Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

Determining the standard enthalpy change of a reaction relies on the principle that the total enthalpy is calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products. Because the enthalpy of formation for pure elements in their standard states, such as calcium and aluminum, is defined as zero, the expression simplifies by focusing only on the compounds involved.

For the reaction $3\mathrm{CaO}+2\mathrm{Al}\longrightarrow 3\mathrm{Ca}+\mathrm{Al}_2\mathrm{O}_3$, the standard reaction enthalpy is determined by the equation:
$\Delta_{r} H^{\circ} = \left[ \Delta_{f} H^{\circ}(\mathrm{Al}_2\mathrm{O}_3) + 3\Delta_{f} H^{\circ}(\mathrm{Ca}) \right]\$ - \left[ 3\Delta_{f} H^{\circ}(\mathrm{CaO}) + 2\Delta_{f} H^{\circ}(\mathrm{Al}) \right]$
Substituting the provided values into this formula gives $\Delta_{r} H^{\circ} = (-1675\ \mathrm{kJ\ mol}^{-1} + 0) - (3 \times -625\ \mathrm{kJ\ mol}^{-1} + 0)$, which simplifies to $-1675 + 1875$. Carrying out this calculation results in a standard reaction enthalpy of $200\ \mathrm{kJ}$.

Determining the enthalpy change for a chemical reaction requires assessing the energy needed to break the bonds in the reactants and comparing it with the energy released when new bonds form in the products. The net enthalpy change is the difference between these two totals, reflecting the energy stability of the system after the transformation.

In the decomposition of ethane, we break one $\mathrm{C}-\mathrm{C}$ bond and six $\mathrm{C}-\mathrm{H}$ bonds. The total energy required for the reactants is $347 + 6 \times 414 = 2831$ kJ/mol. Simultaneously, the formation of the products releases energy through the creation of one $\mathrm{C}=\mathrm{C}$ bond, four $\mathrm{C}-\mathrm{H}$ bonds, and one $\mathrm{H}-\mathrm{H}$ bond. This total release is $611 + 4 \times 414 + 436 = 2703$ kJ/mol.

Subtracting the energy of the product bonds from the energy of the reactant bonds provides the final enthalpy change of the reaction. This calculation, $2831 - 2703 = 128$ kJ/mol, represents the energy required for the overall process.

The calculation of average bond energy relies on Hess's Law, which relates the enthalpy of formation of a compound to the energy required to atomize its constituent elements and the bond energy of the molecule. The total energy required to dissociate one mole of gaseous $SF_6$ into one mole of gaseous sulfur and six moles of gaseous fluorine atoms is expressed as $E_{\text{total}} = (\Delta_f H^\circ[S, g] + 6 \times \Delta_f H^\circ[F, g]) - \Delta_f H^\circ[SF_6, g]$. Substituting the provided values, the total energy needed to break all bonds corresponds to the expression $(275 + 6 \times 80) - (-1100)$, which simplifies to 1855 kJ/mol. Dividing this total energy by the six $S-F$ bonds present in the molecule yields an average bond energy of approximately 309.16 kJ/mol, which rounds to the final value of 309 kJ/mol.

The relationship between the equilibrium constant and thermodynamic properties is defined by the fundamental equations $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$ and $\Delta G^{\circ} = -RT \ln k$. By equating these two expressions, we obtain $-RT \ln k = \Delta H^{\circ} - T \Delta S^{\circ}$. Dividing both sides by $-RT$ results in the correct expression $\ln k = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$, which demonstrates that the expression in the second option is incorrect because it erroneously omits the negative sign required by the division.

Entropy represents the degree of randomness or disorder within a system, so processes that shift particles into more ordered states generally lead to a decrease in entropy. Freezing water at $0^{\circ}\,\mathrm{C}$ or $-10^{\circ}\,\mathrm{C}$ restricts the movement of molecules into a structured ice lattice, just as the reaction $\mathrm{N}_2(g)+3\,\mathrm{H}_2(g) \rightarrow 2\,\mathrm{NH}_3(g)$ lowers entropy by reducing the number of gaseous moles, and the adsorption of $\mathrm{CO}(g)$ onto a lead surface decreases entropy by anchoring chaotic gas particles to a solid. In contrast, the dissolution of $\mathrm{NaCl}$ in water causes the highly ordered solid lattice to break apart into freely moving aqueous ions, which actually increases the randomness of the system.

The process of sublimation, which involves a direct phase transition from solid to gas, can be understood as the sum of the energy changes involved in melting the solid and subsequently vaporizing the resulting liquid. Since enthalpy is a state function, the total enthalpy change for this direct route equals the sum of the enthalpy of fusion and the enthalpy of vaporization, a principle derived from Hess's Law. This relationship is defined by the equation:

$\Delta H_{\text{sub}} = \Delta H_{\text{fus}} + \Delta H_{\text{vap}}$

To calculate the enthalpy of sublimation for the substance, we sum the given values of $2.8 \, \mathrm{kJ}\,\mathrm{mol}^{-1}$ for fusion and $98.2 \, \mathrm{kJ}\,\mathrm{mol}^{-1}$ for vaporization. Combining these values leads to a total enthalpy of sublimation of $101 \, \mathrm{kJ}\,\mathrm{mol}^{-1}$.

The temperature rise in a neutralization reaction is governed by the heat released during the formation of water from the limiting reactant. Calculating the moles of available ions reveals that the sulfuric acid solution provides 0.16 mole of $\mathrm{H}^+$ ions, whereas the sodium hydroxide solution contributes 0.06 mole of $\mathrm{OH}^-$ ions. Since the hydroxide is the limiting reactant, the reaction generates 0.06 mole of $\mathrm{H}_2\mathrm{O}$. The total heat released is determined by the equation $q = n \times |\Delta_r H|$, resulting in 3426 J of energy. This thermal energy is transferred to the mixture, which has a total mass of 1000 g, assuming the density remains 1.0 g cm^-3. By applying the calorimetry relationship $q = m \cdot s \cdot \Delta T$, where the specific heat capacity $s$ is 4.18 J K^-1 g^-1, the temperature change is calculated using the equation $\Delta T = \frac{3426}{1000 \times 4.18}$. This yields a value of approximately 0.8196 K, and expressing this as 81.96 multiplied by 10^-2 K leads to a rounded integer value of 82.

To calculate the standard Gibbs free energy change, we use the relationship $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$, where the enthalpy change is related to the internal energy change by $\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_{g}RT$. For the dimerization reaction $2\mathrm{A}(g) \rightarrow \mathrm{A}_2(g)$, the change in gaseous moles is $\Delta n_g = 1 - 2 = -1$. Converting the provided internal energy of $-20\ \mathrm{kJ\ mol}^{-1}$ to Joules and adding the work term gives an enthalpy change of $\Delta H^{\ominus} = -20000 + (-1 \times 8.314 \times 298) = -22477.572\ \mathrm{J\ mol}^{-1}$. Finally, substituting this into the Gibbs equation yields $\Delta G^{\ominus} = -22477.572 - (298 \times -30)$, which results in $-13537.57\ \mathrm{J\ mol}^{-1}$.

The internal energy of an ideal gas changes in direct proportion to the number of moles, the temperature variation, and the molar heat capacity at constant volume, a relationship represented by the equation $\Delta U = n C_V \Delta T$. Given an energy change of 5000 J for 4 moles of gas heated from 300 K to 500 K, the temperature difference is 200 K. By rearranging the expression to solve for the unknown, we get $C_V = \frac{\Delta U}{n \Delta T}$. Substituting the known values leads to the calculation $C_V = \frac{5000}{4 \times 200}$, which simplifies to a molar heat capacity of 6.25 J/mol K.

Evaporation is a thermodynamic process where liquid molecules gain sufficient kinetic energy to overcome intermolecular forces and transition into the gas phase, requiring energy both to increase the system's internal energy and to perform work against the atmosphere. This relationship is expressed by the equation $\Delta H = \Delta U + \Delta n_g RT$.

First, determine the number of moles by dividing the mass of 90g by the molar mass of water, 18g/mol, which yields $n = 5$ moles. Since the process converts 5 moles of liquid water into 5 moles of water vapor, the change in the number of gaseous moles is $\Delta n_g = 5$. The total enthalpy change for the evaporation is calculated as $\Delta H = 5 \times 41000 = 205000 \text{ J}$. The work performed during expansion is determined by the term $\Delta n_g RT = 5 \times 8.314 \times 373 = 15505.61 \text{ J}$. Finally, rearranging the thermodynamic equation to solve for the change in internal energy gives $\Delta U = \Delta H - \Delta n_g RT = 205000 - 15505.61 = 189494.39 \text{ J}$.

The enthalpy of atomization represents the total energy required to convert a substance from its standard physical state to isolated gaseous atoms, which for $\mathrm{Br}_{2(l)}$ involves both overcoming intermolecular forces to vaporize the liquid and breaking the covalent bond. Since the total enthalpy of atomization $x$ is defined as the sum of the enthalpy of vaporization and the bond enthalpy $y$, represented by $x = \Delta H_{\text{vap}} + y$, the energy required for vaporization must be explicitly included. Because vaporization is an endothermic process that adds a positive energy value to the bond-breaking component, it necessarily follows that $x > y$.

Entropy is defined as a state function through the relation $dS = \frac{dq_{rev}}{T}$, which directly links the differential change in entropy to the absolute temperature of the system. Because the state of a thermodynamic system is uniquely determined by variables including temperature, the absolute entropy $S$ is inherently a function of temperature. Furthermore, since the change in entropy $\Delta S$ is calculated by integrating this temperature-dependent differential over a process, the resulting value is also explicitly dependent on temperature.

Most chemical transformations involving the removal of an electron or forcing one into a high-energy or repulsive state require an input of energy, rendering them endothermic. Specifically, ionizing sodium, adding an electron to the stable configuration of argon, or overcoming the significant inter-electronic repulsion when adding a second electron to an oxygen ion are all processes that demand energy absorption. The only exception is the reaction $\mathrm{H}_{(g)} + e^{-} \rightarrow \mathrm{H}_{(g)}^{-}$, which releases energy upon the formation of the hydride ion, making it the sole exothermic process in this set.

The relationship between enthalpy change and internal energy change in a chemical reaction is governed by the expression $\Delta H = \Delta U + \Delta n_g RT$. In the combustion of ethanol, the balanced chemical reaction is $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}_{(l)} + 3\,\mathrm{O}_{2(g)} \rightarrow 2\,\mathrm{CO}_{2(g)} + 3\,\mathrm{H}_2\mathrm{O}_{(l)}$. We determine the change in the number of gaseous moles, $\Delta n_g$, by subtracting the total moles of gaseous reactants from the gaseous products, resulting in $\Delta n_g = 2 - 3 = -1$.

The heat of combustion at constant pressure is -327000 cal, and we solve for the internal energy change at constant volume, $\Delta U$, by rearranging the thermodynamic formula to $\Delta U = \Delta H - \Delta n_g RT$. Substituting the known values, where R is 2 and T is 300 K, the calculation becomes $\Delta U = -327000 - (-1)(2)(300)$. This simplifies to $\Delta U = -326400$ cal. Since the question asks for the heat evolved, the final magnitude is 326400 cal.

The spontaneity of a reaction is dictated by the Gibbs free energy change, which integrates internal energy, the pressure-volume work associated with gas expansion, and the entropy of the system. The relationship between these thermodynamic quantities is given by $\Delta G = \Delta U + \Delta n_g RT - T \Delta S$. For this reaction, the change in gaseous moles is $\Delta n_g = 2$. With the gas constant taken as 2 cal K⁻¹ mol⁻¹, the work term $\Delta n_g RT$ evaluates to 1.2 kcal, while the entropy term $T \Delta S$ is 6.0 kcal. Substituting the given internal energy of 2.1 kcal into the formula, we have $\Delta G = 2.1 + 1.2 - 6.0$, which yields a final value of -2.70 kcal.

The thermodynamic stability of a metal oxide is determined by the Gibbs free energy of formation, which is given by the relation:
$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$
The oxide is thermodynamically stable only when $\Delta G^\circ < 0$. On an Ellingham diagram, the temperature below which the oxide is stable is identified at the equilibrium point where $\Delta G^\circ = 0$, as the free energy change transitions from a negative to a positive value with increasing temperature.

For 5 moles of an ideal gas expanding into a vacuum from 1 bar at 298 K, the work done is determined by the external pressure acting on the system, defined by the formula $W = -\int p_{ext} dV$. Since the expansion occurs into a vacuum, there is no external pressure to oppose the gas, meaning $p_{ext} = 0$. This absence of resistance indicates that no work is performed during the expansion to double the volume, regardless of the initial state.

The work done by a gas during a reversible expansion process is equivalent to the area enclosed beneath its trajectory on a pressure-volume diagram, defined by the relationship $W = \int P \, dV$. The magnitude of work for the path ABC corresponds to the total geometric area between this path and the volume axis. Based on the provided graph, this area evaluates to a total work done of 48 Joules.

The dissolution of an ionic solid involves an energy balance where the enthalpy of solution is equal to the sum of the lattice enthalpy and the hydration enthalpy. Breaking the ionic crystal lattice into gaseous ions requires an input of energy, whereas the interaction of these ions with solvent molecules releases energy. Consequently, the observed enthalpy of solution represents the combined result of these two opposing energetic contributions.

Mathematically, this relationship is expressed by adding the lattice enthalpy to the hydration enthalpy. Using the values of 4 kJ mol$^{-1}$ for the enthalpy of solution and 788 kJ mol$^{-1}$ for the lattice enthalpy, the energy balance is:

$4\ \text{kJ mol}^{-1} = 788\ \text{kJ mol}^{-1} + \Delta H_{\text{hyd}}$

Determining the hydration enthalpy involves subtracting the lattice enthalpy from the enthalpy of solution, which yields:

$\Delta H_{\text{hyd}} = 4\ \text{kJ mol}^{-1} - 788\ \text{kJ mol}^{-1} = -784\ \text{kJ mol}^{-1}$

An ideal gas is characterized by the absence of intermolecular interactions, which implies that its internal energy $U$ and enthalpy $H$ are exclusively functions of temperature, making statement (a) true. Because the compressibility factor for an ideal gas is strictly defined as $Z = 1$, statement (b) is incorrect, whereas the relationship $C_{P,m} - C_{V,m} = R$ remains a fundamental identity for one mole of such a gas. Additionally, since the internal energy of an ideal gas depends solely on temperature, the expression $dU = C_{V} dT$ holds valid for any process, confirming that statements (a), (c), and (d) are accurate.