Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

The fundamental principle of bomb calorimetry is that since combustion occurs at a constant volume, all heat released by the reaction is absorbed by the calorimeter as described by the equation $q = C \Delta T$. Using the given heat capacity of 5 kJ K⁻¹ and a temperature increase of 5 K, the total energy released by the combustion process is 25 kJ. Because the molar mass of n-octane is 114 g mol⁻¹, the 1.14 g sample corresponds to 0.01 mol of the substance. Dividing the total energy released by the number of moles yields a molar heat of combustion of 2500 kJ mol⁻¹.

Internal energy is a state function, which means its value depends strictly on the current state of the system and is independent of the path taken to reach that state. Because any cyclic process starts and finishes at the exact same point, the final internal energy is identical to the initial internal energy. Therefore, the net change in internal energy $\Delta U$ must be zero for any closed cycle, making it equal across all three given cases.

Hess's Law implies that the total enthalpy change for a chemical process remains consistent regardless of the pathway, permitting the manipulation of thermochemical equations through addition or subtraction. Finding the enthalpy of formation for sulfur dioxide requires representing the reaction $\mathrm{S}(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{SO}_2(\mathrm{g})$ using the provided thermochemical data.

Subtracting the equation $\mathrm{SO}_2(\mathrm{g}) + \frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{SO}_3(\mathrm{g})$ from $\mathrm{S}(\mathrm{g}) + \frac{3}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{SO}_3(\mathrm{g})$ eliminates $\mathrm{SO}_3(\mathrm{g})$ from the system and leaves the desired formation reaction. As the first reaction releases $2x$ and the second releases $y$, subtracting the second from the first effectively reverses the second step and modifies the associated enthalpy. Applying this arithmetic to the energy values, the final enthalpy change is calculated as $y - 2x$ kcal.

The average bond enthalpy of a covalent bond is defined as the energy required to cleave that bond in the gas phase, resulting in isolated gaseous atoms. To determine this value for the O-H bonds in water, we calculate the total enthalpy required to atomize one mole of water vapor into two gaseous hydrogen atoms and one gaseous oxygen atom, which is determined by the difference between the sum of the formation enthalpies of the gaseous atoms and the formation enthalpy of the reactant.

The total energy for atomization is expressed as $\Delta H_{atomization} = [2 \Delta H_f(H_{(g)}) + \Delta H_f(O_{(g)})] - \Delta H_f(H_2O_{(g)})$. Substituting the provided data gives $2 \times 220.0 + 250.0 - (-242.0) = 440.0 + 250.0 + 242.0 = 932.0$. Because this total energy corresponds to breaking two identical O-H bonds, the average bond enthalpy is found by the expression $\frac{932.0}{2} = 466$ kJ mol⁻¹.

The magnitude of work exchanged with the surroundings depends on the path taken, specifically whether the process proceeds reversibly or irreversibly, as reversible paths are inherently more efficient. Since the magnitude of reversible work is invariant between expansion and compression, denoted as $|W_{\text{reversible expansion}}| = |W_{\text{reversible compression}}|$, and given that $|W_{\text{irreversible compression}}|$ is greater than its reversible counterpart while $|W_{\text{irreversible expansion}}|$ is smaller than $|W_{\text{reversible expansion}}|$, the resulting order is $d > c = a > b$.

The enthalpy of formation of a compound is defined by the net energy change resulting from the breaking of reactant bonds and the formation of product bonds. Converting 2 moles of graphite into free gaseous carbon atoms requires $2 \times 710 = 1420$ kJ, and breaking the covalent bonds in 2 moles of hydrogen molecules demands an additional $2 \times 436 = 872$ kJ. Combining these requirements, the total energy input to reach the atomic state is $1420 + 872 = 2292$ kJ.

Conversely, the formation of the ethylene molecule releases energy as the atoms bond. Creating one carbon-carbon double bond releases 611 kJ, while establishing four carbon-hydrogen bonds releases $4 \times 414 = 1656$ kJ. The sum of these released energies is $611 + 1656 = 2267$ kJ. The difference between the total energy absorbed to break reactants and the total energy released to form products results in an enthalpy of formation of $2292 - 2267 = 25$ kJ mol⁻¹.

Argon is a monatomic ideal gas, and its behavior during heat transfer is governed by its molar heat capacities at constant volume and constant pressure, specifically $C_v = \frac{3}{2}R$ and $C_p = \frac{5}{2}R$. When heat is transferred at constant pressure, the energy added relates to the temperature change through the equation:
$Q = n C_p \Delta T$
Substituting the given values, $Q = 500$ J, $n = 0.5$ mol, and $R = 8.3$ J K$^{-1}$ mol$^{-1}$, the temperature change is calculated as $\Delta T = 48.2$ K, resulting in a final temperature of $346.2$ K when added to the initial $298$ K.

The change in internal energy, which depends only on the temperature change for an ideal gas, is determined by the equation:
$\Delta U = n C_v \Delta T$
Using the previously calculated temperature difference and the specific heat capacity $C_v = \frac{3}{2}R$, the internal energy change is $\Delta U = 300$ J. These results confirm the values that align with the conditions presented in option D.

The behavior of this system is governed by the principle that increased pressure shifts equilibrium toward the phase with a smaller volume, and water is unique because its liquid form is denser than its solid form. When the pressure is increased to 2 atm, the equilibrium $H_2O(s) \rightleftharpoons H_2O(l)$ shifts to favor the liquid state, which effectively lowers the freezing point of ice below the maintained temperature of 273.15 K. Because the ice is now effectively at a temperature above its pressure-adjusted melting point, the solid phase cannot persist and disappears completely into liquid water.

The spontaneity of a chemical reaction is governed by the Gibbs free energy change, expressed as $\Delta G = \Delta H - T\Delta S$, where a process is spontaneous only when $\Delta G < 0$. For a reaction with $\Delta H = +$ and $\Delta S = +$, $\Delta G$ becomes negative only when the temperature is high enough for the $T\Delta S$ term to exceed $\Delta H$, meaning the entry claiming spontaneity at low temperature is incorrect. Similarly, for the combination where $\Delta H = -$ and $\Delta S = -$, the reaction is actually spontaneous at low temperatures because the negative $\Delta H$ term dominates the positive $T\Delta S$ term, identifying the entry labeled non-spontaneous as incorrect. In contrast, the cases where $\Delta H = +$ with $\Delta S = -$ always yield a non-spontaneous result, and where $\Delta H = -$ with $\Delta S = +$ always yield a spontaneous result, making those descriptors accurate.

The spontaneity of a reaction is determined by the Gibbs free energy equation, defined by the relationship $\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$, which balances the energy change and entropy at a given temperature. Converting the enthalpy from 55.0 kJ/mol into 55000 J/mol ensures that all terms use consistent units. The reaction occurs at 25°C, which corresponds to 298 K on the absolute temperature scale. Multiplying this temperature by the entropy value of 175 J/K/mol results in a product of 52150 J/mol. Subtracting this entropy contribution from the enthalpy of 55000 J/mol yields a final free energy change of 2850 J/mol.

Determining the standard enthalpy of formation involves calculating the energy change required to synthesize one mole of a substance from its constituent elements in their standard states. According to Hess's Law, since enthalpy is a state function, we can algebraically combine the given combustion reactions to construct the target equation: $2\mathrm{C}(\text{graphite}) + 3\mathrm{H}_2(\text{g}) \rightarrow \mathrm{C}_2\mathrm{H}_6(\text{g})$.

To build this equation, we require two moles of the carbon combustion reaction and three moles of the hydrogen combustion reaction, while reversing the ethane combustion reaction to place ethane on the product side. The calculation follows the arithmetic path of doubling the enthalpy of carbon combustion, adding three times the enthalpy of hydrogen combustion, and finally subtracting the enthalpy of ethane combustion. Using the provided values, the enthalpy of formation is determined by the expression $2(-393.5) + 3(-286) - (-1550) = -95$. The magnitude of this value is 95 kJ mol^-1.

Entropy is a state function, which allows the total change for a complex process to be calculated by summing the entropy changes of each reversible constituent step. For heating a single phase, the entropy change is given by the integral of $\frac{C_p}{T}$ with respect to temperature, whereas for an isothermal phase transition, it is expressed as the heat of transition divided by the equilibrium temperature.

The transformation of ice at 268 K to vapour at 383 K requires five distinct stages. First, warming the ice from 268 to 273 results in $\int_{268}^{273} \frac{C_{p,m}}{T} dT$. Melting the ice at 273 yields $\frac{\Delta H_{m, \text{ fusion}}}{273}$. Heating the resulting water from 273 to 373 contributes $\int_{273}^{373} \frac{C_{p,m}}{T} dT$. Vaporizing the water at 373 adds $\frac{\Delta H_{m, \text{ vaporisation}}}{373}$. Lastly, warming the steam from 373 to 383 provides $\int_{373}^{383} \frac{C_{p,m}}{T} dT$. Summing these individual values accounts for the total entropy change of the entire process.

Free expansion occurs when gas moves into a vacuum, allowing the system to undergo the process without performing work or exchanging heat, which results in no change to the internal energy. Given that the temperature remains constant, shown by $dT = 0$, the gas must have expanded into a space containing no opposing pressure. Consequently, the second vessel must have been empty, meaning its pressure was zero before the stopcock was opened.

The standard enthalpy of combustion for one mole of benzene, governed by the reaction $\mathrm{C}_6\mathrm{H}_6(l) + \frac{15}{2}\mathrm{O}_2(g) \rightarrow 6\mathrm{CO}_2(g) + 3\mathrm{H}_2\mathrm{O}(l)$, can be determined using Hess's Law by calculating the difference between the enthalpies of formation of the products and the reactants. Following the stoichiometric coefficients, the enthalpy of reaction is calculated by summing six times the formation enthalpy of $\mathrm{CO}_2(g)$ and three times that of $\mathrm{H}_2\mathrm{O}(l)$, then subtracting the formation enthalpy of benzene. Plugging in the given values, the calculation becomes $-48.5 + 6 \times (-393.5) + 3 \times (-286)$, which equals $-3267.5 \mathrm{kJ/mol}$ for one mole of benzene. Since the standard enthalpy of combustion for two moles is required, doubling this value yields $-6535 \mathrm{kJ}$, making the value of $x$ equal to 6535.

The total energy required for vaporization is determined by multiplying the molar enthalpy of vaporization by the total number of moles of the substance. First, determine the molar mass of CCl4 by calculating the equation $12 + 4 \times 35.5 = 154 \text{ g mol}^{-1}$. Dividing the given mass of 284 g by this molar mass yields 1.844 mol. Finally, the total heat is calculated using the equation $1.844 \times 30.5 = 56.242 \text{ kJ}$, which rounds to 56 kJ of heat required for the process.

The equilibrium state of a reaction is fundamentally tied to the standard Gibbs free energy change, which serves as a measure of the driving force behind a process at constant temperature and pressure. We determine the standard Gibbs free energy change using the expression $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$. Converting the enthalpy change to 77200 J mol^-1, we subtract the product of the temperature 400 K and the entropy change 122 J K^-1, resulting in a standard Gibbs free energy change of 28400 J mol^-1.

With this value, we apply the relationship $\Delta G^{\circ} = -2.303 RT \log K$ to determine the equilibrium constant. Substituting the universal gas constant 8.314 J K^-1 mol^-1, the temperature 400 K, and the previously calculated free energy into the equation $28400 = -2.303 \times 8.314 \times 400 \times \log K$ allows us to isolate the equilibrium term. Solving this equation yields $\log K = -3.708$. In terms of the requested format, this value is equivalent to $\log K = -37.08 \times 10^{-1}$, which identifies the required integer value as 37.

The enthalpy of neutralization depends entirely on the reaction between $\mathrm{H}^{+}$ ions and $\mathrm{OH}^{-}$ ions to form water. Hydrochloric acid ($\mathrm{HCl}$) is a monoprotic acid that provides one $\mathrm{H}^{+}$ ion per molecule, whereas sulfuric acid ($\mathrm{H}_2\mathrm{SO}_4$) is a diprotic acid that provides two $\mathrm{H}^{+}$ ions per molecule. When taking equal volumes of 1 M acid solutions, the sulfuric acid solution contains twice the number of moles of $\mathrm{H}^{+}$ available for reaction compared to the hydrochloric acid solution. Because the heat released is directly proportional to the moles of $\mathrm{H}^{+}$ neutralized, the amount of heat $y$ liberated by the sulfuric acid is twice the amount of heat $x$ liberated by the hydrochloric acid, which is expressed as:
$y = 2x$
This relationship results in a ratio of 2.

For an ideal gas undergoing a reversible isothermal expansion, the mechanical work performed is determined by the ideal gas constant, the absolute temperature, and the logarithmic ratio of the final volume to the initial volume. Because the process occurs at a constant temperature of $18^{\circ}\text{C}$, we first convert this value to the Kelvin scale, resulting in an absolute temperature of $291.15\text{ K}$.

The work done by the system is calculated using the following equation:
$w = -nRT \ln\left( \frac{V_2}{V_1} \right)$
Substituting the known values of $n = 1\text{ mol}$, $R = 0.08206\text{ L atm mol}^{-1}\text{K}^{-1}$, $T = 291.15\text{ K}$, and the volume ratio of $100/10 = 10$, the calculation becomes $w = -1 \times 0.08206 \times 291.15 \times \ln(10)$. Evaluating this product yields approximately $-55.0128\text{ L atm}$, indicating that energy is transferred out of the system as work. Taking the magnitude of this result and rounding to the nearest integer leads to $55$.

The enthalpy change of a reaction is calculated by determining the difference between the total energy absorbed to break all bonds in the reactants and the total energy released upon the formation of all bonds in the products. In the hydrogenation of ethylene, the process involves breaking one $\mathrm{C=C}$ bond, four $\mathrm{C-H}$ bonds, and one $\mathrm{H-H}$ bond, while simultaneously forming one $\mathrm{C-C}$ bond and six $\mathrm{C-H}$ bonds.

The enthalpy of reaction is expressed as $\Delta H = [\text{BE}(\text{C=C}) + 4\;\text{BE}(\text{C-H}) + \text{BE}(\text{H-H})] - [\text{BE}(\text{C-C}) + 6\;\text{BE}(\text{C-H})]$. This equation simplifies to $\Delta H = \text{BE}(\text{C=C}) + \text{BE}(\text{H-H}) - \text{BE}(\text{C-C}) - 2\;\text{BE}(\text{C-H})$. Substituting the provided bond energy values gives $\Delta H = 615 + 435 - 347 - 2(414)$, which simplifies to $1050 - 1175$, yielding a result of -125 kJ.

When a substance reaches its boiling point at one atmosphere, it exists in a state of phase equilibrium where the liquid and vapor phases are energetically balanced. This state implies that the Gibbs free energy change is zero, defined by the relationship $\Delta G = \Delta H_{\text{vap}} - T \Delta S_{\text{vap}}$. Since equilibrium dictates that $\Delta G$ vanishes, the equation simplifies to $\Delta H_{\text{vap}} = T \Delta S_{\text{vap}}$.

To find the temperature, express the enthalpy of vaporization as 30,000 J/mol to match the units of the entropy of vaporization, which is 75 J mol^-1 K^-1. Dividing the enthalpy value by the entropy value provides the temperature in Kelvin:
$T = \frac{30,000}{75} = 400$
The temperature of vaporization is therefore 400 K.