.5) \times R \times 300$, which simplifies to $\Delta H = -321.30 - 150R$, meaning the value of $x$ is 150.

The thermodynamic criteria for spontaneity are dictated by the Gibbs Free Energy change, denoted as $\Delta G$. A spontaneous process naturally proceeds with a negative $\Delta G$, whereas a reversible process at equilibrium is characterized by $\Delta G = 0$. Conversely, non-spontaneous processes require an input of work and result in a positive $\Delta G$. Therefore, stating that a spontaneous reaction produces a positive $\Delta G$ contradicts the fundamental principles of thermodynamics.

For an ideal gas undergoing an isothermal process, the internal energy remains constant, which implies $\Delta U = 0$. By the first law of thermodynamics, $\Delta U = Q + W$, so the heat exchange is equivalent to the negative of the work, $Q = -W$. In this case, we have $3\ \mathrm{moles}$ of gas undergoing an irreversible compression against a constant external pressure of $5\ \mathrm{atm}$, which dictates that the work done on the system follows the expression $W = -P_{\text{ext}}(V_2 - V_1)$. Substituting the provided values of $60\ \mathrm{L}$ for the initial volume and $20\ \mathrm{L}$ for the final volume into the formula gives $W = -5 \times (20 - 60)$, resulting in a work value of $200\ \mathrm{L\cdot atm}$. Consequently, because $Q = -W$, the heat exchange is $-200\ \mathrm{L\cdot atm}$.

The standard Gibbs free energy change relates to the equilibrium constant through the thermodynamic equation:

$\Delta_{\mathrm{r}}G^{\Theta} = -RT \ln K_P$

With the gas constant R at $8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}$ and the temperature at 298 K, substituting these values into the equation gives $\Delta_{\mathrm{r}}G^{\Theta} = -8.314 \times 10^{-3} \times 298 \times \ln(2.47 \times 10^{-29})$. Evaluating the logarithmic term results in -65.87, and performing the multiplication yields 163.19 kJ. Rounding this provides 163 kJ.

The relationship between the enthalpy of vaporization and the internal energy of vaporization is expressed by the equation $\Delta H_{\text{vap}}^{\ominus} = \Delta U_{\text{vap}}^{\ominus} + \Delta n_g RT$, which accounts for the work done by the system during the expansion phase change. For the vaporization of water, the change in gaseous moles $\Delta n_g$ is 1.

At 373.15 K, substituting the enthalpy change of 40.79 kJ/mol into the equation gives:
$40.79 = \Delta U_{\text{vap}}^{\ominus} + \frac{1 \times 8.3 \times 373.15}{1000}$
The energy consumed as expansion work is 3.0971 kJ/mol. Subtracting this expansion work from the total enthalpy yields $\Delta U_{\text{vap}}^{\ominus} = 40.79 - 3.0971 = 37.6929 \;\mathrm{kJ} \;\mathrm{mol}^{-1}$. When rounded to the nearest integer, the change in internal energy is 38 kJ/mol.

When a substance undergoes a phase transition at constant pressure and temperature, the process occurs at thermodynamic equilibrium, where the change in Gibbs free energy is equal to zero. This relationship is defined by the equation $\Delta G = \Delta H - T\Delta S = 0$, which allows the melting point to be calculated using the expression $T = \frac{\Delta H}{\Delta S}$. To perform this calculation, first convert the enthalpy of fusion from 30.4 kJ/mol into 30400 J/mol to maintain consistency with the entropy units. Dividing this enthalpy value by the entropy change of 28.4 J K⁻¹ mol⁻¹ yields a melting temperature of 1070.42 K. Rounding this result to the nearest integer, the melting point of sodium chloride is 1070 K.

The heat released by the reaction of solid rocket fuel depends on the enthalpy difference between the products and the reactants. By analyzing the balanced chemical equation $\mathrm{Fe}_2\mathrm{O}_3 + 2\mathrm{Al} \rightarrow \mathrm{Al}_2\mathrm{O}_3 + 2\mathrm{Fe}$, we determine the total energy change using standard enthalpies of formation. Subtracting the reactant enthalpy from the product enthalpy gives the total energy released per mole of the reaction, which is expressed by the equation $\Delta H_r = -1700 - (-840) = -860\ \mathrm{KJ}$.

Determining the energy per unit mass requires calculating the total mass of the fuel mixture in the stated ratio. One mole of $\mathrm{Fe}_2\mathrm{O}_3$ has a molar mass of 160 g, while two moles of Al account for $2 \times 27 = 54\ \mathrm{g}$, resulting in a combined mass of 214 g. Dividing the total heat evolved by the total mixture mass leads to $\frac{860}{214} \approx 4.01\ \mathrm{KJ/g}$, which rounds to the nearest integer of 4.

Hess's Law states that the enthalpy change of a reaction remains constant regardless of the reaction pathway. To determine the enthalpy change for the reaction $\mathrm{H}_2\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g}) + \mathrm{OH}(\mathrm{g})$, we begin by reversing the provided formation equation for water, $\mathrm{H}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{g})$, which changes the sign of the enthalpy from -242 kJ mol⁻¹ to +242 kJ mol⁻¹. Since we require the breakdown of two moles of water to match the stoichiometry of the radical formation data, we double this value to get 484 kJ mol⁻¹.

Next, we identify the enthalpy required to produce the radical fragments by adding the enthalpy of the reaction $\mathrm{H}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{OH}(\mathrm{g})$, which is 78 kJ mol⁻¹, to the enthalpy of the reaction $\mathrm{H}_2(\mathrm{g}) \rightarrow 2 \mathrm{H}(\mathrm{g})$, which is 436 kJ mol⁻¹. This addition results in an enthalpy of 514 kJ mol⁻¹ for the process where elements form 2 moles of hydrogen atoms and 2 moles of hydroxyl radicals. By summing the energy for the reversed water formation, 484 kJ mol⁻¹, and the energy for radical formation, 514 kJ mol⁻¹, we find the total enthalpy change for the stoichiometric process $\mathrm{2H}_2\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{2H}(\mathrm{g}) + \mathrm{2OH}(\mathrm{g})$ is 998 kJ mol⁻¹. Dividing this result by two provides the enthalpy change of 499 kJ mol⁻¹ for the desired single-molecule reaction.

Standard Gibbs free energy change is linked to enthalpy and entropy through the expression $\Delta G^0 = \Delta H^0 - T \Delta S^0$, while its relationship with the equilibrium constant $K$ is defined by $\Delta G^0 = -2.303 RT \log K$. Converting the enthalpy change to Joules gives $\Delta H^0 = -54070 \text{ J mol}^{-1}$. By substituting the temperature of $298 \text{ K}$ and the entropy change of $10 \text{ J K}^{-1} \text{ mol}^{-1}$, the free energy is calculated as $\Delta G^0 = -54070 - (298 \times 10)$, which equals $-57050 \text{ J mol}^{-1}$.

Substituting this result into the equilibrium expression leads to $-57050 = -2.303 \times 8.314 \times 298 \times \log K$. Utilizing the provided value $2.303 \times 8.314 \times 298 = 5705$, the expression simplifies to $-57050 = -5705 \times \log K$. Solving for $\log K$ by dividing $-57050$ by $-5705$ results in $10$.

The internal energy of an ideal gas depends exclusively on its temperature, which is expressed as $U=f(T)$. Because the expansion is described as isothermal, the temperature of the gas remains constant as it transitions from 2 litre to 6 litre, implying $\Delta U=0$. Consequently, the change in internal energy is 0 J.