Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

The maximum work obtained during an isothermal reversible expansion is determined by the relationship $W = -2.303 nRT \log_{10} \left( \frac{V_2}{V_1} \right)$ where the negative sign signifies work done by the system on its surroundings. Calculating the magnitude of this work requires taking the absolute value of this expression, effectively removing the negative sign.

Given 5 moles, a gas constant of 8.3 J K⁻¹ mol⁻¹, a temperature of 300 K, and a volume ratio of 20 litre divided by 10 litre, which simplifies to 2, the calculation involves multiplying these constants by the logarithm of 2. Using 0.3010 for the value of the logarithm, the expression for the magnitude becomes $2.303 \times 5 \times 8.3 \times 300 \times 0.3010$. Performing this multiplication yields 8630.0415, which rounds to a nearest integer of 8630.

The enthalpy change of a chemical reaction can be determined by the difference between the sum of the combustion enthalpies of the reactants and the sum of the combustion enthalpies of the products. This principle is governed by the following equation:

$\Delta H = \Sigma \Delta H_{\text{comb}}(\text{reactants}) - \Sigma \Delta H_{\text{comb}}(\text{products})$

For the reaction converting three moles of acetylene gas into one mole of liquid benzene, the reactants consist of three moles of acetylene, each with an enthalpy of combustion of -1300 kJ mol^-1, while the product is one mole of benzene with an enthalpy of combustion of -3268 kJ mol^-1. Substituting these values into the expression yields $\Delta H = 3 \times (-1300) - (-3268)$. Calculating this result, -3900 + 3268 leads to a final enthalpy change for the reaction of -632 kJ mol^-1.

The process is governed by the First Law of Thermodynamics, $\Delta U = q + w$, where $\Delta U$ represents the change in internal energy, $q$ is heat, and $w$ is work. For an ideal gas undergoing an isothermal process, $\Delta U = 0$, and because the gas expands into a vacuum from $4.0 \text{ L}$ to $2.0 \text{ L}$, the opposing external pressure $P_{\text{ext}}$ is zero. The work done is calculated as:
$w = -P_{\text{ext}} \Delta V = -(0) \Delta V = 0$
Substituting these values into the first law equation, $0 = q + 0$, confirms that the heat absorbed during this process is $0 \text{ L atm}$.

The thermodynamic feasibility of reducing a metal oxide is dictated by the change in standard Gibbs free energy, represented by the relationship $\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}$, where a more negative value indicates a more spontaneous process. Because entropy, which measures molecular disorder, is intrinsically higher for a substance in the liquid state compared to the solid state, forming a liquid metal product during the reduction process enhances the entropy of the system. This higher entropy in the liquid state increases the entropy change, $\Delta S^{\ominus}$, for the reaction, which in turn causes the $-T\Delta S^{\ominus}$ term to become more negative. Consequently, the standard Gibbs free energy change, $\Delta G^{\ominus}$, shifts toward a more negative value, rendering the reduction of the metal oxide significantly easier to achieve.

The conversion between enthalpy change and internal energy change for a reaction involving gaseous species is defined by the relationship $\Delta H = \Delta U + \Delta n_g RT$. In the combustion of magnesium, represented by the chemical equation $\text{Mg}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{MgO}(s)$, the change in gaseous moles $\Delta n_g$ is $0 - 0.5 = -0.5$ because only oxygen exists in the gas phase. Substituting the provided values leads to the expression $-601.70\ \text{kJ} = \Delta U + (-0.5) \times (8.3 \times 10^{-3}\ \text{kJ K}^{-1} \text{mol}^{-1}) \times 300\ \text{K}$. Calculating the work term yields $-1.245\ \text{kJ}$, which results in $\Delta U = -601.70 + 1.245 = -600.455\ \text{kJ}$. The magnitude of this value, when rounded to the nearest integer, is 600.

The principle of calorimetry holds that the heat released during an exothermic neutralization reaction is entirely absorbed by the resulting solution, causing its temperature to increase according to the equation $q = mc\Delta T$ Calculating the reactant amounts shows that the $\mathrm{HNO}_3$ solution contains 120 mmol of acid, while the $\mathrm{NaOH}$ solution contains 40 mmol of base. Since the acid and base react \in a 1:1 ratio, 40 mmol of $\mathrm{NaOH}$ is the limiting reagent, resulting in the formation of 40 mmol of water.

With an enthalpy of neutralization of 57 kJ/mol, the total energy released is 2280 J, found by multiplying 0.04 mol by 57000 J/mol. The total volume of the mixture is 1000 mL, which corresponds to a mass of 1000 g. Given the specific heat capacity of 4.2 J K$^{-1}$ g$^{-1}$, the temperature change is obtained from the expression $2280 = 1000 \times 4.2 \times \Delta T$ which leads to $\Delta T = 54 × 10^{-2}\ {}^\circ\mathrm{C}$.

The connection between the change in enthalpy and the change in internal energy for gas-phase reactions is expressed by the equation $\Delta H = \Delta U + \Delta n_g RT$ Determine the change \in gaseous moles $\Delta n_g$ by subtracting the \sum of the stoichiometric coefficients of gaseous reactants from those of the gaseous products, which results \in $\Delta n_g = 1$. Convert the reaction temperature of 27°C to 300 K and calculate the expansion work term $\Delta n_g RT$ by multiplying the change in gaseous moles, the gas constant 8.314 J K-1 mol-1, and the temperature, yielding 2494.2 J mol-1 or 2.4942 kJ mol-1. Finally, add this work value to the internal energy change of -59.6 kJ mol-1 to obtain an enthalpy change of -57.1058 kJ mol-1, which corresponds to an integer value of 57.

The thermodynamic relationship $\Delta H = \Delta U + \Delta n_g RT$ bridges the gap between the internal energy change measured in a constant volume bomb calorimeter and the enthalpy change of a reaction. Because the combustion reaction releases heat, the internal energy change is $\Delta U = -726\,\mathrm{kJ}\,\mathrm{mol}^{-1}$.

The difference in gaseous moles between products and reactants dictates the work term, $\Delta n_g RT$. For the combustion of methanol, one mole of gaseous $\mathrm{CO}_2$ is produced while $1.5$ moles of gaseous $\mathrm{O}_2$ are consumed, resulting in $\Delta n_g = 1 - 1.5 = -0.5\,\mathrm{mol}$. By converting the temperature of $27^\circ\mathrm{C}$ to $300\,\mathrm{K}$ and expressing the gas constant as $8.3 \times 10^{-3}\,\mathrm{kJ}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$, the work correction is calculated as $-0.5 \times 8.3 \times 10^{-3} \times 300$, which equals $-1.245\,\mathrm{kJ}\,\mathrm{mol}^{-1}$. Applying this correction to the internal energy yields $\Delta H = -726 - 1.245 = -727.245\,\mathrm{kJ}\,\mathrm{mol}^{-1}$, which corresponds to a nearest integer value of 727.

The heat absorbed by a calorimeter during combustion effectively equals the energy released by the gas, as the internal energy change is approximately equal to the enthalpy of combustion for this reaction. The calorimeter absorbs heat according to its capacity and the temperature rise, so the total energy released is $2.5 \times 0.45 = 1.125$ kJ, based on a temperature increase from 298.0 K to 298.45 K. With an enthalpy of combustion of 9 kJ mol⁻¹, the amount of gas burnt is $1.125 / 9 = 0.125$ mol. Applying the molar mass of 280 g mol⁻¹, the total mass of the gas is $0.125 \times 280 = 35$ g.

The change in internal energy for an ideal gas is determined by the equation $\Delta U = n C_v \Delta T$, involving the number of moles $n$, the molar heat capacity at constant volume $C_v$, and the temperature change $\Delta T$. Given the temperature increases from $300\ \mathrm{K}$ to $500\ \mathrm{K}$, the change $\Delta T$ is $200\ \mathrm{K}$. Applying the provided internal energy change of $5000\ \mathrm{J}$, we find $5000 = n C_v \times 200$, which results in the product $n C_v = 25\ \mathrm{J\ K^{-1}}$.

We relate the constant volume and constant pressure heat capacities using Mayer's relation, $C_p - C_v = R$. Substituting $C_v = C_p - R$ into the expression $n C_v = 25$, we isolate $n$ with the following algebraic steps:
$n = \frac{25}{C_p - R}$
Substituting the specified values $C_p = 20.785\ \mathrm{J\ K^{-1}\ mol^{-1}}$ and $R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$, the denominator becomes $20.785 - 8.314 = 12.471$. Dividing $25$ by this result yields $n = 2$.

The heat of neutralization for a strong acid with a strong base is consistently $-57.3 \mathrm{kJ\,mol^{-1}}$ because it represents the standard energy released when fully dissociated $\mathrm{H}^+$ and $\mathrm{OH}^-$ ions combine to form water. When a weak acid like $\mathrm{CH}_3\mathrm{COOH}$ reacts with a strong base, the observed enthalpy of $-55.3 \mathrm{kJ\,mol^{-1}}$ is less negative because some of the energy released is consumed to ionize the undissociated acid molecules. The enthalpy of ionization is calculated by determining the difference between the experimental enthalpy of the weak acid neutralization and the standard enthalpy of a strong acid neutralization. By subtracting the strong acid value from the weak acid value, we obtain $-55.3 - (-57.3) = 2 \mathrm{kJ\,mol^{-1}}$.

For an ideal gas undergoing an isothermal expansion, the temperature remains constant, which means the process follows Boyle's Law where the product of pressure and volume is constant. This relationship allows us to determine the work done during a reversible expansion using the expression $w = -2.303 nRT \log\left(\frac{V_2}{V_1}\right)$. Since the gas expands from state 1 to state 2 at a constant temperature, the ratio of the volumes can be derived from the ratio of the pressures as $\frac{V_2}{V_1} = \frac{P_1}{P_2} = \frac{6}{3} = 2$.

Substituting the given values of $n = 1.0\;\mathrm{mol}$, $R = 8.3\;\mathrm{J}\;\mathrm{K}^{-1}\;\mathrm{mol}^{-1}$, $T = 300\;\mathrm{K}$, and $\log(2) = 0.30$ into the equation, we calculate the work as $w = -2.303 \times 1.0 \times 8.3 \times 300 \times 0.30$. This calculation yields a result of $w = -1718.1\;\mathrm{J}$. Because the magnitude of the work is required, the final value is $1718\;\mathrm{J}$.

The enthalpy of formation of a compound represents the energy change that occurs when one mole of that substance is synthesized from its constituent elements in their standard physical states. Because the enthalpies of combustion are provided for all the species involved in the formation of ethane, we can determine the standard enthalpy of formation by applying Hess's Law, which dictates that the energy change of a reaction is equal to the sum of the combustion enthalpies of the reactants minus the combustion enthalpy of the product.

The balanced equation for the formation of ethane is $2\mathrm{C}(\text{graphite}) + 3\mathrm{H}_2(\text{g}) \rightarrow \mathrm{C}_2\mathrm{H}_6(\mathrm{g})$. Using the relationship that the heat of formation is the difference between the combustion enthalpies of the reactants and the product, the expression is \Delta_f H^\ominus = \[2 × \Delta_c H^\ominus(\mathrm{C}) + 3 × \Delta_c H^\ominus(\mathrm{H}_2)]$ - \Delta_c H^\ominus(\mathrm{C}_2\mathrm{H}_6)$. Substituting the numerical values from the data provided, we find the calculation$\Delta_f H^\ominus = [2(-394.0) + 3(-286.0)] - (-1560.0)$, which simplifies to$-788.0 - 858.0 + 1560.0 = -86.0,\mathrm{kJ,mol}^{-1}$.

The transition from liquid water to gaseous steam involves enthalpy that accounts for both the rise in internal energy and the expansion work performed against external pressure. This relationship is defined by the thermodynamic equation $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ denotes the change in moles of gas produced per mole of liquid vaporized. Since the process turns one mole of liquid water into one mole of water vapor, the value of $\Delta n_g$ is exactly $1$.

While the problem mentions a specific mass of water on the fish, the internal energy of vaporization is an intensive property and remains constant per mole regardless of the total amount present. To determine $\Delta U$, the expression is rearranged to $\Delta U = \Delta H - \Delta n_g RT$. Substituting the provided enthalpy of $41.1\,\mathrm{kJ/mol}$, a temperature of $373\,\mathrm{K}$, and a gas constant of $8.31 \times 10^{-3}\,\mathrm{kJ K^{-1} mol^{-1}}$, the work term evaluates to approximately $3.1\,\mathrm{kJ/mol}$. Subtracting this expansion work from the enthalpy of $41.1\,\mathrm{kJ/mol}$ yields an internal energy change of $38\,\mathrm{kJ/mol}$.

The Ellingham diagram depicts the change in $\Delta G^{\circ}$ with temperature based on the equation $\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}$, where the entropy change determines the slope of the line. In the formation of magnesium oxide, $2\mathrm{Mg}(\mathrm{s}) + \mathrm{O}_2(\mathrm{g}) \rightarrow 2\mathrm{MgO}(\mathrm{s})$, the consumption of oxygen gas leads to a negative entropy change, making $-T \Delta S^{\circ}$ positive and causing $\Delta G^{\circ}$ to increase as temperature rises. Consequently, the plot shows a positive slope, and any phase change occurring during the heating process results in an abrupt increase in this slope.

Enthalpy is defined as the sum of internal energy and the product of pressure and volume, given by the expression $H = U + PV$. For any process occurring at constant pressure, this definition necessitates that the change in enthalpy is represented by the equation $\Delta H = \Delta U + P \Delta V$. The relation in the first option incorrectly introduces a negative sign, failing to satisfy the fundamental thermodynamic definition of enthalpy.

The internal energy change of a gas is governed by the first law of thermodynamics, represented by the formula $\Delta U = q + w$. In this constant pressure process, the heat exchanged is calculated using $q = n C_P \Delta T$. First, the number of moles of nitrous oxide is determined by dividing the mass of 2.2 g by its molar mass of 44 g/mol, which yields 0.05 mol. With a temperature drop from 310 K to 270 K, the temperature change is -40 K, resulting in a heat exchange of $q = 0.05 \times 100 \times (-40) = -200\ \text{J}$.

The work done during compression is determined by $w = -P \Delta V$. Applying the pressure of 1 atm and the specified change in volume from 217.1 mL to 167.75 mL, the work done evaluates to $w = 5\ \text{J}$ after converting the units from liters and atmospheres to joules. Summing these values according to the first law, the total change in internal energy is $\Delta U = -200\ \text{J} + 5\ \text{J} = -195\ \text{J}$. Since the problem defines this change as $-x\ \text{J}$, the value of x is 195.

The combustion of coal in a bomb calorimeter releases heat that is fully absorbed by the apparatus, causing a measurable rise in temperature. Since the calorimeter operates at a constant volume, the measured heat release is directly proportional to the enthalpy change of the reaction. We determine this value by calculating the total heat energy released and normalizing it per mole of reactant.

The number of moles of carbon consumed is found by dividing the mass of the sample by its molar mass, yielding $n = 2.4 / 12 = 0.2$ mol. The total heat evolved during the reaction is calculated using the heat capacity of the bomb calorimeter and the change in temperature:
$q = 20.0 \times (300 - 298) = 40$
kJ.

Because combustion is an exothermic process, the enthalpy change per mole is obtained by dividing the total heat energy by the number of moles, resulting in $\Delta H = -40 / 0.2 = -200$ kJ mol$^{-1}$. By comparing this result to the given expression $-x$ kJ mol$^{-1}$, we find $x = 200$.

The equilibrium state of a chemical reaction is characterized by the condition where the change in Gibbs free energy is equal to zero. By utilizing the fundamental thermodynamic relationship $\Delta G = \Delta H - T \Delta S$, we can determine the temperature of equilibrium by setting $\Delta G$ to zero, which simplifies the equation to $T = \Delta H / \Delta S$. To ensure consistency, we convert the given standard enthalpy change of -165 kJ mol⁻¹ into -165000 J mol⁻¹ to match the units of the entropy change of -550 J K⁻¹. Substituting these values into the expression provides the following calculation:

$T = \frac{-165000}{-550}$

Performing this division results in a temperature of 300 K.

For a process to occur spontaneously under constant temperature and pressure conditions, the Gibbs free energy change must be negative, expressed as $\Delta G_{T,P} < 0$. When describing system constraints, a process occurring with no pressure change signifies an isobaric condition, while one with no temperature change is termed isothermal, making the combination of $\Delta P = 0$ and $\Delta T = 0$ represent an isothermal and isobaric process. The enthalpy change of a reaction is calculated by taking the sum of the bond energies of the reactant molecules and subtracting the sum of the bond energies of the product molecules, given by $\Delta H_{\text{reaction}} = \sum (\text{Bond energies of reactants}) - \sum (\text{Bond energies of products})$. Lastly, reactions that release energy, known as exothermic processes, are distinguished by a decrease in enthalpy, denoted as $\Delta H < 0$.