Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

The defining characteristic of an adiabatic process is the absence of heat exchange with the surroundings, which implies $q = 0$. By applying the First Law of Thermodynamics, written as $\Delta U = q + W$, the change in internal energy simplifies to being equal to the work done, or $\Delta U = W$.

Because the ideal gas undergoes expansion, its internal energy decreases, which is calculated using the relationship $\Delta U = n C_v \Delta T$. With 1 mole of gas and a heat capacity of 20 J mol^-1 K^-1, the energy lost is 3000 J, which is represented as a negative value for work in this thermodynamic context. Converting the initial temperature of 27 °C to 300 K, the relationship is expressed by the equation $1 \times 20 \times (T_2 - 300) = -3000$. Solving for the final temperature by dividing -3000 by 20 results in -150, which leads to $T_2 = 150 \mathrm{K}$ after adding 300 to the initial temperature.

The enthalpy change for a chemical reaction is determined by subtracting the total enthalpy of formation of the reactants from the total enthalpy of formation of the products. For the reaction $\mathrm{CCl}_4(\mathrm{g}) + 2\mathrm{H}_2\mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g}) + 4\mathrm{HCl}(\mathrm{g})$, we first calculate the total enthalpy of the products by summing the enthalpy of formation of one mole of $\mathrm{CO}_2(\mathrm{g})$ and four moles of $\mathrm{HCl}(\mathrm{g})$, which equals -394 + 4 × (-92) = -762 kJ mol$^{-1}$. Next, we calculate the total enthalpy of the reactants by summing the enthalpy of formation of one mole of $\mathrm{CCl}_4(\mathrm{g})$ and two moles of $\mathrm{H}_2\mathrm{O}(\mathrm{g})$, which equals -105 + 2 × (-242) = -589 kJ mol$^{-1}$. Subtracting the reactant enthalpy sum from the product enthalpy sum yields -762 - (-589) = -173 kJ mol$^{-1}$. As the problem requests the magnitude of this enthalpy change, the final result is 173 kJ mol$^{-1}$.

Methane combustion is an exothermic process that releases significant thermal energy as it proceeds. Within an adiabatic system, this generated heat is trapped because the boundary prevents any energy exchange with the surroundings, leading to a direct rise in temperature. Conversely, a diathermic system allows for the transfer of thermal energy, meaning the heat released by the reaction dissipates into the environment to maintain a constant temperature.

The total work done in a cyclic process corresponds to the algebraic sum of the work performed during each distinct phase, expressed as $W = W_{1 \to 2} + W_{2 \to 3} + W_{3 \to 1}$. For the isobaric transition from 1 to 2, the work is calculated as $W_{1 \to 2} = -P \Delta V$. Given the pressure and volume values from the graph, this becomes $W_{1 \to 2} = -1 \times (40 - 20) = -20$ bar L. During the isochoric process from 2 to 3, the volume remains constant, which results in zero work done, so $W_{2 \to 3} = 0$.

The final path from 3 to 1 is an isothermal process where the work is determined by $W_{3 \to 1} = -nRT \ln(V_{final} / V_{initial})$, which is equivalent to $-P_1 V_1 \ln(V_1 / V_3)$. Using the constant product $P_1 V_1 = 20$ bar L and the volume values $V_1 = 20$ and $V_3 = 40$, this calculation becomes $-20 \ln(20 / 40)$, which simplifies to $20 \ln(2)$. Using the provided conversions $\ln 2 = \ln 10 \times \log 2 = 2.3 \times 0.3 = 0.69$, we find $W_{3 \to 1} = 20 \times 0.69 = 13.8$ bar L. Combining these components, the total work is $W = -20 + 0 + 13.8 = -6.2$ bar L. To obtain the magnitude in Joules, given that $1$ bar L equals $100$ J, the final result is $|W| = 6.2 \times 100 = 620$ J.

When standard enthalpies of combustion are provided, the enthalpy of formation for a substance is determined through Hess's Law by finding the difference between the total heat of combustion of the reactants and the heat of combustion of the product. This approach effectively treats the formation reaction as a specific thermal pathway, allowing the standard heat of formation to be calculated directly from known combustion data.

The formation of ethanol follows the equation $2\mathrm{C}(\mathrm{s}) + 3\mathrm{H}_2(\mathrm{g}) + 0.5\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{C}_2\mathrm{H}_5\mathrm{OH}(\mathrm{l})$. Using the relationship $\Delta H_f = \sum \Delta H_{\text{comb}}(\text{reactants}) - \sum \Delta H_{\text{comb}}(\text{products})$, the calculation becomes $\Delta H_f = [2 \times (-393.5) + 3 \times (-241.8)] - (-1234.7)$. The combustion contributions of the reactants total $-787.0 - 725.4 = -1512.4\ \mathrm{kJ}\ \mathrm{mol}^{-1}$. Combining this result with the combustion enthalpy of ethanol yields $-1512.4 + 1234.7 = -277.7\ \mathrm{kJ}\ \mathrm{mol}^{-1}$, which rounds to 278.

Hess's Law dictates that the total enthalpy change of a chemical transformation is equal to the sum of the enthalpy changes for each individual step in the process, regardless of the path taken. To determine the enthalpy change for the conversion of gaseous chlorine into aqueous chloride ions, we must account for three specific energy contributions: bond dissociation, electron gain, and hydration. First, the dissociation of $\frac{1}{2} \mathrm{Cl}_2(\mathrm{g})$ into $\mathrm{Cl}(\mathrm{g})$ requires an energy input of $\frac{1}{2} \times 240 = 120$ kJ mol⁻¹. Next, the chlorine atom gains an electron to form a gaseous chloride ion, releasing -350 kJ mol⁻¹, and the subsequent hydration of this ion releases an additional -380 kJ mol⁻¹. Summing these energetic values provides the net enthalpy change for the entire process:

$120 + (-350) + (-380) = -610 \text{ kJ mol}^{-1}$

The resulting negative sign indicates energy is released, and the magnitude of this enthalpy change is 610 kJ mol⁻¹.

The law of conservation of energy implies that heat energy generated by metabolism which is not converted into physical work must be dissipated to maintain thermal equilibrium. Since the athlete gains 1800 kJ from glucose and utilizes 50% for activity, the remaining 900 kJ of excess heat must be released.

This heat is removed through the evaporation of water with an enthalpy of 45 kJ mol$^{-1}$, leading to a required molar amount calculated by $n = 900 / 45 = 20$ mol. Given that water has a molar mass of 18 g mol$^{-1}$, the final weight of water needed is determined by the equation $20 \times 18 = 360$ g.

Thermodynamic properties are categorized based on whether they depend on the system's size; extensive properties like Volume, Gibbs free energy change, and Mole grow with the amount of matter, while intensive properties remain constant regardless of the sample size. Applying this distinction to the provided list reveals that Molar heat capacity, Molarity, $E^{\theta}$ cell, and Molar mass are all intensive, resulting in a total count of 4.

Endothermic processes are characterized by the net absorption of energy from the environment, which is required when chemical bonds are broken, phase transitions from liquid to gas occur, or solid lattice structures are dismantled during dissolution. Dissociation of gaseous iodine ($I_2(g) \rightarrow 2I(g)$) and hydrogen chloride ($HCl(g) \rightarrow H(g) + Cl(g)$) involves breaking stable covalent bonds, which always requires energy input, thus classifying both as endothermic. Similarly, the conversion of liquid water to steam ($H_2O(l) \rightarrow H_2O(g)$) consumes heat to overcome intermolecular attractions, while the dissolution of ammonium chloride in water is an endothermic process because disrupting its solid lattice demands more energy than is released upon hydration. Because the combustion of carbon ($C(s) + O_2(g) \rightarrow CO_2(g)$) releases energy and is thus exothermic, there are four endothermic processes total.

A bomb calorimeter measures the heat released at constant volume, which is equivalent to the internal energy change, $\Delta U$, of the reaction. The total heat absorbed by the calorimeter is calculated from the heat capacity and the temperature rise, giving $20 \text{ kJ K}^{-1} \times 0.5 \text{ K} = 10 \text{ kJ}$. Given the molar mass of ethane is $30 \text{ g mol}^{-1}$, the $0.3 \text{ g}$ sample corresponds to $0.01 \text{ moles}$. Since this heat was evolved from $0.01 \text{ moles}$ of ethane, the molar internal energy change is $\Delta U = -10 \text{ kJ} / 0.01 \text{ mol} = -1000 \text{ kJ mol}^{-1}$, with the negative sign indicating an exothermic reaction.

To determine the heat of combustion at constant pressure, which corresponds to the enthalpy change $\Delta H$, we use the relation $\Delta H = \Delta U + \Delta n_g RT$. For the combustion reaction $\text{C}_2\text{H}_6(g) + \frac{7}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$, the change in gaseous moles is calculated by subtracting the moles of gaseous reactants from the moles of gaseous products: $\Delta n_g = 2 - (1 + 3.5) = -2.5$. Using $R = 8.3 \text{ J K}^{-1} \text{mol}^{-1} = 0.0083 \text{ kJ K}^{-1} \text{mol}^{-1}$ and $T = 300 \text{ K}$, the work term $\Delta n_g RT$ becomes $-2.5 \times 0.0083 \times 300 = -6.225 \text{ kJ mol}^{-1}$. Therefore, the enthalpy of combustion is $\Delta H = -1000 \text{ kJ mol}^{-1} - 6.225 \text{ kJ mol}^{-1} = -1006.225 \text{ kJ mol}^{-1}$, making the magnitude of the heat evolved $1006 \text{ kJ mol}^{-1}$ to the nearest integer.

Thermodynamic state functions describe a system through specific identities, such as the Gibbs free energy relation, which states that $\Delta G = \Delta H - T \Delta S$ at constant temperature. Similarly, entropy change for a reversible process is fundamentally related to heat and absolute temperature, expressed as $\Delta S = \frac{q_{\text{rev}}}{T}$ under isothermal conditions.

The first law of thermodynamics dictates that the change in internal energy for a system doing expansion work is $\Delta U = q - P \Delta V$, meaning the positive sign in the first relation is incorrect. Finally, the enthalpy change for an ideal gas is derived from the ideal gas law to be $\Delta H = \Delta U + \Delta nRT$, which contradicts the negative sign presented in the final relation.

Because the container has adiabatic walls, no heat is exchanged with the surroundings, meaning the entire energy output from the electric heater is absorbed by the gas within. The total energy delivered is calculated by the formula $Q = P \cdot t$, and because this heating occurs at a constant volume, the relationship between this energy, the heat capacity at constant volume, and the change in temperature is expressed as $Q = C_v \Delta T$. Equating these two expressions allows us to derive the formula $C_v = \frac{P \cdot t}{\Delta T}$. Substituting the given values of 60, 100, and 5 into this equation results in a heat capacity of 1200 J K^-1.

The enthalpy change for a reaction is determined by calculating the difference between the total bond enthalpy of the reactants and the products using the relation $\Delta H = \sum BE_{\text{reactants}} - \sum BE_{\text{products}}$. For the reaction $A_2 + B_2 \rightarrow 2 AB$, the total enthalpy change is -400 kJ mol⁻¹ because the reaction produces two moles of $AB$ with an enthalpy of formation of -200 kJ mol⁻¹. Defining the bond enthalpies as $A_2 = x$, $B_2 = 0.5x$, and $AB = x$ based on the given ratio allows us to form the equation:
$\Delta H = (x + 0.5x) - 2x = -400$
Solving for $x$ leads to -0.5x = -400, which results in a bond enthalpy of 800 kJ mol⁻¹ for $A_2$.

Because gas volumes are directly proportional to their mole counts under identical temperature and pressure conditions, the combustion of the gaseous mixture can be analyzed using stoichiometric ratios. Methane burns according to the equation $\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}$, producing one volume of carbon dioxide for every volume of methane, while ethene combustion follows $\mathrm{C}_2\mathrm{H}_4 + 3\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}$, generating two volumes of carbon dioxide per volume of ethene. By establishing that the mixture consists of 16.8 L total volume, the relationship between these reactions indicates that 11.2 L of ethene and 5.6 L of methane are required to yield the specified 28.0 L of carbon dioxide.

To determine the total heat released, these gas volumes must be converted into molar quantities using the ideal gas law, $n = \frac{PV}{RT}$. Using a pressure of 1 atm, a temperature of 298 K, and a gas constant of 0.082 L atm K$^{-1}$ mol$^{-1}$, the volumes equate to 0.458 moles of ethene and 0.229 moles of methane. Applying the enthalpies of combustion by multiplying 0.458 moles by 1400 kJ mol$^{-1}$ and 0.229 moles by 900 kJ mol$^{-1}$ results in energy contributions of 641.2 kJ and 206.1 kJ, respectively. Summing these values gives a total heat evolution of 847.3 kJ.

The spontaneity of a reaction at constant temperature and pressure is dictated by the change in Gibbs free energy, which is defined by the following expression:
$\Delta G = \Delta H - T \Delta S$
A negative $\Delta G$ indicates a spontaneous process, whereas a positive $\Delta G$ indicates a non-spontaneous one. To evaluate the processes at a temperature $T = 300 \text{ K}$, enthalpy values are converted to $\text{J mol}^{-1}$ for unit consistency. Applying this to each, process A gives $\Delta G = -25000 - (300 \times -80) = -1000 \text{ J mol}^{-1}$, and process B gives $\Delta G = -22000 - (300 \times 40) = -34000 \text{ J mol}^{-1}$, both of which are negative and spontaneous. Conversely, process C yields $\Delta G = 25000 - (300 \times -50) = 40000 \text{ J mol}^{-1}$, and process D yields $\Delta G = 22000 - (300 \times 20) = 16000 \text{ J mol}^{-1}$; because both outcomes are positive, they are non-spontaneous, meaning there are two non-spontaneous processes in total.

In chemical thermodynamics, the state of a reaction relative to equilibrium is determined by the slope of the Gibbs free energy versus extent of reaction graph, where a negative slope ($dG < 0$) indicates spontaneity, a zero slope ($dG = 0$) signifies equilibrium, and a positive slope ($dG > 0$) indicates a non-spontaneous condition. Given this, point (a) occupies the spontaneous descending region, point (b) rests at the minimum energy equilibrium point, and point (c) sits on the non-spontaneous ascending portion of the curve. It follows that the statements describing the reaction as at equilibrium at point (b) with point (c) being non-spontaneous, as well as the reaction being spontaneous at (a) and non-spontaneous at (c), are valid, totaling two true statements.

Chemical equilibrium is reached when the change in Gibbs free energy is zero, which is expressed by the equation $\Delta G = \Delta H - T\Delta S = 0$. At this condition, the term $T\Delta S$ is equivalent to the enthalpy change of the reaction, $\Delta H$. The enthalpy change is calculated from the internal energy change measured in the bomb calorimeter, $\Delta U$, using the relationship $\Delta H = \Delta U + \Delta n_g RT$. For this combustion reaction, the change in the number of gaseous moles $\Delta n_g$ is the difference between the moles of gaseous products and gaseous reactants, calculated as $2 - 4 = -2$. Substituting the known values where $\Delta U = -1406$ kJ mol$^{-1}$, temperature $T = 300$ K, and the gas constant $R = 8.3$ J K$^{-1}$ mol$^{-1}$ (as $8.3 \times 10^{-3}$ kJ K$^{-1}$ mol$^{-1}$), the enthalpy change is $\Delta H = -1406 + (-2 \times 8.3 \times 300 \times 10^{-3})$. Performing this arithmetic gives $\Delta H = -1406 - 4.98 = -1410.98$ kJ mol$^{-1}$, which means the minimum value of $T\Delta S$ required for equilibrium is 1411 kJ mol$^{-1}$.

Hess's Law dictates that the net enthalpy change of a reaction is constant regardless of the specific pathway taken, allowing for the algebraic manipulation of thermochemical equations. To find the enthalpy of the reaction $\mathrm{C}(\text{graphite}) + \frac{1}{2} \mathrm{O}_2(g) \rightarrow \mathrm{CO}(g)$, we can construct the target equation by mathematically combining the provided reactions.

The second given reaction, $\mathrm{C}(\text{graphite}) + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g)$ with $\Delta H_2^{\circ} = -y\,\mathrm{kJ}\,\mathrm{mol}^{-1}$, provides the required carbon atom as a reactant. To produce carbon monoxide, we must reverse the first reaction, $2\mathrm{CO}(g) + \mathrm{O}_2(g) \rightarrow 2\mathrm{CO}_2(g)$, which flips the sign of its enthalpy change to $+x\,\mathrm{kJ}\,\mathrm{mol}^{-1}$. Since the target requires only one mole of carbon monoxide, dividing this reversed equation by two results in $\mathrm{CO}_2(g) \rightarrow \mathrm{CO}(g) + \frac{1}{2} \mathrm{O}_2(g)$ with an associated enthalpy change of $\frac{x}{2}\,\mathrm{kJ}\,\mathrm{mol}^{-1}$. Summing these two adjusted equations leads to the cancellation of common species on the reactants and products sides, yielding the target equation. The final enthalpy change is the sum of these individual changes, $-y + \frac{x}{2}$, which simplifies to $\frac{x - 2y}{2}$.

The behavior of black body radiation is fundamentally governed by Wien's Displacement Law, which dictates that the product of the peak wavelength and the temperature of the black body is a constant. Consequently, as the temperature increases, the peak of the emission spectrum shifts toward shorter wavelengths, which corresponds to higher frequencies. This intensity distribution cannot be described by classical models assuming particles undergo simple harmonic motion, as such models fail to account for the observed energy drop-off at higher frequencies; instead, the spectrum is correctly explained by Planck's hypothesis of energy quantization. Given the temperature relationship for the graph is $T_1 > T_2 > T_3 > T_4$, only the statements describing the shift of the spectral peak and the role of energy quantization are accurate.

In an isothermal expansion of an ideal gas, the entropy change is determined by the heat exchange with the environment, which is directly related to the expansion volume. For one mole of gas, the entropy change is calculated using the product of the number of moles, the gas constant, and the natural logarithm of the volume ratio. Substituting the specific values into the standard expression provides the magnitude of this change:

$\Delta S = n R \ln \left( \frac{V_2}{V_1} \right)$

With one mole of gas, a gas constant of 8.314 J K⁻¹ mol⁻¹, and a volume ratio of 1.5, the calculation proceeds as:

$\Delta S = 1 \times 8.314 \times \ln(1.5) \approx 3.37$

Because the process involves thermally conducting walls, the entropy change for the surroundings reflects this value, which rounds to 3 J K⁻¹ as the nearest integer.