Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

The enthalpy of formation of methane, $\Delta_f H^{\circ}$, is determined by the energy required to atomize the reactants minus the energy released during bond formation: $\Delta_f H^{\circ} = \Delta_{\text{sub}} H(\text{C}) + 2\Delta_{\text{diss}} H(\text{H}_2) - 4E_{\text{C-H}}$, where $E_{\text{C-H}}$ represents the bond enthalpy of the $\text{C}-\text{H}$ bond. Substituting the provided values yields $-X = Y + 2Z - 4E_{\text{C-H}}$, which describes the energy change of the reaction $\text{C (graphite)} + 2\text{H}_2 (g) \rightarrow \text{CH}_4 (g)$. Rearranging the equation to solve for $E_{\text{C-H}}$ results in $4E_{\text{C-H}} = X + Y + 2Z$, simplifying to $\frac{X+Y+2Z}{4}$.

Molar heat capacity measures the energy required to raise the temperature of one mole of a substance by one degree, a property fundamentally determined by how many ways a substance can store thermal energy through its internal degrees of freedom. Liquids, such as $\mathrm{Br}_2(\ell)$, typically exhibit higher molar heat capacities than solids because their greater degrees of freedom allow them to absorb more energy without a significant increase in temperature. In contrast, monoatomic gases like $\mathrm{He}(g)$ have the lowest molar heat capacities as their energy storage is largely restricted to simple translational motion, resulting in the distinct order $\mathrm{Br}_2(\ell) > \mathrm{Cu}(s) > \mathrm{He}(g)$.

Chemical equilibrium is achieved when the Gibbs free energy change of a reaction is zero, which connects the enthalpy and entropy changes through the relationship $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$. By rearranging this condition, the equilibrium temperature is determined by the ratio of the standard enthalpy change to the standard entropy change of the reaction.

To calculate the change in standard entropy, subtract the total entropy of the reactants from the entropy of the product:
$\Delta S^\circ = 110 - \left( \frac{1}{2} \times 70 + \frac{5}{2} \times 50 \right) = -50 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$

The enthalpy change of $-35$ kJ mol⁻¹ must be converted to $-35000$ J mol⁻¹ to ensure units are consistent. Applying the equilibrium condition $T = \frac{\Delta H^\circ}{\Delta S^\circ}$ and substituting these values yields the temperature:
$T = \frac{-35000}{-50} = 700 \, \mathrm{K}$

Hess's Law of constant heat summation provides the framework here, establishing that the total enthalpy change of a chemical transformation remains constant regardless of the steps taken to reach the final product. By treating the energy released in each process as a negative enthalpy value-specifically $\Delta H = -X$ for the diamond-to-graphite conversion, $\Delta H = -Y$ for diamond combustion, and $\Delta H = -Z$ for graphite combustion-we can manipulate these equations to relate the variables. To derive the equation for the conversion of diamond into graphite, one must subtract the combustion equation for graphite from the combustion equation for diamond. This subtraction cancels out the common $\mathrm{O}_2(\text{g})$ and $\mathrm{CO}_2(\text{g})$ terms, isolating the desired transition $\mathrm{C}(\text{diamond}) \rightarrow \mathrm{C}(\text{graphite})$. Applying the same arithmetic to the enthalpy terms yields the relationship $-X = -Y - (-Z)$, which simplifies to $-X = -Y + Z$, or equivalently $X = Y - Z$.

For an isothermal process involving an ideal gas, the change in internal energy is $\Delta U = n C_v \Delta T = 0$ because the temperature remains constant. According to the first law of thermodynamics, $\Delta U = q + w$, which implies $q = -w$ for an isothermal process. The reversible work done by the gas is determined using the equation $w = -nRT \ln \left( \frac{V_2}{V_1} \right)$, which can be expressed in terms of log base 10 as $w = -nRT \times 2.3 \times \log \left( \frac{V_2}{V_1} \right)$ given the provided conversion $\ln 10 = 2.3$.

Substituting the given values $n = 1$, $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$, $T = 300 \text{ K}$, and the ratio $V_2 / V_1 = 20 / 10 = 2$, the work done calculation is $w = -1 \times 8.3 \times 300 \times 2.3 \times \log 2$. Using $\log 2 = 0.30$, the equation becomes $w = -2490 \times 2.3 \times 0.30 = -1718.1 \text{ J}$, which is equivalent to $-1.718 \text{ kJ}$. Consequently, the heat exchanged is $q = -w = 1.718 \text{ kJ}$. Thus, the values are $\Delta U = 0$, $q = 1.718 \text{ kJ}$, and $w = -1.718 \text{ kJ}$.

The total enthalpy change for this phase transition process is calculated by summing the enthalpy changes of the sequential steps: cooling liquid water from $10^{\circ}\mathrm{C}$ to $0^{\circ}\mathrm{C}$, freezing it at $0^{\circ}\mathrm{C}$, and cooling the resulting ice from $0^{\circ}\mathrm{C}$ to $-10^{\circ}\mathrm{C}$. The enthalpy change for cooling the liquid water is $\Delta H_1 = C_p(\mathrm{H}_2\mathrm{O}(\ell)) \Delta T = y(0 - 10) = -10y \text{ J/mol}$, and the enthalpy change for cooling the ice is $\Delta H_3 = C_p(\mathrm{H}_2\mathrm{O}(s)) \Delta T = z(-10 - 0) = -10z \text{ J/mol}$. The phase change at $0^{\circ}\mathrm{C}$ corresponds to the negative of the fusion enthalpy, which is $\Delta H_2 = -\Delta_{\text{fus}}H = -x \text{ kJ/mol} = -1000x \text{ J/mol}$. Combining these individual contributions, the total enthalpy change is $\Delta H = -1000x - 10y - 10z = -10(100x + y + z) \text{ J/mol}$.

The spontaneity of a reaction is determined by the Gibbs free energy change given by the equation $\Delta G = \Delta H - T\Delta S$, where the process is spontaneous when $\Delta G < 0$ and reaches equilibrium when $\Delta G = 0$. Given that both $\Delta H$ and $\Delta S$ are positive, the equilibrium temperature is defined by $T_e = \frac{\Delta H}{\Delta S}$. For the reaction to be spontaneous, the inequality $\Delta H - T\Delta S < 0$ must hold, which rearranges to $T > \frac{\Delta H}{\Delta S}$, confirming that the reaction is spontaneous when $T > T_e$.

Hess's Law of Constant Heat Summation dictates that the enthalpy change of a reaction is independent of the path taken, meaning the heat of dilution can be determined by finding the difference between the enthalpies of solution at two different concentrations. To find the enthalpy of dilution for $\mathrm{HCl} \cdot 10 \mathrm{H}_2\mathrm{O}$ to $\mathrm{HCl} \cdot 40 \mathrm{H}_2\mathrm{O}$, subtract the enthalpy of dissolution for equation (a) from equation (b). This calculation is represented by the equation $\Delta H_{\text{dilution}} = \Delta H_{(b)} - \Delta H_{(a)}$. Substituting the given values, $\Delta H_{\text{dilution}} = -72.79 \text{ kJ mol}^{-1} - (-69.01 \text{ kJ mol}^{-1})$, which results in $-3.78 \text{ kJ mol}^{-1}$. The magnitude of this energy change confirms the specified value for the dilution process.

The First Law of Thermodynamics establishes that the change in internal energy is determined by the heat exchanged and the work performed on the system, represented by the equation $\Delta U = q + w$. Because the vessel is thermally insulated, no heat enters or escapes the system, meaning the heat interaction is $q = 0$. Mechanical stirring adds energy to the liquid by performing work on it, which is defined as positive work, resulting in $w > 0$. Consequently, the internal energy of the liquid increases, confirming that $\Delta U > 0$.

The spontaneity of a chemical process is governed by the Gibbs free energy equation, $\Delta G = \Delta H - T \Delta S$, where a reaction proceeds spontaneously only if $\Delta G < 0$. Since the reaction is endothermic, the enthalpy change $\Delta H$ is positive, which creates a barrier to spontaneity. Because the reaction becomes spontaneous as the temperature increases, the entropy change $\Delta S$ must be positive, allowing the $T \Delta S$ term to grow sufficiently to overcome the positive enthalpy and drive the overall free energy below zero.

Thermodynamic potentials offer a fundamental method for describing how properties like energy and enthalpy respond to variations in temperature, pressure, or volume. Starting with the differential of Gibbs free energy, $dG = V dP - S dT$, one finds that keeping pressure constant leads to $(\frac{\partial G}{\partial T})_P = -S$, whereas keeping temperature constant yields $(\frac{\partial G}{\partial P})_T = V$. The specific heat capacities are similarly defined by examining how enthalpy changes with temperature at constant pressure, giving $(\frac{\partial H}{\partial T})_P = C_P$, and how internal energy changes with temperature at constant volume, leading to $(\frac{\partial U}{\partial T})_V = C_V$.

The heat of formation for any compound is determined by applying Hess's Law, which relates the enthalpy change of a reaction to the heats of formation of all substances involved. The combustion of benzene, which releases energy, follows the reaction equation
$\mathrm{C}_6\mathrm{H}_6 + \frac{15}{2} \mathrm{O}_2 \rightarrow 6 \mathrm{CO}_2 + 3 \mathrm{H}_2\mathrm{O}$
with an enthalpy change of -3267.0 kJ mol⁻¹. Since oxygen is in its standard state, its heat of formation is zero. Consequently, the enthalpy of combustion is equal to the sum of the heats of formation of the products minus the heat of formation of benzene.

Substituting the standard heats of formation for carbon dioxide (-393.5 kJ mol⁻¹) and water (-286.0 kJ mol⁻¹) results in the expression
$-3267.0 = [6 \times (-393.5) + 3 \times (-286.0)] - \Delta H_f(\mathrm{C}_6\mathrm{H}_6)$
Performing the arithmetic gives $6 \times (-393.5) = -2361.0$ and $3 \times (-286.0) = -858.0$, which sum to -3219.0. Substituting this total yields $-3267.0 = -3219.0 - \Delta H_f(\mathrm{C}_6\mathrm{H}_6)$. Rearranging to solve for the heat of formation of benzene, $\Delta H_f(\mathrm{C}_6\mathrm{H}_6) = 3267.0 - 3219.0$, we find the value is 48 kJ mol⁻¹.

Resonance energy characterizes the energetic stabilization of a molecule as the absolute difference between its experimentally observed enthalpy of formation and the theoretical enthalpy calculated for a structure with localized, non-resonant bonds. By subtracting the total bond energy of the formed product bonds from the energy required to break the reactant bonds, the theoretical enthalpy of formation is determined as follows:

$\Delta H_{\text{theoretical}} = 940 + \frac{1}{2}(500) - (410 + 602) = 178 \text{ kJ mol}^{-1}$

The magnitude of the resonance energy is then found by calculating the difference between this theoretical value and the actual enthalpy of formation of $80 \text{ kJ mol}^{-1}$, resulting in $|80 - 178| = 98 \text{ kJ mol}^{-1}$.

Haber's process involves the equilibrium $\mathrm{N}_2(\mathrm{g}) + 3\mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2\mathrm{NH}_3(\mathrm{g})$, which is characterized by a negative enthalpy change $\Delta H_r < 0$ and a decrease in entropy $\Delta S_r < 0$. Because the enthalpy of reaction is negative, the function $\frac{-\Delta H_r^{\circ}}{T}$ displays an inverse dependency on temperature, resulting in a trend where it first decreases before stabilizing at a constant value. The same thermodynamic behavior is observed for the function $\frac{-\Delta G_R^{\circ}}{T}$, confirming that both properties exhibit an initial decline followed by a constant region.

The Born-Haber cycle relies on Hess's Law, which states that the total enthalpy change for the formation of an ionic solid from its elements is the sum of the enthalpy changes of the individual energetic steps. This cycle accounts for the sublimation of the metal, its ionization, the dissociation of the diatomic gas, the electron gain of the nonmetal, and the assembly of the gaseous ions into a crystal lattice.

The overall energy balance is expressed by the equation:
$\Delta H_f^{\circ} = \Delta H_{\text{sub}}^{\circ} + \Delta H_{i}^{\circ} + \frac{1}{2} \Delta H_{\text{bond}}^{\circ} + \Delta H_{eg}^{\circ} - \Delta H_{\text{lattice}}^{\circ}$
Substituting the provided experimental values into this relationship gives:
$-400 = 100 + 500 + \frac{1}{2} \Delta H_{\text{bond}}^{\circ} - 300 - 800$
Simplifying the constants on the right side results in $-400 = \frac{1}{2} \Delta H_{\text{bond}}^{\circ} - 500$. Solving for the dissociation term yields $\frac{1}{2} \Delta H_{\text{bond}}^{\circ} = 100$, confirming that the bond dissociation enthalpy of the molecule is 200 kJ/mol.

Phase transition processes, such as the melting of ice, occur at a constant temperature because the heat energy supplied is utilized exclusively to convert the solid phase into the liquid phase rather than increasing the thermal motion of the particles. For the equilibrium defined by $\mathrm{H_2O}\,(\text{s}) \rightleftharpoons \mathrm{H_2O}\,(\text{l})$, the absorbed energy is entirely invested in overcoming the intermolecular forces that bind water molecules within the rigid ice structure. Because the temperature remains steady during this process, the average kinetic energy of the molecules does not increase until all the ice has completely melted.

The standard enthalpy of formation, denoted as $\Delta_{\mathrm{f}} H^{\circ}$, is defined as the enthalpy change when one mole of a substance is formed from its constituent elements in their most stable forms under standard conditions. For oxygen, the most stable allotrope at 298 K is $\mathrm{O}_2(\mathrm{g})$, which means $\Delta_{\mathrm{f}} H_{298}^{\circ}$ for $\mathrm{O}_2(\mathrm{g})$ is zero, whereas for the atomic form $\mathrm{O}(\mathrm{g})$, it is non-zero. Regarding the definition of a standard state, while often associated with standard temperature and pressure conventions, it is important to understand that the term does not inherently mandate a specific temperature like $0^{\circ}\mathrm{C}$. By evaluating the options, we identify that the definitions provided in other choices regarding the stability of $\mathrm{O}(\mathrm{g})$ or the assignment of standard temperature conditions do not align with established thermodynamic principles, leading us to select the statement regarding the standard state definition as the correct choice in this context.

Dissolving an ionic solid like KCl in water is a two-step energetic process governed by Hess's Law. First, the crystal lattice must be broken to separate the ions into their gaseous state, which is an endothermic process requiring energy equal to the negative of the lattice energy; since the lattice energy is given as $-z$, this step contributes $+z$ to the total enthalpy change. Second, these gaseous ions are hydrated by the solvent, a process that releases energy equal to the sum of the individual ion hydration energies, expressed as $(-x) + (-y)$ or $-(x+y)$. Combining these two energetic contributions, the final heat of solution is:
$\Delta H_{\text{solution}} = z-(x+y)$

In a reversible thermodynamic process, the magnitude of the work done is represented by the area under the curve on a pressure-volume diagram. For this specific transformation, the process occurs at a constant pressure of 3 atm, and the volume expands from 1000 mL to 2000 mL, which is calculated using the formula $|w| = P \Delta V$ The change in volume is 1000 mL, which equals 1 L, and given the constant pressure of 3 atm, the work magnitude is 3 L atm. Converting this value into SI units by multiplying 3 by 101.3 gives 303.9 J. Rounding this result to the nearest integer yields 304 J, and since the gas expands, the work is done by the system, represented as -304 J.

The temperature rise in a neutralization reaction is proportional to the heat released, which is maximized when a strong acid reacts with a strong base because no energy is consumed for the dissociation of the acid. Mixing 30 mL of 1 M HCl and 30 mL of 1 M NaOH generates 30 mmol of neutralized product within a 60 mL total volume, yielding the highest heat density for the reaction $HCl + NaOH \rightarrow NaCl + H_2O$. In contrast, reactions involving acetic acid require additional energy for dissociation, which reduces the net enthalpy released and leads to a lower temperature increase. Furthermore, mixing 50 mL of 1 M HCl and 20 mL of 1 M NaOH results in only 20 mmol of neutralized product distributed over a larger 70 mL total volume, which further dilutes the heat released and results in a smaller rise in temperature.