Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

Hess's Law dictates that the total enthalpy change of a chemical reaction is consistent regardless of the pathway taken, allowing us to determine the heat of formation by manipulating known combustion equations. Given the combustion reactions for carbon and carbon monoxide:

$\mathrm{C} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 \quad \Delta H_1 = -393.5\ \mathrm{kJ\ mol}^{-1}$
$\mathrm{CO} + \frac{1}{2}\mathrm{O}_2 \rightarrow \mathrm{CO}_2 \quad \Delta H_2 = -283.5\ \mathrm{kJ\ mol}^{-1}$

To find the enthalpy of formation for carbon monoxide, which is represented by the equation $\mathrm{C} + \frac{1}{2}\mathrm{O}_2 \rightarrow \mathrm{CO}$, we subtract the second equation from the first. This operation yields the enthalpy change as:

$\Delta_f H = \Delta H_1 - \Delta H_2 = -393.5 - (-283.5) = -110.5\ \mathrm{kJ\ mol}^{-1}$

The Gibbs free energy equation is defined as $\Delta G = \Delta H - T \Delta S$, where a negative $\Delta G$ value signifies a spontaneous reaction. A process that shifts from non-spontaneous at low temperatures to spontaneous at high temperatures requires a positive entropy change $\Delta S$, ensuring that the $-T \Delta S$ term becomes increasingly negative as the temperature rises. This condition can only occur when the enthalpy change $\Delta H$ is also positive, meaning the reaction is endothermic and relies on the positive entropy component to drive spontaneity at higher thermal energies.

The equilibrium constant $K_p$ provides a direct measure of the standard Gibbs free energy change for a reaction occurring at a constant temperature, governed by the relation $\Delta G^{\circ} = -RT \ln K_p$. This value reflects the overall thermodynamic energy difference between the products and reactants under standard conditions.

To determine the standard free energy of formation for $\mathrm{NO}_2(g)$, we apply the fundamental principle that the standard free energy change of a reaction is equal to the sum of the free energies of formation of the products minus the reactants. For the balanced reaction $2\mathrm{NO}(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}_2(g)$, this is represented as $\Delta G^{\circ} = 2\Delta G^{\circ}_f(\mathrm{NO}_2) - 2\Delta G^{\circ}_f(\mathrm{NO}) - \Delta G^{\circ}_f(\mathrm{O}_2)$. Because the standard free energy of formation for the element $\mathrm{O}_2$ is defined as zero, the expression simplifies to $\Delta G^{\circ} = 2\Delta G^{\circ}_f(\mathrm{NO}_2) - 2\Delta G^{\circ}_f(\mathrm{NO})$. By substituting $-RT \ln K_p$ for the reaction free energy and converting the given formation energy of $\mathrm{NO}$ from $86.6 \text{ kJ/mol}$ to $86600 \text{ J/mol}$, the equation can be rearranged to isolate $\Delta G^{\circ}_f(\mathrm{NO}_2)$ as $0.5[2 \times 86600 - R(298) \ln(1.6 \times 10^{12})]$.

When measuring the heat of combustion in a bomb calorimeter, the value obtained corresponds to the change in internal energy, $\Delta E$, because the reaction occurs at a constant volume. To determine the enthalpy of combustion, $\Delta H$, we apply the thermodynamic relationship $\Delta H = \Delta E + \Delta n_g RT$, which accounts for the work associated with any changes in the volume of gaseous species.

The balanced equation for the complete combustion of ethanol shows that two moles of gaseous products are formed while three moles of gaseous reactants are consumed, resulting in a $\Delta n_g$ value of -1. Given an internal energy change of -1364.47 kJ mol⁻¹ at 298 K, and using a gas constant $R$ of 0.008314 kJ K⁻¹ mol⁻¹ to maintain unit consistency, the calculation follows the expression $\Delta H = \Delta E + \Delta n_g RT$. Substituting these values leads to $\Delta H = -1366.95$ kJ mol⁻¹.

The internal energy of an ideal gas depends exclusively on its temperature, so for any isothermal process, the change in internal energy is $\Delta U = 0$. According to the first law of thermodynamics, $\Delta U = q + w$, which implies that the heat exchanged must be exactly equal and opposite to the work done, expressed as $q = -w$. Given that the system absorbs $208$ J of heat, $q$ is defined as $+208$ J. Therefore, to satisfy the thermodynamic balance during this expansion, the work done $w$ must be $-208$ J.

The thermodynamic relationship between the equilibrium constant $K$ and standard Gibbs free energy is defined by the equation $\Delta G^{\circ} = -RT \ln K$. By substituting the state function definition $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$ into this expression, we arrive at $-RT \ln K = \Delta H^{\circ} - T\Delta S^{\circ}$. Solving this equation for $\ln K$ yields $\ln K = -\frac{\Delta H^{\circ} - T\Delta S^{\circ}}{RT}$, which confirms that the expression provided in option C is incorrect because it lacks the necessary negative sign.

The entropy change for an ideal gas during an isothermal reversible expansion is determined by the number of moles of the gas and the ratio of the final volume to the initial volume. This relationship is defined by the formula $\Delta S = n R \ln\left(\frac{V_2}{V_1}\right)$, where $n$ is the number of moles and $R$ is the universal gas constant.

Substituting the given values, where $n = 2$ and the volume ratio is $V_2 / V_1 = 100 / 10 = 10$, the equation becomes $\Delta S = 2 \times 8.314 \times 2.303 \times \log(10)$. Because the common logarithm of 10 is 1, the expression simplifies to $2 \times 8.314 \times 2.303 = 38.3$. This results in a final value of 38.3 J mol⁻¹ K⁻¹.

Gibbs free energy dictates the spontaneity of a chemical process, where a negative value signals that the reaction will proceed forward naturally. At equilibrium, the system is perfectly balanced with $\Delta G = 0$, and substituting this into the equation $\Delta G = \Delta H - T\Delta S$ reveals that the equilibrium temperature is defined by $T_e = \frac{\Delta H}{\Delta S}$. To achieve a spontaneous reaction, the condition $\Delta G < 0$ must be met, which rearranges to $\Delta H < T\Delta S$ or $T > \frac{\Delta H}{\Delta S}$. Because both enthalpy and entropy changes are positive, the system becomes spontaneous only when the operating temperature $T$ rises above the equilibrium threshold $T_e$.

The total enthalpy change for a reaction is determined by the difference between the energy required to break the bonds of the reactants and the energy released upon forming the bonds of the products. For the synthesis of two moles of $\mathrm{NH}_3$ from $\mathrm{N}_2$ and $\mathrm{H}_2$, the overall enthalpy change is $-92\ \text{kJ mol}^{-1}$, calculated as $2 \times -46.0$. We equate this to the difference between the energy absorbed to break one $\mathrm{N}\equiv\mathrm{N}$ bond and three $\mathrm{H-H}$ bonds, totaling $2020\ \text{kJ mol}^{-1}$, and the energy released by forming six $\mathrm{N-H}$ bonds, represented as $6x$. Solving the expression
$-92 = 2020 - 6x$
leads to $6x = 2112$, which identifies the average bond enthalpy of the $\mathrm{N-H}$ bond as $352\ \text{kJ mol}^{-1}$.

The enthalpy of formation of an ion is determined using Hess's Law, which states that the standard enthalpy change of a reaction is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. Given that the enthalpy of formation of elements in their standard states is zero, and the enthalpy of formation of $\mathrm{H}^{+}(\mathrm{aq})$ is specified as zero, the enthalpy of formation for the hydroxide ion $\mathrm{OH}^{-}(\mathrm{aq})$ can be derived by combining the provided thermochemical data.

The formation of liquid water, represented by $\mathrm{H}_2(\mathrm{g}) + \frac{1}{2}\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{l})$, involves an enthalpy change of $-286.20$ kJ, which corresponds directly to the enthalpy of formation of $\mathrm{H}_2\mathrm{O}(\mathrm{l})$. For the ionization reaction $\mathrm{H}_2\mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{OH}^{-}(\mathrm{aq})$, the enthalpy change of $57.32$ kJ follows the relationship $\Delta_f H^\circ(\mathrm{H}^{+}) + \Delta_f H^\circ(\mathrm{OH}^{-}) - \Delta_f H^\circ(\mathrm{H}_2\mathrm{O})$. Substituting zero for $\mathrm{H}^{+}(\mathrm{aq})$ and the value $-286.20$ kJ for the water term, the calculation becomes $57.32 = 0 + \Delta_f H^\circ(\mathrm{OH}^{-}) - (-286.20)$. Solving for the unknown, we find $\Delta_f H^\circ(\mathrm{OH}^{-}) = 57.32 - 286.20$, which results in a value of $-228.88$ kJ.

The oxidizing power of chlorine in an aqueous solution is determined by the total enthalpy change of the process, which can be evaluated using Hess's Law. This principle states that the total energy involved in the conversion of chlorine gas into aqueous chloride ions is the sum of the enthalpy changes of the individual stages, specifically the dissociation of chlorine molecules into atoms, the gain of electrons to form chloride ions, and the hydration of those ions.

This relationship is expressed as $\Delta H = \frac{1}{2} \Delta_{diss} H^{\ominus} + \Delta_{eg} H^{\ominus} + \Delta_{hyd} H^{\ominus}$. Substituting the provided values of 240 kJ/mol for dissociation, -349 kJ/mol for electron gain, and -381 kJ/mol for hydration into the expression, the calculation becomes $\Delta H = 120 - 349 - 381$. Performing this arithmetic yields a final value of -610 kJ/mol for the total enthalpy change of the process.

Chemical systems reach a state of equilibrium when the Gibbs free energy change, $\Delta G$, is zero. This relationship is defined by the equation $\Delta G = \Delta H - T \Delta S$, which implies that at equilibrium, the enthalpy change balances the product of absolute temperature and entropy change, expressed as $\Delta H = T \Delta S$. To determine the temperature required, we must first calculate the total change in entropy for the reaction by subtracting the weighted entropies of the reactants from the entropy of the product: 50 - (0.5 * 60 + 1.5 * 40), which simplifies to -40 J K^-1 mol^-1.

Because the enthalpy change is given as -30 kJ, we must convert this value into -30,000 J to ensure unit consistency with the entropy. By rearranging the equilibrium condition into the formula $T = \frac{\Delta H}{\Delta S}$, we can substitute our known values to solve for the unknown temperature. Dividing -30,000 J by -40 J K^-1 mol^-1 yields a result of 750 K.

Thermodynamic spontaneity at constant pressure is governed by the sign of the Gibbs free energy change, expressed by the equation $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$. A reaction proceeds spontaneously when $\Delta G^{\circ}$ is negative, which requires the temperature to satisfy the condition $\Delta H^{\circ} - T\Delta S^{\circ} < 0$. This implies that the temperature must exceed the ratio of the enthalpy change to the entropy change, represented by the inequality $T > \frac{\Delta H^{\circ}}{\Delta S^{\circ}}$. Substituting the values 179.1 kJ mol^-1 and 160.2 J K^-1 mol^-1, the expression $T > \frac{179100}{160.2}$ shows that the reaction becomes spontaneous above approximately 1118 K.

When a substance changes from liquid to gas, the energy supplied as enthalpy of vaporization is used both to increase the internal energy and to perform work against the external pressure. This relationship is defined by $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in the number of gaseous moles. For the vaporization of 1 mole of water, the change in gaseous moles is $1 - 0 = 1$. Converting the enthalpy of vaporization to Joules gives $41000 \text{ J mol}^{-1}$, and using the temperature of $373 \text{ K}$ and the gas constant of $8.3 \text{ J mol}^{-1} \text{ K}^{-1}$, the equation rearranges to $\Delta U = \Delta H - \Delta n_g RT$. Inserting the specific values leads to $\Delta U = 41000 - (1 \times 8.3 \times 373)$, resulting in $37904.1 \text{ J mol}^{-1}$, which is $37.904 \text{ kJ mol}^{-1}$.

Spontaneity in chemical and physical processes is governed by the simultaneous competition between the drive to reach a minimum energy state and the drive to achieve maximum randomness or disorder. In an isolated system, where no energy exchange with the environment is possible, the influence of energy minimization is effectively restricted, leaving the inherent tendency toward disorder as the primary determinant of directionality. As a result, a process within an isolated system can only be spontaneous if it leads to an increase in the total entropy of the system, represented as $\Delta S > 0$.

Determining the average bond energy of a carbon-hydrogen bond requires analyzing the conversion of methane into its constituent atoms in the gaseous state, starting from its standard state of graphite carbon and gaseous hydrogen molecules. Because the standard enthalpy of formation, which is -74.8 kJ mol^-1 for methane, describes the formation of the molecule from $\mathrm{C}(\mathrm{s})$ and $\mathrm{H}_2(\mathrm{g})$, we must first account for the energy change required to convert these reactants into free gaseous atoms. Specifically, we require the enthalpy of sublimation to vaporize solid carbon into $\mathrm{C}(\mathrm{g})$ and the bond dissociation energy to split $\mathrm{H}_2(\mathrm{g})$ molecules into $\mathrm{H}(\mathrm{g})$ atoms, which then allows us to calculate the total energy associated with the four C-H bonds in $\mathrm{CH}_4(\mathrm{g})$.

The standard enthalpy of formation measures the enthalpy change when one mole of a substance is created from its elements in their standard states. For $\mathrm{ICl}(g)$, the formation reaction is represented as $\frac{1}{2}\mathrm{I}_2(s) + \frac{1}{2}\mathrm{Cl}_2(g) \rightarrow \mathrm{ICl}(g)$. We apply Hess's Law to determine this value by constructing a path that breaks the reactants down into gaseous atoms before forming the product.

$\Delta H_f^\circ = \frac{1}{2} \left[ \Delta H_{\text{sub}}(\mathrm{I}_2) + \Delta H_{\text{diss}}(\mathrm{I}_2) + \Delta H_{\text{diss}}(\mathrm{Cl}_2) \right] - \Delta H_{\text{diss}}(\mathrm{ICl})$

To find the energy required, we first add the enthalpy of sublimation for iodine, 62.76 kJ mol$^{-1}$, to the bond dissociation enthalpies of 151.0 kJ mol$^{-1}$ and 242.3 kJ mol$^{-1}$. Taking half of this total sum gives 228.03 kJ mol$^{-1}$ as the energy needed to create the constituent gaseous atoms. Finally, we subtract the dissociation energy of the $\mathrm{I}-\mathrm{Cl}$ bond, 211.3 kJ mol$^{-1}$, because this energy is released when the bond forms. The final enthalpy calculation is $228.03 - 211.3 = 16.73$ kJ mol$^{-1}$, which represents the standard enthalpy of formation.

The relationship between the enthalpy change and internal energy change for a chemical reaction is governed by the ideal gas constant and the change in the number of gaseous moles:
$\Delta H - \Delta U = \Delta n_g RT$
For the formation of carbon monoxide, the chemical reaction is $\mathrm{C}_{(s)} + \frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}$. We determine the value of $\Delta n_g$ by subtracting the stoichiometric coefficient of the gaseous reactant from the coefficient of the gaseous product, noting that solid carbon is excluded from the calculation, which gives $\Delta n_g = 1 - 0.5 = 0.5$. Substituting the gas constant 8.314 and temperature 298 into the relationship, we use the following equation:
$\Delta H - \Delta U = 0.5 \times 8.314 \times 298$
Calculating this product yields a final value of 1238.78 J mol-1.

For an ideal gas expanding within an isolated system, the first law of thermodynamics simplifies to $\Delta U = w$ because no heat is exchanged with the surroundings. A reversible expansion performs the maximum possible work, meaning the system loses more internal energy to the surroundings compared to an irreversible expansion, as the magnitude of work done satisfies $|w_{\text{rev}}| > |w_{\text{irrev}}|$. Given that internal energy is directly proportional to temperature through $\Delta U = n C_V \Delta T$, this greater reduction in energy in the reversible process results in a lower final temperature, confirming that $(T_f)_{\text{irrev}} > (T_f)_{\text{rev}}$.

The enthalpy change for any chemical reaction is determined by the difference between the activation energies of the forward and backward processes, represented as $\Delta H = E_f - E_b$. Since endothermic reactions are characterized by a positive enthalpy change, the net energy must be greater than zero, requiring that $E_f - E_b > 0$. This inequality mathematically necessitates that the activation energy for the forward reaction is greater than the activation energy for the reverse process, leading to the condition $E_b < E_f$.