Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

Enthalpy change is an extensive property, meaning the total energy required depends directly on the amount of substance undergoing phase change. Determining the enthalpy change for 85 g of ammonia requires first calculating the number of moles by dividing the mass by the molar mass of 17 g/mol, which yields $n = \frac{85}{17} = 5$ moles. Because the vaporization enthalpy is provided per mole, the total energy change is found by multiplying this value by the total number of moles, giving $23.4 \times 5 = 117$ kJ.

State variables are thermodynamic properties defined exclusively by the current condition of a system, independent of the specific process or path taken to arrive there. In this set, internal energy ($U$), volume ($V$), and enthalpy ($H$) are determined solely by the system's state, classifying them as state functions. Conversely, heat ($q$) is defined as a path function because its value depends on the specific mechanism of energy transfer rather than the system's current state alone. Therefore, the list contains three state variables.

The enthalpy of formation of a compound is determined using Hess's Law, which states that the total enthalpy change for a chemical process is independent of the pathway taken. For the formation of propane, the reaction is defined as $3 \mathrm{C}(\text{graphite}) + 4 \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{C}_3\mathrm{H}_8(\mathrm{g})$. We calculate the enthalpy of formation by summing the combustion enthalpies of the reactants and subtracting the combustion enthalpy of the product.

Using the provided combustion values, the calculation is represented as:
$\Delta H_f = 3(-393.5) + 4(-285.8) - (-2220.0)$
Evaluating these terms, we find:
$\Delta H_f = -1180.5 - 1143.2 + 2220.0 = -103.7 \text{ kJ mol}^{-1}$
Taking the magnitude of this value and rounding to the nearest integer, the final result is 104 kJ mol⁻¹.

The conversion between internal energy change and enthalpy change for a chemical reaction involving gases is governed by the relation:
$\Delta H = \Delta U + \Delta n_g R T$
In this reaction, the change in the number of gaseous moles, $\Delta n_g$, is calculated by subtracting the stoichiometric coefficients of gaseous reactants from those of the gaseous products, which results in $2 - 1.5 = 0.5$. Because $\Delta U$ is provided in $\mathrm{kJ\ mol^{-1}}$, the gas constant $R$ must be converted to $8.314 \times 10^{-3}\ \mathrm{kJ\ mol^{-1}\ K^{-1}}$ to maintain unit consistency. Substituting the values $\Delta U = -742.24\ \mathrm{kJ\ mol^{-1}}$, $\Delta n_g = 0.5$, $R = 8.314 \times 10^{-3}\ \mathrm{kJ\ mol^{-1}\ K^{-1}}$, and $T = 298\ \mathrm{K}$ into the expression gives $\Delta H = -742.24 + (0.5 \times 8.314 \times 10^{-3} \times 298)$, which simplifies to $-741.001\ \mathrm{kJ\ mol^{-1}}$. The magnitude of this result, rounded to the nearest integer, is 741.

The neutralization reaction between hydrochloric acid and sodium hydroxide is determined by the limiting reagent, which is the reactant present in the smaller amount. Calculating the millimoles for both species shows that 200 mL of 0.2 M HCl provides 40 mmol, while 300 mL of 0.1 M NaOH provides 30 mmol. Because the reaction stoichiometry is 1:1, the 30 mmol of NaOH limits the extent of the reaction, which results in the production of 0.03 mol of water.

With the amount of reaction known, the total heat released can be determined by using the molar enthalpy of neutralization. Multiplying the 0.03 mol by 57.1 kJ per mole yields a heat release of 1713 J. Since the density is 1.00 g per cubic centimeter, the total volume of 500 mL corresponds to a mass of 500 g for the mixture.

The temperature rise is found by applying the relationship $q = mc\Delta T$. Substituting the values into the equation gives $1713 = 500 \times 4.18 \times \Delta T$. Solving for $\Delta T$ results in 0.8196. Representing this change as $81.96 \times 10^{-2}$, which rounds to 82 as the nearest integer.

The thermodynamic connection between free energy, enthalpy, and entropy is governed by the Gibbs free energy equation, $\Delta G = \Delta H - T \Delta S$. To determine the entropy change, we isolate the entropy term from the expression to solve for it directly:

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

By substituting the provided values into this formula, the numerator becomes the difference between 51.4 kJ mol⁻¹ and -49.4 kJ mol⁻¹, which equals 100.8 kJ mol⁻¹. Converting this value into joules by multiplying by 1000 yields 100800 J mol⁻¹. When this result is divided by the reaction temperature of 300 K, the calculated entropy change is 336 J K⁻¹ mol⁻¹.

The equilibrium constant $K$ for a chemical reaction is related to the standard Gibbs free energy change $\Delta G^{\circ}$ via the equation $\Delta G^{\circ} = -2.303 RT \log K$. To determine $\Delta G^{\circ}$ for the reaction where $3\mathrm{HC} \equiv \mathrm{CH}(g)$ forms $\mathrm{C}_{6}\mathrm{H}_{6}(l)$, we find the difference between the total Gibbs free energy of formation of the products and that of the reactants using the expression $\Delta G^{\circ} = \Delta_f G^{\circ}(\mathrm{C}_{6}\mathrm{H}_{6}) - 3 \times \Delta_f G^{\circ}(\mathrm{HC} \equiv \mathrm{CH})$. Substituting the provided values yields $\Delta G^{\circ} = (-1.24 \times 10^5) - 3 \times (-2.04 \times 10^5)$, which simplifies to $\Delta G^{\circ} = -1.24 \times 10^5 + 6.12 \times 10^5 = 4.88 \times 10^5 \mathrm{~J~mol}^{-1}$.

Applying the temperature of $25^{\circ}\mathrm{C}$, which corresponds to $298 \mathrm{~K}$, we use the relation $\Delta G^{\circ} = -2.303 RT \log K$ to find that $4.88 \times 10^5 = -2.303 \times 8.314 \times 298 \times \log K$. Solving for the magnitude of $\log K$ gives $\left| \log K \right| = \frac{4.88 \times 10^5}{2.303 \times 8.314 \times 298}$, which evaluates to approximately $85.5$. Expressing this magnitude as $x \times 10^{-1}$ gives $855 \times 10^{-1}$, confirming that the value of $x$ is 855.

Combustion energy is directly proportional to the amount of substance consumed, meaning the molar enthalpy change corresponds to the heat released by the molar mass of that specific material. Since graphite is elemental carbon with a molar mass of 12 g/mol, the standard enthalpy change of $2.48 \times 10^{2} \, \text{kJ/mol}$ indicates that this amount of energy is liberated for every 12 g of graphite. Determining the heat generated by a single gram requires dividing the total enthalpy change per mole by the molar mass.

$\frac{248}{12} \approx 20.67$

Rounding the resulting 20.67 kJ/g to the nearest integer provides a final heat generation of 21 kJ.

Spontaneity in a chemical reaction is determined by the change in Gibbs free energy, $\Delta G^\circ$, which is related to enthalpy and entropy through the equation $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$. A reaction becomes spontaneous when the free energy change drops below zero, so the minimum temperature at which this process initiates is found by solving for the state where $\Delta G^\circ = 0$, leading to the condition $T = \frac{\Delta H^\circ}{\Delta S^\circ}$.

To find the values for this equation, first calculate the standard enthalpy change of the reaction by subtracting the sum of the enthalpies of formation of the reactants from those of the products: $\Delta H^\circ = (0 + (-110.5)) - (-266.3 + 0) = 155.8 \text{ kJ mol}^{-1}$. Next, calculate the standard entropy change using the provided molar entropy values: $\Delta S^\circ = (27.28 + 197.6) - (57.49 + 5.74) = 161.65 \text{ J K}^{-1} \text{mol}^{-1}$.

With these values, convert the enthalpy into joules to ensure the units remain consistent across the calculation, yielding $155800 \text{ J mol}^{-1}$. Dividing this by the entropy change gives the temperature $T = \frac{155800 \text{ J mol}^{-1}}{161.65 \text{ J K}^{-1} \text{mol}^{-1}}$, which results in approximately $963.8 \text{ K}$. Rounding this to the nearest integer provides the threshold temperature of $964 \text{ K}$.

The relationship between the enthalpy of vaporization and the change in internal energy is derived from the thermodynamic expression $\Delta H = \Delta U + \Delta(PV)$. For the phase transition $\mathrm{H}_2\mathrm{O}(l) \rightarrow \mathrm{H}_2\mathrm{O}(g)$, treating the vapor as an ideal gas and ignoring the negligible volume of the liquid allows us to represent the work term $\Delta(PV)$ as $\Delta n_g RT$. Since one mole of liquid water converts to one mole of gaseous water, the change in the number of gaseous moles is $\Delta n_g = 1$. Substituting the absolute temperature of $373\ \mathrm{K}$ and the gas constant $8.31\ \mathrm{J}\ \mathrm{mol}^{-1}\ \mathrm{K}^{-1}$ into the equation $\Delta_{\text{vap}} H - \Delta_{\text{vap}} U = \Delta n_g RT$ yields $1 \times 8.31 \times 373$, which equals $3099.63\ \mathrm{J}\ \mathrm{mol}^{-1}$. Expressing this value in the form $X \times 10^2\ \mathrm{J}\ \mathrm{mol}^{-1}$ results in $30.9963 \times 10^2\ \mathrm{J}\ \mathrm{mol}^{-1}$, which rounds to 31.

The activation energy of a reverse reaction is fundamentally linked to the enthalpy change of the process and the activation energy of the forward reaction through the relationship $\Delta H = E_a^f - E_a^b$. Because the enthalpy change represents the difference between the forward and reverse activation barriers, one can rearrange this expression to isolate the unknown reverse activation energy. Given the forward activation energy is 30 kJ mol⁻¹ and the enthalpy change is -20 kJ mol⁻¹, the calculation is:

$E_a^b = E_a^f - \Delta H$
$E_a^b = 30 - (-20) = 50 \text{ kJ mol}^{-1}$

A reaction becomes spontaneous when the standard Gibbs free energy change, $\Delta G^{\circ}$, drops below zero. The relationship between the enthalpy change, $\Delta H^{\circ}$, and the entropy change, $\Delta S^{\circ}$, is defined by the Gibbs-Helmholtz equation $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$. Using the provided values of $\Delta H^{\circ} = 80,000 \text{ J mol}^{-1}$ and $\Delta S^{\circ} = 2T \text{ J K}^{-1} \text{mol}^{-1}$, the condition $\Delta G^{\circ} < 0$ becomes:

$80,000 - T(2T) < 0$

Applying algebraic rearrangement leads to $2T^2 > 80,000$, which simplifies further to $T^2 > 40,000$. Taking the square root of both sides reveals that the temperature must be greater than $200 \text{ K}$ for the reaction to occur spontaneously. Therefore, the minimum temperature required is $200 \text{ K}$.

For an ideal gas undergoing an isothermal process, the change in internal energy is zero because internal energy depends solely on temperature. By the first law of thermodynamics, this implies that the heat transferred to the gas during the process is equal in magnitude to the work it performs, but with the opposite sign. Since the gas expands against a constant external pressure, this is an irreversible process, and the work done is defined by the product of the external pressure and the change in volume.

To determine the work performed, we use the equation $W = -P_{\text{ext}} \Delta V$, where the change \in volume is derived from the ideal gas law as $\Delta V = nRT \left( \frac{1}{P_2} - \frac{1}{P_1} \right)$. Combining these, the expression for work is $W = -P_{\text{ext}} \cdot nRT \left( \frac{1}{P_2} - \frac{1}{P_1} \right)$. Substituting the given values, with an external pressure $P_{\text{ext}} = 4.3$ MPa, $n = 5$ mol, temperature $T = 293$ K, gas constant $R = 8.314$ J mol$^{-1}$ K$^{-1}$, initial pressure $P_1 = 2.1$ MPa, and final pressure $P_2 = 1.3$ MPa, we calculate the work as:

$W = -4.3 \times 5 \times 8.314 \times 293 \left( \frac{1}{1.3} - \frac{1}{2.1} \right)$

This calculation yields a result of approximately $-15347.7$ J mol$^{-1}$, which is equivalent to $-15.347$ kJ mol$^{-1}$. Because the heat transferred in the process is equal to $-W$, the magnitude of the heat transferred is approximately $15$ kJ mol$^{-1}$ when rounded to the nearest integer.

The Born-Haber cycle utilizes Hess's Law to construct a thermodynamic pathway that forms an ionic solid from its constituent elements. By equating the total enthalpy of formation to the energy changes of each individual step-sublimation of the metal, dissociation of the non-metal molecule, ionization of the metal, electron gain by the non-metal, and the formation of the crystal lattice-the lattice energy can be derived.

Calculating the lattice enthalpy involves summing the energy contributions from each intermediate step. The energy required to sublimate potassium is 89.2 kJ mol-1, and ionizing the gaseous potassium atom requires 419.0 kJ mol-1. For chlorine, the bond dissociation enthalpy is 243.0 kJ mol-1; taking half of this value, 121.5 kJ mol-1, and combining it with the electron gain enthalpy of -348.6 kJ mol-1 yields a total of 281.1 kJ mol-1 before the lattice formation step.

The thermodynamic cycle relates these quantities through the following equation:
$\Delta_f H^{\circ} = \Delta_{\text{sub}} H^{\circ} + \frac{1}{2} \Delta_{\text{bond}} H^{\circ} + \Delta_{\text{ionisation}} H^{\circ} + \Delta_{\text{electron gain}} H^{\circ} + \Delta H_{\text{lattice}}$
Substituting the formation enthalpy of -436.7 kJ mol-1 and the calculated sum of 281.1 kJ mol-1 results in the relation -436.7 = 281.1 + $\Delta H_{\text{lattice}}$. Solving for the unknown term yields -717.8 kJ mol-1, which corresponds to a magnitude of 718 kJ mol-1.

The relationship between enthalpy and internal energy during a phase transition is defined by the thermodynamic equation $\Delta H = \Delta U + \Delta n_g RT$. When water evaporates, the phase changes from liquid to gas, resulting in the formation of one mole of vapor from zero moles of liquid, which means $\Delta n_g = 1$. Since the volume of the liquid water is negligible compared to the vapor, we can effectively treat the expansion work as the pressure-volume work term $\Delta n_g RT$.

Calculating this work term requires multiplying the number of moles by the gas constant and the absolute temperature, resulting in $1 \times 8.3 \times 373 = 3095.9 \text{ J}$, or $3.0959 \text{ kJ mol}^{-1}$. To isolate the internal energy change, subtract this expansion work from the given enthalpy of vaporization. Using the provided enthalpy value, the expression becomes $\Delta U = 41 \text{ kJ mol}^{-1} - 3.0959 \text{ kJ mol}^{-1}$, yielding $37.9041 \text{ kJ mol}^{-1}$. Rounding this value to the nearest integer provides a final internal energy change of $38 \text{ kJ mol}^{-1}$.

The formation of the ionic solid $\mathrm{NaBr}$ is analyzed through the Born-Haber cycle, which sums the energetic contributions of the individual processes required to transform the elements into a stable crystalline lattice. We begin by considering the ionization enthalpy of gaseous sodium, requiring $495.8\ \mathrm{kJ\ mol^{-1}}$ of energy, and the electron gain enthalpy of gaseous bromine, which releases $325.0\ \mathrm{kJ\ mol^{-1}}$, represented as $-325.0\ \mathrm{kJ\ mol^{-1}}$. The final step involves the lattice enthalpy, which is the energy released when the gaseous ions combine into the solid state, contributing $-728.4\ \mathrm{kJ\ mol^{-1}}$ to the total enthalpy change.

The total enthalpy of formation is found by summing these individual values:
$\Delta H_{\text{formation}} = 495.8 + (-325.0) + (-728.4) = -557.6\ \mathrm{kJ\ mol^{-1}}$
Converting this total into the required format, the energy for the formation of the $\mathrm{NaBr}$ ionic solid is $-5576 \times 10^{-1}\ \mathrm{kJ\ mol^{-1}}$.

When a chemical reaction releases gas into an environment under constant pressure, it performs work on its surroundings by expanding. This energy transfer is determined by the thermodynamic expression $W = -\Delta n_g RT$, where $\Delta n_g$ represents the change in gaseous moles between products and reactants.

Determining the molar quantity of hydrogen generated starts by reacting 50 g of iron. Given the stoichiometric relationship where iron reacts with hydrochloric acid to produce one mole of hydrogen gas per mole of iron, the number of moles consumed is calculated as 50 divided by 55.85, resulting in 0.8952 mol. Since no gaseous substances are present on the reactant side, the net change in gaseous moles $\Delta n_g$ is exactly 0.8952 mol.

Substituting these parameters into the equation using a temperature of 298 K and an ideal gas constant of 8.314 J mol^-1 K^-1 leads to the result of -2217.92 J. The expansion of the gas against constant pressure performs this work on the surroundings, so taking the magnitude of the calculated value provides the final result of 2218 J when rounded to the nearest integer.

When two flasks are connected by a valve, the system reaches thermal equilibrium through heat exchange between the gases. Since the process is isolated, the internal energy lost by the gas at the higher temperature must be equal to the internal energy gained by the gas at the lower temperature, as dictated by the conservation of energy.

To find the equilibrium temperature, we first calculate the number of moles for each gas using the molar mass of 28.0 g/mol for nitrogen. This gives $n_A = \frac{2.8}{28} = 0.1$ moles and $n_B = \frac{0.2}{28} \approx 0.00714$ moles. Given that the gases are diatomic, the molar heat capacity at constant volume is identical, simplifying the energy balance equation to $n_A(T_1 - T) = n_B(T - T_2)$, where the initial temperatures are 300 K and 60 K. Substituting these values into the equation yields the final equilibrium temperature $T = 284$ K.

We determine the final pressure using the ideal gas law $P = \frac{n_{total}RT}{V}$. Combining the quantities of gas gives a total number of moles $n_{total} \approx 0.10714$ moles. Substituting this value along with $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$, the temperature $T = 284$ K, and the volume $V = 3 \times 10^{-3} \text{ m}^3$ into the equation $P = \frac{0.10714 \times 8.31 \times 284}{3 \times 10^{-3}}$ results in a pressure of approximately $84180$ Pa. Converting this to bar using the relation $1 \text{ bar} = 10^5$ Pa gives $0.8418$ bar, which is equal to $84.18 \times 10^{-2}$ bar. Therefore, the value of x is 84.

The spontaneity of a reaction at constant temperature and pressure is determined by the Gibbs free energy relation $\Delta G = \Delta H - T \Delta S$. To perform this calculation, convert the entropy change from -176 J/K mol to -0.176 kJ/K mol so that all terms share consistent energy units.

Substitute the values at 298 K to find the product of temperature and entropy, which equals -52.448 kJ/mol. Subtracting this value from the enthalpy change of -57.8 kJ/mol yields a Gibbs free energy change of -5.352 kJ/mol. The magnitude of this result is 5.352, which rounds to the nearest integer 5.

The First Law of Thermodynamics dictates that the change in internal energy is determined by the sum of heat absorbed and work performed, expressed as $\Delta U = q + w$. Following convention, heat absorbed by a system is positive, while work done by a system is negative. Substituting the given values, we subtract 200 J from 150 J to obtain a change in internal energy of -50 J. Since the problem asks for the magnitude of this change, we take the absolute value of the result to arrive at 50 J.