Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

The standard enthalpy change for a chemical reaction is determined by finding the difference between the sum of the standard enthalpies of formation of the products and the sum of the standard enthalpies of formation of the reactants, with each value multiplied by its respective stoichiometric coefficient. Since the standard enthalpy of formation for elemental oxygen is defined as zero, only the compounds contribute to the energy calculation.

The total enthalpy change is given by the expression \Delta_r H^\ominus = \[2 \Delta_f H^\ominus (\mathrm{SO}_2) + 2 \Delta_f H^\ominus (\mathrm{H}_2 \mathrm{O})]$ - [2 \Delta_f H^\ominus (\mathrm{H}_2 \mathrm{S})]$. Substituting the provided values into this equation results in$\Delta_r H^\ominus = 2(-297.0) + 2(-286.0) - 2(-20.1)$. Carrying out the multiplication and addition leads to$\Delta_r H^\ominus = -594.0 - 572.0 + 40.2$, which simplifies to -1125.8 kJ/mol. The magnitude of this enthalpy change, when rounded to the nearest integer, is 1126.

The standard free energy change $\Delta G^{\ominus}$ for the reaction $N_2O_4 \rightleftharpoons 2NO_2$ is defined as the difference between the Gibbs free energy of the products and the reactants in their standard states, given by $\Delta G^{\ominus} = G_{2NO_2} - G_{N_2O_4}$. The values provided indicate that the change in free energy when 1 mole of $N_2O_4$ converts to the equilibrium mixture is $G_{eq} - G_{N_2O_4} = -0.84\ \text{kJ mol}^{-1}$, and the change when 2 moles of $NO_2$ convert to the equilibrium mixture is $G_{eq} - G_{2NO_2} = -6.24\ \text{kJ mol}^{-1}$. Using these, the standard free energy change is calculated as $\Delta G^{\ominus} = (G_{2NO_2} - G_{eq}) - (G_{N_2O_4} - G_{eq}) = 6.24\ \text{kJ mol}^{-1} - 0.84\ \text{kJ mol}^{-1} = 5.40\ \text{kJ mol}^{-1}$.

Since $\Delta G^{\ominus}$ is positive, the forward reaction is non-spontaneous under standard conditions, but this does not imply that the dissociation does not occur at all, as the system will still reach a state of minimum Gibbs free energy (equilibrium), making statement B incorrect. Furthermore, equilibrium reactions never proceed to complete conversion in either direction, as the forward and reverse reaction rates become equal, rendering statement C false. Statement A is also incorrect because the calculated value is $+5.40\ \text{kJ mol}^{-1}$ rather than $-5.40\ \text{kJ mol}^{-1}$. Consequently, only statements D and E are correct descriptions of the thermodynamic state of the equilibrium mixture relative to the pure substances.

The Born-Haber cycle relates the standard enthalpy of formation ($\Delta H_f^\circ$) of an ionic compound to the sum of the enthalpy changes for sublimation ($\Delta H_{sub}$), dissociation ($\Delta H_{diss}$), ionization ($\Delta H_{IE}$), electron gain ($\Delta H_{EG}$), and lattice enthalpy ($\Delta H_{lattice}$):

$\Delta H_f^\circ = \Delta H_{sub} + \frac{1}{2}\Delta H_{diss} + \Delta H_{IE} + \Delta H_{EG} + \Delta H_{lattice}$

Substituting the given values, where $\Delta H_f^\circ = -594\,\mathrm{kJ\,mol^{-1}}$, $\Delta H_{sub} = 155\,\mathrm{kJ\,mol^{-1}}$, $\frac{1}{2}\Delta H_{diss} = 75\,\mathrm{kJ\,mol^{-1}}$, $\Delta H_{IE} = 520\,\mathrm{kJ\,mol^{-1}}$, and $\Delta H_{EG} = -313\,\mathrm{kJ\,mol^{-1}}$, gives:

$-594 = 155 + 75 + 520 - 313 + \Delta H_{lattice}$

$-594 = 437 + \Delta H_{lattice}$

$\Delta H_{lattice} = -1031\,\mathrm{kJ\,mol^{-1}}$

The magnitude of the lattice enthalpy is $1031\,\mathrm{kJ\,mol^{-1}}$.

The relationship between molar heat capacities for an ideal gas is defined by the equation $C_P = C_V + R$, where $C_P$ and $C_V$ denote the heat capacity at constant pressure and volume, respectively. Since the gas constant $R$ is a positive value, $C_P$ must be greater than $C_V$, which contradicts the claim in the first statement.

In a constant volume process, the system performs no expansion or compression, meaning work done is zero, represented by the condition $w = 0$. According to the first law of thermodynamics, $\Delta U = q + w$, all heat added to the system effectively increases the internal energy, which manifests as an increase in the chaotic kinetic motion and temperature of the gas molecules, which aligns with the second statement.

To determine the standard Gibbs free energy change for the reaction, utilize the thermodynamic relationship $\Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus}$. First, calculate the standard reaction enthalpy by subtracting the enthalpy of formation of the reactants from the enthalpy of formation of the products, which is given by the expression $\Delta_r H^{\ominus} = 2(42) - (8 + 80) = -4$ kJ mol⁻¹. Next, find the standard reaction entropy by subtracting the molar entropy of the reactants from the molar entropy of the products, calculated as $\Delta_r S^{\ominus} = 2(200) - (140 + 250) = 10$ J mol⁻¹ K⁻¹.

Convert the entropy change to 0.01 kJ mol⁻¹ K⁻¹ so that the units are consistent with the enthalpy value. Finally, substitute the temperature of 600 K into the Gibbs free energy expression $\Delta_r G^{\ominus} = -4 - (600 \times 0.01)$, which results in a final value of -10 kJ mol⁻¹.

The relationship between the standard Gibbs free energy change and the equilibrium constant is given by $\Delta_r G^{\ominus} = -RT \ln K = -2.303 RT \log K$. First, determine the standard enthalpy change of the reaction, $\Delta_r H^{\ominus} = \sum \nu_p \Delta_f H^{\ominus}_p - \sum \nu_r \Delta_f H^{\ominus}_r$, which is $\Delta_r H^{\ominus} = 2(32) - (6 + x) = 58 - x$ kJ/mol. Next, calculate the standard entropy change of the reaction, $\Delta_r S^{\ominus} = \sum \nu_p S^{\ominus}_p - \sum \nu_r S^{\ominus}_r$, which is $\Delta_r S^{\ominus} = 2(222) - (146 + 280) = 444 - 426 = 18$ J K$^{-1}$ mol$^{-1}$, or $0.018$ kJ K$^{-1}$ mol$^{-1}$.

Using the relation $\Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus}$, equate this to the expression for equilibrium constant: $-2.303 RT \log K = \Delta_r H^{\ominus} - T \Delta_r S^{\ominus}$. Substituting the given values at $T = 500$ K, $\Delta_r G^{\ominus} = -2.303 \times 8.3 \times 10^{-3} \times 500 \times 2.2 \approx -21.026$ kJ/mol. Substituting this into the thermodynamic equation yields $-21.026 = (58 - x) - (500 \times 0.018)$, which simplifies to $-21.026 = 58 - x - 9$. Solving for $x$ gives $x = 58 - 9 + 21.026 = 70.026$. Thus, the value of $x$ is approximately 70.

The work done in a reversible, isothermal process is determined by the equation $W = -nRT \ln(V_2/V_1)$. Substituting the given values $n = 2 \text{ mol}$, $T = 300 \text{ K}$, $V_1 = 2 \text{ dm}^3$, and $V_2 = 20 \text{ dm}^3$ gives the magnitude $|W| = |2 \times 8.314 \times 300 \times \ln(10)| \approx 11.5 \text{ kJ}$, which corresponds to II. For an irreversible isothermal expansion against constant external pressure, the work is calculated using $W = -P_{\text{ext}}(V_2 - V_1)$. With $P_{\text{ext}} = 3 \text{ kPa}$ and a change in volume of $2 \text{ m}^3$, the magnitude is $|W| = |3 \text{ kPa} \times 2 \text{ m}^3| = 6 \text{ kJ}$, which matches III.

The change in internal energy for an ideal gas process is $\Delta U = n \overline{C}_V \Delta T$. Using $n = 1 \text{ mol}$, $\overline{C}_V = 1.5R$, and $\Delta T = 320 \text{ K}$, the magnitude is $|\Delta U| = |1 \times 1.5 \times 8.314 \times 320| \approx 4 \text{ kJ}$, matching I. The change in enthalpy at constant pressure is determined by $\Delta H = n \overline{C}_p \Delta T$. Applying $\overline{C}_p = 2.5R$ and $\Delta T = 337 \text{ K}$, the result is $|\Delta H| = |1 \times 2.5 \times 8.314 \times 337| \approx 7 \text{ kJ}$, which corresponds to IV.

The relationship between the equilibrium constant $K$, enthalpy change $\Delta H^{\theta}$, and entropy change $\Delta S^{\theta}$ is governed by the thermodynamic definition of Gibbs free energy change, expressed as $\Delta G^{\theta} = \Delta H^{\theta} - T \Delta S^{\theta} = -RT \ln K$ Rearranging this fundamental identity allows us to isolate the entropy term as $\Delta S^{\theta} = \frac{\Delta H^{\theta}}{T} + R \ln K$

To calculate this value, we use the given parameters $T=300\ \text{K}$, $R=8.3\ \text{J K}^{-1}\ \text{mol}^{-1}$, and the enthalpy change converted to joules as $\Delta H^{\theta}=28400\ \text{J mol}^{-1}$. Utilizing the conversion $\ln K = 2.3 \log K$, we evaluate the logarithmic term by expressing $K=1.8 \times 10^{-7}$ as $18 \times 10^{-8}$, which simplifies to $\log K = \log 18 - 8$. Applying the provided logarithmic values $\log 3 = 0.48$ and $\log 2 = 0.30$, we find $\log 18 = \log 2 + 2 \log 3 = 0.30 + 0.96 = 1.26$, resulting in $\log K = 1.26 - 8 = -6.74$.

Substituting these components into our expression yields $\Delta S^{\theta} = \frac{28400}{300} + (8.3)(2.3)(-6.74)$. Simplifying the terms gives $\Delta S^{\theta} = 94.67 - 128.66$, which equals approximately $-34\ \text{J K}^{-1}\ \text{mol}^{-1}$. Taking the magnitude as requested, we arrive at 34.

The first law of thermodynamics relies on the principle of conservation of energy, which dictates that the change in internal energy of a system is the sum of the heat transferred to it and the work performed on it. Because internal energy is a state function, the energy change for a reverse path is simply the negative of the energy change for the forward path. For the forward process from state X to Y, the change in internal energy is calculated using the relation $\Delta U = q + w$ as 10 J - 18 J, resulting in -8 J.

Returning from state Y back to X requires an internal energy change of 8 J. Given that 6 J of heat is evolved during this reverse process, we represent the heat transfer as $q = -6$ J. Substituting these values into the first law equation $\Delta U = q + w$, we get $8 = -6 + w$, which results \in $w = 14$ J. A positive value for work in this convention indicates that 14 J of work is performed on the gas by the surroundings.

The heat of atomization of methane accounts for the breaking of four $C-H$ bonds, establishing the bond energy of a single $C-H$ bond as $x/4$ kJ mol$^{-1}$. For ethane, the heat of atomization $y$ represents the energy required to break one $C-C$ bond and six $C-H$ bonds, such that $y = BE_{C-C} + 6(x/4)$, which simplifies to $BE_{C-C} = y - 1.5x$ kJ mol$^{-1}$. Converting this bond energy into Joules per molecule requires multiplying by $1000$ to convert kilojoules to Joules and dividing by the Avogadro constant $N_A$, resulting in $E = \frac{1000(y - 1.5x)}{N_A} = \frac{1000((2y - 3x)/2)}{N_A} = \frac{250(4y - 6x)}{N_A}$ J. Since the energy of a photon is related to wavelength by $E = hc/\lambda$, the longest wavelength $\lambda$ capable of breaking the bond is determined by $\lambda = hc/E$, which gives $\lambda = \frac{N_A hc}{250(4y - 6x)}$.

Thermodynamic processes dictate specific energy and entropy changes based on the constraints of the system: a reversible isothermal expansion requires heat exchange defined by $q = nRT \ln \frac{V_{2}}{V_{1}}$, whereas free expansion occurs in a vacuum against zero external pressure, resulting in $q = 0$. Irreversible compression work is evaluated using the external pressure opposing volume change, corresponding to $w = -p_{\text{ext}} (V_{1} - V_{2})$, and because entropy is a state function, its net change over a reversible cyclic process must be zero, matching the expression $\frac{q_{\text{rev}}}{T} = 0$.

Hess's Law states that the total enthalpy change for a chemical reaction is independent of the path taken and can be determined by summing the enthalpy changes of individual steps. To find the enthalpy of formation for solid $\mathrm{Al}_2\mathrm{Cl}_6$, we manipulate the four provided thermochemical equations to construct the target reaction $2\mathrm{Al}(\mathrm{s}) + 3\mathrm{Cl}_2(\mathrm{g}) \rightarrow \mathrm{Al}_2\mathrm{Cl}_6(\mathrm{s})$.

We retain the first reaction as given, multiply the second reaction by 3, multiply the third reaction by 6, and add the fourth reaction to produce the target solid. Summing these contributions gives the enthalpy change as $\Delta_{\mathrm{f}}H = -1200 - 492 - 498 + 663$ kJ/mol. This calculation results in -1527 kJ/mol.

The work done by an ideal gas in an isothermal process is determined by the external pressure $p_{ex}$ and the volume change, governed by the integral $w = -\int_{V_i}^{V_f} p_{ex} dV$. Reversible expansion occurs where the internal and external pressures remain balanced throughout, leading to the work formula $w = -nRT \ln \frac{V_f}{V_i}$, whereas free expansion occurs against zero external pressure, resulting in $w = 0$. For irreversible processes involving a constant external pressure, the work done during expansion is represented by $w = -p_{ex}(V_f - V_i)$, while the expression $w = -p_{ex}(V_i - V_f)$ corresponds to the work associated with irreversible compression to maintain consistency with the provided list.

Bomb calorimeters operate at constant volume, meaning the heat released during combustion reflects the change in internal energy, $\Delta U$, rather than the change in enthalpy, $\Delta H$. Since the reaction involves a change in the number of gaseous moles, we convert this internal energy measurement into the standard enthalpy of combustion, $\Delta H_c^{\circ}$, using the relationship $\Delta H = \Delta U + \Delta n_g RT$. With the ethanol mass of 3.365 g and its molar mass of 46 g mol$^{-1}$, the combustion of one mole of ethanol releases an internal energy of $-1359.8$ kJ mol$^{-1}$. For the balanced combustion equation $\mathrm{C_2H_5OH}(\ell) + 3\mathrm{O_2}(g) \rightarrow 2\mathrm{CO_2}(g) + 3\mathrm{H_2O}(\ell)$, the change in gaseous moles $\Delta n_g$ is $2 - 3 = -1$. Applying this with the gas constant $R = 8.314$ J K$^{-1}$ mol$^{-1}$ and temperature $298.15$ K, the enthalpy of combustion calculates to $-1362.27$ kJ mol$^{-1}$.

Hess's Law allows us to link this combustion enthalpy to the standard enthalpies of formation for the reactants and products. The enthalpy of combustion for the reaction is defined as the sum of the standard enthalpies of formation of the products minus those of the reactants, where the enthalpy of formation for oxygen is zero and for carbon dioxide it is equivalent to the enthalpy of combustion of graphite, $-393.5$ kJ mol$^{-1}$. By substituting the known values into the equation $-1362.27 = [2 \times (-393.5) + 3 \times (-285.8)] - \Delta H_f^{\circ}[\mathrm{C_2H_5OH}(\ell)]$, we determine the standard enthalpy of formation of ethanol to be approximately $-2.82 \times 10^2$ kJ mol$^{-1}$. Calculating the absolute value gives $2.82 \times 10^2$ kJ mol$^{-1}$, which rounds to 3.

The first law of thermodynamics is defined by the equation $\Delta U = q + w$, where $\Delta U$ represents the change in internal energy, $q$ is the heat exchanged with the surroundings, and $w$ is the work done on the system. Microwave radiation transfers heat into the water, meaning $q$ is positive, while the expansion of water during boiling constitutes work done by the system against the external pressure, rendering $w$ negative. Since the net energy added to the system as heat significantly exceeds the work done by expansion, the internal energy of the system increases, resulting in a positive value for $\Delta U$.

The magnitude of work done during an isothermal process corresponds to the area under the pressure-volume curve. For an expansion process, a single-stage step involves working against a constant external pressure equal to the lower final pressure, which results in a smaller area compared to a multi-stage reversible process, meaning $|w_B| > |w_A|$. Conversely, in a compression process, a single-stage step requires working against a constant external pressure equal to the higher final pressure, creating a rectangle larger than the area under the reversible curve, thus confirming $|w_C| > |w_D|$. Because the magnitude of work in compression is generally higher than in expansion due to the higher external pressure required to reduce volume, the overall order of magnitudes is $|w_C| > |w_D| > |w_B| > |w_A|$.

For an ideal gas undergoing isothermal reversible expansion, the internal energy change $\Delta U$ and enthalpy change $\Delta H$ are zero because both depend solely on temperature. The work done during a reversible isothermal process is expressed as $w = -nRT \ln(\frac{P_1}{P_2})$, where the value of $nRT$ is derived from initial conditions as $P_1V_1 = (0.5 \times 10^6 \text{ Pa}) \times (20.0 \times 10^{-3} \text{ m}^3) = 10,000 \text{ J}$. Substituting the pressure terms, the expression becomes $w = -10,000 \ln(\frac{0.5}{0.2}) = -10,000 \ln(2.5)$. Utilizing the relation $\ln(x) = 2.303 \log(x)$ alongside the provided values yields $\ln(2.5) = 2.303 (\log 5 - \log 2) = 2.303 (0.6989 - 0.3010) \approx 0.916$. The work done is calculated as $w = -10,000 \times 0.916 \text{ J} = -9160 \text{ J} = -9.16 \text{ kJ}$. Applying the first law of thermodynamics $\Delta U = q + w$, the heat exchange $q = -w$ equals $9.16 \text{ kJ}$, which approximates to $9.1 \text{ kJ}$.

At the transition temperature, the system exists in equilibrium, where the standard Gibbs free energy change is $\Delta G^{\circ} = 0$. Equating the expression $105 - 35 \log T$ to zero establishes the relationship
$105 - 35 \log T = 0$
which simplifies to $\log T = 3$. Solving for $T$ results in an absolute temperature of $1000 \mathrm{K}$. Subtracting 273 converts this value to the Celsius scale, yielding $727 ^\circ\mathrm{C}$.

The work done in any thermodynamic process is geometrically equivalent to the area under the pressure-volume curve, which is mathematically expressed by the integral $W = \int P \, dV$. Because the magnitude of work is directly proportional to the total area enclosed between the curve and the volume axis, the graph that encompasses the largest region represents the maximum work performed. Option A depicts the curve that covers the greatest area within the $P-V$ plane, thereby indicating the highest amount of work done by the system.

The relationship between the equilibrium constant $K$ and temperature $T$ is derived from the Gibbs free energy equation $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$ combined with the thermodynamic relation $\Delta G^{\circ} = -RT \ln K$. Equating these expressions yields $-RT \ln K = \Delta H^{\circ} - T\Delta S^{\circ}$, which rearranges to $\ln K = -\frac{\Delta H^{\circ}}{R} \left(\frac{1}{T}\right) + \frac{\Delta S^{\circ}}{R}$. Converting the natural logarithm to base 10 by dividing by $2.303$ gives the linear equation $\log_{10} K = -\frac{\Delta H^{\circ}}{2.303 R} \left(\frac{1}{T}\right) + \frac{\Delta S^{\circ}}{2.303 R}$, which follows the form $y = mx + c$, where the intercept $c$ is $\frac{\Delta S^{\circ}}{2.303 R}$ and the slope $m$ is $-\frac{\Delta H^{\circ}}{2.303 R}$.