Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

The thermodynamic relationship between standard Gibbs energy, enthalpy, and entropy is defined by the fundamental equation $\Delta_rG^{\circ} = \Delta_rH^{\circ} - T\Delta_rS^{\circ}$. By comparing this expression with the provided equation $\Delta_rG^{\circ} = A - BT$, we can directly identify that the constant $A$ corresponds to the standard reaction enthalpy, $\Delta_rH^{\circ}$, and the constant $B$ corresponds to the standard reaction entropy, $\Delta_rS^{\circ}$. Since an endothermic reaction is defined by the absorption of heat resulting in a positive change in enthalpy, the reaction must be endothermic whenever the constant $A$ is greater than zero.

The change in entropy for a system with constant pressure heat capacity $C_p$ undergoing a temperature change from an initial temperature $T_i$ to a final temperature $T_f$ is given by the integral $\Delta S = \int_{T_i}^{T_f} C_p \frac{dT}{T} = C_p \ln\left(\frac{T_f}{T_i}\right)$. When two identical blocks of mass $m$ are brought together, energy conservation dictates that the heat lost by the hotter block equals the heat gained by the cooler block, leading to an equilibrium temperature $T_f = \frac{T_1+T_2}{2}$. The total entropy change is the sum of the changes for each block, expressed as $\Delta S_{total} = C_p \ln\left(\frac{T_f}{T_1}\right) + C_p \ln\left(\frac{T_f}{T_2}\right) = C_p \ln\left(\frac{T_f^2}{T_1 T_2}\right)$. Substituting $T_f = \frac{T_1+T_2}{2}$ into this expression yields $\Delta S_{total} = C_p \ln\left(\frac{[(T_1+T_2)/2]^2}{T_1 T_2}\right)$, which simplifies to $C_p \ln\left[\frac{(T_1+T_2)^2}{4T_1T_2}\right]$.

The standard Gibbs free energy change for a reaction is defined by the relationship between the equilibrium constant and temperature, expressed by the following equation:
$\Delta G^\circ = -RT \ln K$
For the auto-ionization of water at 298 K, the equilibrium constant is 10^-14. By substituting the gas constant of 8.314/1000 kJ/mol K along with the given temperature and equilibrium constant into the equation, we compute the result. This calculation produces a value of approximately 80 kJ/mol, indicating the energy requirements for the process under standard conditions.

Hess's Law allows the total Gibbs free energy change of a chemical reaction to be determined by algebraically summing the free energy changes of individual steps involved in the process. To derive the target reaction $2 \mathrm{Fe}_2\mathrm{O}_3(s) + 6 \mathrm{CO}(g) \longrightarrow 4 \mathrm{Fe}(s) + 6 \mathrm{CO}_2(g)$, the first given equation is taken as is, while the second equation is multiplied by 3 to match the stoichiometry of carbon monoxide and eliminate the oxygen gas. The resulting free energy change is calculated as:

$\Delta_r G^\circ = \Delta_r G^\circ_1 + 3(\Delta_r G^\circ_2) = 1487.0 \text{ kJ mol}^{-1} + 3(-514.4 \text{ kJ mol}^{-1})$

$\Delta_r G^\circ = 1487.0 \text{ kJ mol}^{-1} - 1543.2 \text{ kJ mol}^{-1} = -56.2 \text{ kJ mol}^{-1}$

The relationship between enthalpy change and internal energy change is given by the equation $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ represents the change in the total number of moles of gaseous products and reactants. For the equality $\Delta H = \Delta U$ to hold, the condition $\Delta n_g = 0$ must be satisfied. In the reaction $2\mathrm{HI}(g)\rightarrow\mathrm{H}_2(g)+\mathrm{I}_2(g)$, the calculation yields $\Delta n_g = (1 + 1) - 2 = 0$, whereas the reactions in options A, C, and D result in non-zero values of $\Delta n_g = -2$, $\Delta n_g = -1$, and $\Delta n_g = -1$, respectively.

The First Law of Thermodynamics states that the change in internal energy is given by $\Delta U = q + W$, and for any cyclic process, the total change in internal energy is zero, such that $\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 0$. Using the given values for process AB, the internal energy change is $\Delta U_{AB} = q_{AB} + W_{AB} = 2 + (-5) = -3 \text{ kJ mol}^{-1}$. Substituting this result and the given $\Delta U_{BC} = -5 \text{ kJ mol}^{-1}$ into the cyclic process equation, $-3 - 5 + \Delta U_{CA} = 0$, yields $\Delta U_{CA} = 8 \text{ kJ mol}^{-1}$. Applying the First Law to process CA, where $\Delta U_{CA} = q_{CA} + W_{CA}$ and $W_{CA} = 3 \text{ kJ mol}^{-1}$, results in $8 = q_{CA} + 3$, which simplifies to $q_{CA} = 5 \text{ kJ mol}^{-1}$.

The standard free energy change is related to the equilibrium constant by the equation $\Delta G^\circ = -RT \ln K_p$. For the dissociation reaction $A_2(g) \rightleftharpoons 2A(g)$, starting with 1 mole of $A_2$ and a dissociation degree $\alpha = 0.2$, the equilibrium amounts are $1 - \alpha = 0.8$ moles of $A_2$ and $2\alpha = 0.4$ moles of $A$, with a total of $1.2$ moles. At a total pressure of $1 \text{ atm}$, the partial pressures are $p_{A_2} = \frac{0.8}{1.2} = \frac{2}{3} \text{ atm}$ and $p_A = \frac{0.4}{1.2} = \frac{1}{3} \text{ atm}$. Substituting these into the expression $K_p = \frac{(p_A)^2}{p_{A_2}}$, we obtain $K_p = \frac{(1/3)^2}{2/3} = \frac{1}{9} \times \frac{3}{2} = \frac{1}{6}$. The free energy change is therefore $\Delta G^\circ = -8.314 \times 320 \times \ln(1/6) = 8.314 \times 320 \times (\ln 2 + \ln 3) = 2660.48 \times (0.693 + 1.098) = 2660.48 \times 1.791 \approx 4764.9 \text{ J mol}^{-1}$.

The equilibrium between liquid and vapour phases is established when the Gibbs free energy change of the vaporization process satisfies the relationship $\Delta G^\circ = -RT \ln P$. The standard Gibbs free energy change of vaporization is calculated as the difference between the formation energies of the gaseous and liquid states, represented by the equation $103 - 100.7 = 2.3$ kcal mol-1, which converts to 2300 cal mol-1. Substituting the given values into the equilibrium equation gives $2300 = -2 \times 500 \times \ln P$. Simplifying this expression results in the equation $\ln P = -2.3$, and since the base of the natural logarithm raised to -2.3 is approximately 0.1, the vapour pressure $P$ is approximately 0.1 atm.

The relationship between the equilibrium constant $K$ and absolute temperature $T$ is governed by the van 't Hoff equation, expressed as $\ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}$. When plotted as $\ln K$ against $\frac{1}{T}$, the slope of the line is defined by $-\frac{\Delta H^\circ}{R}$. Because exothermic reactions are characterized by a negative enthalpy change $\Delta H^\circ$, this calculation results in a positive slope for the graphical representation. Lines A and B correctly illustrate this behavior by exhibiting positive slopes.

The thermodynamic relationship between the enthalpy of combustion at constant pressure and the internal energy of combustion at constant volume is given by $\Delta H = \Delta U + \Delta n_g RT$. First, analyze the balanced combustion reaction for benzene, $\mathrm{C_6H_6}(l) + \frac{15}{2}\mathrm{O_2}(g) \rightarrow 6\mathrm{CO_2}(g) + 3\mathrm{H_2O}(l)$, to determine the change in gaseous moles, calculated as $\Delta n_g = 6 - 7.5 = -1.5$. Using the internal energy of -3263.9 kJ mol$^{-1}$, the gas constant $8.314 \times 10^{-3}$ kJ K$^{-1}$ mol$^{-1}$, and the temperature of 298 K, substitute these values into the thermodynamic equation as $\Delta H = -3263.9 + (-1.5 \times 8.314 \times 10^{-3} \times 298)$. Solving this expression yields $\Delta H = -3263.9 - 3.716$, which results in -3267.6 kJ mol$^{-1}$.

Entropy is a measure of the randomness or disorder of a system, which decreases when the substance undergoes compression or moves toward a more ordered state. For an ideal gas undergoing isothermal compression, the change in entropy is given by the expression $\Delta S = nR \ln\left(\frac{P_1}{P_2}\right)$, where $P_1$ and $P_2$ represent the initial and final pressures. Since the pressure of the nitrogen gas increases from $1\text{ atm}$ to $5\text{ atm}$ in this process, the ratio $\frac{P_1}{P_2}$ is less than one, resulting in a negative value for the natural logarithm and consequently a negative change in entropy.

The total enthalpy change for the process is the sum of the enthalpy changes for cooling the liquid water, the phase change at $0^\circ\text{C}$, and cooling the resulting ice. Although the question specifies $58^\circ\text{C}$, calculations consistent with the provided options indicate a temperature change of $\Delta T = 5^\circ\text{C}$ is used.

The enthalpy change for cooling 1 mole of liquid water from $5^\circ\text{C}$ to $0^\circ\text{C}$ is $\Delta H_1 = C_p(\text{water}, l) \times \Delta T = 75.3\,\text{J}\,\text{mol}^{-1}\,\text{K}^{-1} \times 5\,\text{K} = 376.5\,\text{J}\,\text{mol}^{-1}$. The enthalpy change for the freezing process at $0^\circ\text{C}$ is $\Delta H_2 = 6000\,\text{J}\,\text{mol}^{-1}$. Finally, the enthalpy change for cooling the ice from $0^\circ\text{C}$ to $-5^\circ\text{C}$ is $\Delta H_3 = C_p(\text{ice}, s) \times \Delta T = 36.8\,\text{J}\,\text{mol}^{-1}\,\text{K}^{-1} \times 5\,\text{K} = 184\,\text{J}\,\text{mol}^{-1}$. Summing the magnitudes of these contributions gives the total enthalpy change as $\Delta H = 376.5\,\text{J}\,\text{mol}^{-1} + 6000\,\text{J}\,\text{mol}^{-1} + 184\,\text{J}\,\text{mol}^{-1} = 6560.5\,\text{J}\,\text{mol}^{-1}$, which is $6.56\,\text{kJ}\,\text{mol}^{-1}$.

Thermochemical problems rely on Hess's Law, which dictates that the total enthalpy change of a chemical process is independent of the pathway taken. This allows us to manipulate chemical equations algebraically, adding or multiplying them to derive a target equation from known experimental data. Our objective is to generate the reaction for the formation of methane, $\mathrm{C}_{\text{graphite}} + 2\mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{CH}_4(\mathrm{g})$. We begin with the first equation, $\mathrm{C}_{\text{graphite}} + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g})$, providing an enthalpy of -393.5 kJ/mol. To align with our target stoichiometry, we double the second reaction, $\mathrm{H}_2(\mathrm{g}) + \frac{1}{2}\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2\mathrm{O}(\mathrm{l})$, which gives $\mathrm{2H}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{2H}_2\mathrm{O}(\mathrm{l})$ and results in an enthalpy change of -571.6 kJ/mol. Finally, we include the third equation, $\mathrm{CO}_2(\mathrm{g}) + \mathrm{2H}_2\mathrm{O}(\mathrm{l}) \rightarrow \mathrm{CH}_4(\mathrm{g}) + \mathrm{2O}_2(\mathrm{g})$, which has an enthalpy of +890.3 kJ/mol. By summing these three equations, the intermediate species-carbon dioxide, water, and oxygen-cancel out completely, leaving only our target reactants and product. The total enthalpy change for the reaction is the sum of these manipulated values: -393.5 + -571.6 + 890.3, which equals -74.8 kJ/mol.

For an ideal gas, enthalpy is a state function that depends solely on temperature, expressed as $\Delta H = n C_p \Delta T$, which implies that $\Delta H = 0$ during any isothermal process. Because the gas undergoes expansion, the volume increases, leading to a greater spatial distribution of gas molecules and a corresponding positive change in entropy, represented by the relationship $\Delta S = nR \ln(V_2/V_1) > 0$.

The relationship between the enthalpy change ($\Delta H$) and the internal energy change ($\Delta U$) for a reaction is expressed by the equation $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ represents the difference between the number of moles of gaseous products and gaseous reactants. For the reaction $\mathrm{A(g)} \rightarrow \mathrm{A(l)}$, the change in gaseous moles is calculated as $\Delta n_g = 0 - 1 = -1$. Substituting the given enthalpy change of $\Delta H = -3RT$ into the thermodynamic equation yields $-3RT = \Delta U - RT$, which simplifies to $\Delta U = -2RT$. Comparing the magnitudes of these values, $|\Delta H| = |-3RT| = 3RT$ and $|\Delta U| = |-2RT| = 2RT$, it follows that $3RT > 2RT$, confirming that $|\Delta H| > |\Delta U|$.

The first law of thermodynamics states that the internal energy change of a system is given by $\Delta U = q + w$, accounting for both heat transfer and work interaction. Because an adiabatic process takes place without any heat exchange with the surroundings, the value of $q$ is zero. This constraint simplifies the energy balance equation to $\Delta U = w$, indicating that any change in the internal energy of the system is entirely equivalent to the adiabatic work.

The First Law of Thermodynamics is expressed as $\Delta U = Q - W$, where $\Delta U$ denotes the change in internal energy, $Q$ is the heat absorbed by the gas, and $W$ is the work done by the gas. For the process $A \to B$, the internal energy change is given by $\Delta U_{AB} = 5 - 8 = -3$ J. Since the system returns to its initial state $A$ in a cycle, the net change in internal energy must be zero, implying $\Delta U_{BA} = -\Delta U_{AB} = 3$ J.

For the reverse process $B \to A$, where heat is evolved such that $Q = -3$ J, the equation $\Delta U_{BA} = Q - W$ becomes $3 = -3 - W$. Solving for $W$ results in $W = -6$ J. The negative sign indicates that work is done by the surrounding on the gas, confirming that 6 J of work is performed on the system.

The work done by a system during gas expansion against a constant external pressure is defined by the relationship $W = -P_{ext}\Delta V$, which can be calculated using the ideal gas behavior as $W = -\Delta n_g RT$. The decomposition reaction $2\mathrm{H}_2\mathrm{O}_2(\mathrm{l})\rightleftharpoons 2\mathrm{H}_2\mathrm{O}(\mathrm{l})+\mathrm{O}_2(\mathrm{g})$ indicates that 100 mol of hydrogen peroxide will produce 50 mol of $O_2(g)$. Consequently, the work is determined by the equation $W = -(50\,\mathrm{mol})(8.3\,\mathrm{J\,K^{-1}\,mol^{-1}})(300\,\mathrm{K}) = -124500\,\mathrm{J} = -124.5\,\mathrm{kJ}$, and since work done by the gas during expansion is positive, the magnitude of the work is 124.5 kJ.

The equilibrium constant for a reaction is intrinsically linked to the standard Gibbs free energy change through the relationship $\Delta G^{\circ} = -RT \ln K_{\text{eq}}$. The change in Gibbs free energy itself is calculated using the thermodynamic definition $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$. Inserting the provided values where $\Delta H^{\circ}$ is -29.8 kJ mol⁻¹, $\Delta S^{\circ}$ is -0.100 kJ K⁻¹ mol⁻¹, and the temperature is 298 K results in the evaluation $\Delta G^{\circ} = -29.8 - (298 \times -0.100)$. This calculation simplifies to $\Delta G^{\circ} = 0$, which dictates that $\ln K_{\text{eq}} = 0$ because $\Delta G^{\circ} = -RT \ln K_{\text{eq}}$. Consequently, $K_{\text{eq}} = 1$.

The spontaneity of a reaction at a given temperature depends on the change in Gibbs free energy, which is directly related to the $-\ln K_p$ values plotted on the Ellingham diagram. For the reaction $MO(s) + C(s) \rightarrow M(s) + CO(g)$, the overall process is spontaneous when the free energy change is negative, a condition satisfied in this specific plot when the line for carbon oxidation lies below that of the metal oxidation. Based on the provided data, this condition holds true at temperatures below 1200 K, indicating that carbon effectively acts as a reducing agent to convert the metal oxide into the pure metal within this temperature range.