Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

The standard heat of formation for a chemical compound can be calculated using the enthalpies of combustion of the reactants and the product by applying the principle that the enthalpy change of a reaction is the sum of the enthalpies of combustion of the reactants minus the enthalpy of combustion of the product. The formation of ethane from graphite and hydrogen is represented by the stoichiometric equation $2\mathrm{C}(\text{graphite}) + 3\mathrm{H}_2(\text{g}) \rightarrow \mathrm{C}_2\mathrm{H}_6(\text{g})$, and the relationship for the heat of formation is $\Delta_f H^{\circ} = 2 \Delta H_{\text{comb}}^{\circ}(\text{graphite}) + 3 \Delta H_{\text{comb}}^{\circ}(\mathrm{H}_2) - \Delta H_{\text{comb}}^{\circ}(\mathrm{C}_2\mathrm{H}_6)$.

Given the values of $-286\text{ kJ/mol}$ for graphite, $-393.5\text{ kJ/mol}$ for hydrogen, and $-1560\text{ kJ/mol}$ for ethane, we substitute these into the expression: $\Delta_f H^{\circ} = 2(-286) + 3(-393.5) - (-1560)$. Calculating this yields $-572 - 1180.5 + 1560$, which results in $-192.5\text{ kJ/mol}$.

Entropy is a state function, which allows us to determine the total change by summing the entropy variations of individual, reversible stages: phase transitions at constant temperature and temperature changes in a single phase. Converting 1 kg of ice at 273 K to water vapour at 383 K involves four distinct stages: melting the ice, heating the liquid water, vaporizing the water, and heating the steam.

For the first stage, the entropy change is calculated as $\Delta S_1 = \frac{\Delta H_{fusion}}{T} = \frac{334}{273}$, which equals 1.22 kJ K$^{-1}$ kg$^{-1}$. Warming the liquid water from 273 K to 373 K results in $\Delta S_2 = C_{p, liquid} \ln(\frac{373}{273}) = 4.2 \ln(\frac{373}{273})$, yielding 1.31 kJ K$^{-1}$ kg$^{-1}$. The vaporization of water at 373 K provides $\Delta S_3 = \frac{\Delta H_{vap}}{T} = \frac{2491}{373}$, which equals 6.67 kJ K$^{-1}$ kg$^{-1}$. Finally, heating the steam from 373 K to 383 K contributes $\Delta S_4 = C_{p, vapour} \ln(\frac{383}{373}) = 2.0 \ln(\frac{383}{373})$, adding 0.05 kJ K$^{-1}$ kg$^{-1}$. Summing these individual contributions of 1.22 + 1.31 + 6.67 + 0.05 gives a total entropy change of 9.26 kJ K$^{-1}$ kg$^{-1}$.

A chemical reaction is spontaneous when the change in Gibbs free energy, defined by the equation $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$, becomes negative. For this endothermic reaction with a positive entropy change, feasibility is only achieved when the temperature is high enough for the entropy-driven term to outweigh the enthalpy barrier.

Equilibrium occurs at the point where the Gibbs free energy is exactly zero, which is determined using the relation $T = \frac{\Delta H^\circ}{\Delta S^\circ}$. Converting the enthalpy of 491.1 kJ mol^-1 to 491100 J mol^-1 ensures consistency with the entropy of 198.0 J K^-1 mol^-1, leading to the following calculation:
$T = \frac{491100}{198.0} = 2480.3 \text{ K}$
Temperatures exceeding 2480.3 K ensure the reaction proceeds spontaneously.

For a diatomic ideal gas at temperatures where rotational degrees of freedom are fully active, the molar heat capacities are fixed constants given by $C_P = \frac{7}{2}R$ and $C_V = \frac{5}{2}R$. These values remain entirely independent of pressure and volume, while the internal energy $U$ exhibits a strictly linear dependence on temperature $T$. Any graphical representation suggesting that these heat capacities vary with their respective state variables, or that internal energy does not scale proportionally with temperature, fails to accurately describe these thermodynamic relationships.

Kirchhoff's Law describes how the enthalpy change of a process varies with temperature based on the specific heat difference between the final and initial states. In the case of iodine sublimation, the shift in enthalpy as the temperature increases is governed by the difference in specific heats between the gaseous and solid forms.

Calculate the difference in specific heat, $\Delta C_p$, by subtracting the solid's specific heat from the vapor's specific heat, which yields $\Delta C_p = 0.031 - 0.055 = -0.024 \text{ cal g}^{-1} \text{ K}^{-1}$. The temperature dependence follows the relation:
$\Delta H_2 = \Delta H_1 + \Delta C_p (T_2 - T_1)$
Substituting the known values, the expression becomes $\Delta H_2 = 24 + (-0.024)(50)$. This simplifies to $\Delta H_2 = 24 - 1.2$, providing a final result of $22.8 \text{ cal g}^{-1}$.

The spontaneity of a process is determined by the Gibbs free energy equation $\Delta G = \Delta H - T\Delta S$, which requires $\Delta G < 0$ for a reaction to occur spontaneously. When $\Delta H < 0$ and $\Delta S > 0$, the expression $\Delta H - T\Delta S$ yields a negative value at all temperatures because both the exothermic enthalpy term and the subtracted entropy term are negative. This specific combination guarantees that the free energy change will remain below zero regardless of the thermal conditions.

For an ideal gas undergoing isothermal expansion, the temperature remains constant, meaning the product of pressure and molar volume must stay fixed as determined by the relation $PV_m = K$. Because this requirement dictates that pressure must be inversely proportional to molar volume according to $P = \frac{K}{V_m}$, the correct graphical depictions must show a horizontal line for $PV_m$ versus volume (Graph C) and a rectangular hyperbola for $P$ versus $V_m$ (Graph A). Since these two plots accurately reflect the thermodynamic conditions of an ideal gas, the remaining graphs labeled (B) and (D) contradict these fundamental relationships and therefore fail to represent an isothermal process.

Hess's Law of Constant Heat Summation dictates that the total enthalpy change of a chemical process is independent of the pathway taken, meaning the energy change for a direct reaction is equivalent to the sum of the enthalpy changes of its constituent individual steps. The overall combustion of graphite, represented by $C(\text{graphite}) + O_2(g) \rightarrow CO_2(g)$ with an enthalpy change of $x$, can be decomposed into two sequential steps: the partial combustion of graphite to carbon monoxide with an enthalpy change of $y$, and the subsequent further oxidation of carbon monoxide to carbon dioxide with an enthalpy change of $z$. By adding the two intermediate equations $C(\text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g)$ and $CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)$, the intermediate carbon monoxide cancels out, leaving the original reaction equation and confirming the relationship $x = y + z$.

For a chemical process to occur spontaneously, the change in Gibbs free energy must be negative, as defined by the relationship $\Delta G = \Delta H - T\Delta S < 0$. Rearranging this inequality to $\Delta H < T\Delta S$ reveals that the process becomes spontaneous when the temperature exceeds the ratio of the enthalpy change to the entropy change, $T > \frac{\Delta H}{\Delta S}$. Substituting the provided values of 200 J mol⁻¹ and 40 J K⁻¹ mol⁻¹ into the expression $T = \frac{200}{40}$, the threshold temperature is 5 K. Therefore, the process will be spontaneous only at temperatures strictly above 5 K.

When an ideal gas expands against a constant external pressure, the energy exchanged with the surroundings as work is determined by the formula $W = -P_{ext} \Delta V$. The negative sign is a convention indicating that the system is performing work during the expansion, thereby losing energy.

The change in volume is calculated by subtracting the initial volume of 1 L from the final volume of 10 L, which results in 9 L. Multiplying this volume by the constant external pressure of 1 bar gives a product of -9 L bar. Converting this value into the appropriate energy unit of kilojoules results in a final work value of -0.9 kJ.

In an isothermal process for an ideal gas, the internal energy change is zero, meaning that any work performed on the system is released as an equivalent amount of heat. The work done during compression is calculated using the equation $w = -P_{\text{ext}}(V_2 - V_1)$. Given an external pressure $P_{\text{ext}} = 4 \, \mathrm{Nm}^{-2}$, an initial volume $V_1 = 5 \, \mathrm{m}^3$, and a final volume $V_2 = 1 \, \mathrm{m}^3$, the work done is $w = -4 \times (1 - 5) = 16 \, \mathrm{J}$.

This energy is transferred as heat to $1 \, \text{mole}$ of aluminum, which has a molar heat capacity $C_m = 24 \, \mathrm{J \, mol}^{-1} \mathrm{K}^{-1}$. Using the heat transfer equation $q = n C_m \Delta T$, we substitute these values into $16 = 1 \times 24 \times \Delta T$. Solving for the temperature change gives $\Delta T = \frac{16}{24} \, \mathrm{K}$, resulting in an increase of $\frac{2}{3} \, \mathrm{K}$.

The internal energy change for an ideal gas depends solely on temperature, which allows us to determine the change using the molar heat capacity at constant volume through the formula
$\Delta U = n C_V \Delta T$
Given that the gas undergoes a temperature increase from 100 K to 200 K with 5 mol of gas and a molar heat capacity of 28 J K⁻¹ mol⁻¹, the calculation is 5 times 28 times 100, which yields 14,000 J or 14 kJ.

For the change in the product of pressure and volume, we apply the ideal gas law relation
$\Delta(pV) = nR \Delta T$
Using the values 5 mol for the amount of gas, 8 J K⁻¹ mol⁻¹ for the gas constant, and a temperature difference of 100 K, the calculation results in 5 times 8 times 100, which equals 4,000 J or 4 kJ. These results confirm that the process involves an internal energy change of 14 kJ and a change in the pressure-volume product of 4 kJ.

The relationship between enthalpy change and internal energy change for a chemical process involving gases is defined by the equation $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ represents the difference in the number of moles between gaseous products and gaseous reactants. For the combustion of one mole of liquid heptane, the balanced chemical equation is $\mathrm{C}_7\mathrm{H}_{16(l)} + 11\mathrm{O}_2 \rightarrow 7\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O}_{(l)}$, where the gaseous species are the oxygen reactant and the carbon dioxide product. Calculating the change in gaseous moles from this reaction yields $\Delta n_g = 7 - 11 = -4$. Substituting this value into the rearranged expression $\Delta H - \Delta U = \Delta n_g RT$ results in a final difference of $-4RT$.

The first law of thermodynamics relates the change in internal energy to heat and work through the equation $\Delta U = q + w$. In an adiabatic process, there is no exchange of heat with the surroundings, meaning $q$ is zero, which simplifies the relationship to $\Delta U = w$. Therefore, the equation $\Delta U = -w$ is incorrect because it contradicts the fundamental energy balance required for an adiabatic system.

In thermodynamics, state functions are properties that depend solely on the current state of a system, whereas path functions depend on the specific sequence of steps taken to transition between states. Heat ($q$) and work ($w$) represent energy in transit rather than properties inherent to a system, which makes them path-dependent. In contrast, combinations like $q+w$, which relates to the change in internal energy, or $H-TS$, which defines Gibbs free energy, are determined exclusively by the initial and final states of the system and therefore act as state functions.

Entropy acts as a measure of disorder within a system, and for reactions involving gases, this disorder is directly related to the total number of gaseous moles present. When a chemical process results in a net decrease in the number of gaseous moles, the system transitions to a more ordered state, signifying a negative change in entropy. This is clearly observed in the synthesis of ammonia, where the reaction $N_2 (g) + 3H_2 (g) \rightleftharpoons 2NH_3 (g)$ converts four moles of gaseous reactants into only two moles of product, creating a condition where $\Delta n_g < 0$ and the entropy decreases.

When the temperature of a substance changes at constant pressure, its enthalpy change is calculated by integrating the molar heat capacity function with respect to temperature and scaling the result by the total amount of substance in moles. Because the heat capacity for silver is defined as $C_P = 23 + 0.01T$, we integrate this linear expression across the interval from $300\text{ K}$ to $1000\text{ K}$ and multiply the value by the three moles provided.

The mathematical expression for this process is $\Delta H = n \int_{T_1}^{T_2} C_P \, dT$, which translates to $3 \int_{300}^{1000} (23 + 0.01T) \, dT$. Performing the integration yields the function $3 [23T + 0.005T^2]$ evaluated between the given temperature limits. Inserting these bounds, the calculation becomes $3 [(23 \times 1000 + 0.005 \times 1000^2) - (23 \times 300 + 0.005 \times 300^2)]$, which simplifies to $3 [(23000 + 5000) - (6900 + 450)]$. This results in $3 \times 20650$, totaling $61950\text{ J}$, which is approximately $62\text{ kJ}$.

The first law of thermodynamics establishes that the internal energy of a system changes according to the net balance of work done on it and the heat transferred out. Since compressing a spring constitutes work done on the system, the 10 kJ represents an addition of energy, whereas the 2 kJ of heat escaping the system represents a reduction. By applying the energy balance equation $\Delta U = 10 - 2$, the total change in internal energy is determined to be 8 kJ.

The magnitude of work done during a reversible isothermal expansion of an ideal gas is described by the equation $|w| = nRT \ln(V_f/V_i)$. By distributing the terms, this expression becomes $|w| = nRT \ln V_f - nRT \ln V_i$, which allows the relationship between work and the natural logarithm of volume to be interpreted as a linear function $y = mx + C$. Within this framework, the slope is defined as $nRT$ and the y-intercept is given by $-nRT \ln V_i$. Because $T_2 > T_1$, the slope associated with the higher temperature is steeper, reflecting a more rapid increase in work for a given expansion. Furthermore, this higher temperature results in a y-intercept that is more negative, which, when combined with the steeper slope, ensures that the curve representing $T_2$ remains positioned above the curve for $T_1$ as the volume increases.

The relationship between the standard Gibbs free energy change and the equilibrium constant $K$ is defined by $\Delta_rG^\circ = -RT \ln K$, where $K = \frac{[Y]}{[X]}$. A positive value of $\Delta_rG^\circ$ results in $\ln K < 0$, signifying $K < 1$ and indicating that the reactant X is the major component, whereas a negative value of $\Delta_rG^\circ$ results in $\ln K > 0$, signifying $K > 1$ and indicating that the product Y is the major component.

Calculating the values for the provided temperatures, at $T = 315$ K, $\Delta_rG^\circ = 120 - \frac{3}{8}(315) = 1.875$ kJ $\mathrm{mol}^{-1}$. Since this value is greater than zero, X is the major component, which confirms the validity of option A. At $T = 350$ K, $\Delta_rG^\circ = 120 - 131.25 = -11.25$ kJ $\mathrm{mol}^{-1}$, which is less than zero, meaning Y is the major component and invalidating option B. For $T = 300$ K, $\Delta_rG^\circ = 120 - 112.5 = 7.5$ kJ $\mathrm{mol}^{-1}$, and for $T = 280$ K, $\Delta_rG^\circ = 120 - 105 = 15$ kJ $\mathrm{mol}^{-1}$. Because $\Delta_rG^\circ > 0$ in both these instances, X is the major component, which invalidates both options C and D.