Chemical Thermodynamics – NEET UG ChemistryPractice Questions & PYQs

When monatomic gases are mixed in equal volumes under identical conditions, the resulting mixture retains the fundamental properties of a monatomic gas because the number of translational degrees of freedom remains constant. For any monatomic gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$ and the molar heat capacity at constant pressure is $C_p = \frac{5}{2}R$. The specific heat ratio is determined by dividing these two quantities:

$\frac{C_p}{C_v} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3} = 1.67$

When liquid $\mathrm{H}_2\mathrm{O}$ vaporizes, the heat absorbed represents the enthalpy of vaporization ($\Delta H$), which accounts for both the increase in the internal energy ($\Delta U$) of the substance and the work performed against atmospheric pressure during expansion. This relationship is defined by the equation $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in gaseous moles, $R$ is the gas constant, and $T$ is the absolute temperature in Kelvin. For the process $\mathrm{H}_2\mathrm{O(l)} \rightleftharpoons \mathrm{H}_2\mathrm{O(g)}$, the value of $\Delta n_g$ is calculated as the change from zero moles of gas to one mole of gas, giving $\Delta n_g = 1$.

To calculate the change in internal energy, we rearrange the equation as $\Delta U = \Delta H - \Delta n_g RT$. Converting the enthalpy of vaporization to 40660 J/mol to ensure consistent units and using 373 K for the temperature, we substitute these values into the expression: $\Delta U = 40660 - (1 \times 8.314 \times 373)$. Evaluating this equation gives 37558.878 J/mol, which simplifies to 37.56 kJ/mol.

The relationship between Gibbs energy and temperature is defined by the equation $\Delta G = \Delta H - T\Delta S$, where an increase in temperature results in a sharp decrease in $\Delta G$ specifically when the reaction entropy change $\Delta S$ is positive. In the reaction $\mathrm{C}_{(\text{graphite})}+\frac{1}{2}\mathrm{O}_{2(\text{g})}\rightarrow\mathrm{CO}_{(\text{g})}$, the system proceeds from 1/2 mole of gaseous reactants to 1 mole of gaseous product, which leads to a net increase in disorder and a positive entropy change. Since $\Delta S$ is positive, raising the temperature makes the term $-T\Delta S$ significantly more negative, thereby causing $\Delta G$ to decrease.

In free expansion, an ideal gas expands against zero external pressure, meaning no work is performed by the system such that $w = 0$. Since the process is adiabatic, there is no thermal exchange with the surroundings, which implies $q = 0$. According to the first law of thermodynamics, $\Delta U = q + w$, the total change in internal energy is zero, and because the internal energy of an ideal gas depends exclusively on temperature, the process must also satisfy $\Delta T = 0$.

Bond dissociation energy represents the energy required to break one mole of a specific chemical bond. The reaction $4\mathrm{H}_{g} \rightarrow 2\mathrm{H}_{2,g}$ releases 869.6 kJ while forming two moles of H-H bonds, which means the formation of a single mole of H-H bonds releases half that amount, calculated as $\frac{869.6}{2} = 434.8$ kJ. Because breaking a bond is an endothermic process that absorbs energy from the surroundings, the enthalpy change for the dissociation is positive, resulting in a bond dissociation energy of +434.8 kJ.

Phase transitions occurring at constant temperature and pressure signify a state of equilibrium, where the change in Gibbs free energy is zero. Under these conditions, the fundamental thermodynamic equation $\Delta G = \Delta H - T\Delta S$ rearranges to define entropy change as $\Delta S = \frac{\Delta H}{T}$. Converting the given parameters to standard units, the temperature is 300 K and the enthalpy change is 30,000 J mol^-1. Dividing the enthalpy by the temperature results in an entropy change of 100 J mol^-1 K^-1.

Hess's Law establishes that the total enthalpy change of a reaction is independent of the pathway taken, meaning we can sum intermediate steps to find the energy change for a target process. To obtain the net reaction $B + D \rightarrow E + 2C$, multiply the reaction $\frac{1}{2}A \rightarrow B$ by two to get $A \rightarrow 2B$ with $\Delta H = 300 \text{ kJ/mol}$, keep $3B \rightarrow 2C + D$ as $\Delta H = -125 \text{ kJ/mol}$, and reverse $E + A \rightarrow 2D$ to $2D \rightarrow A + E$ with $\Delta H = -350 \text{ kJ/mol}$. Adding these balanced equations together cancels out the species $A$ and arranges the remaining components into the target reaction. Calculating the sum of these enthalpy changes gives the final result of $\Delta H = 300 - 125 - 350 = -175 \text{ kJ/mol}$.

In an endothermic reaction, the energy level of the products is inherently higher than the energy level of the reactants. Because the reaction path must pass through a transition state that resides at an energy level higher than both the reactants and the products, the activation energy, defined as the energy required to reach this transition state from the reactants, must be greater than the enthalpy of reaction, which represents only the energy difference between the products and the reactants. Consequently, the minimum activation energy is always more than the enthalpy of reaction to ensure the system successfully overcomes the energy barrier required to form the final products.

Hess's Law states that the total enthalpy change for a chemical reaction is independent of the path taken, allowing us to derive the enthalpy of a target reaction by algebraically combining known thermochemical equations. The target reaction $\mathrm{Fe}_2\mathrm{O}_3(\mathrm{s})+\mathrm{CO}(\mathrm{g})\rightarrow\mathrm{2FeO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g})$ is constructed by using the reaction $\mathrm{Fe}_2\mathrm{O}_3(\mathrm{s})+\mathrm{3CO}(\mathrm{g})\rightarrow\mathrm{2Fe}(\mathrm{s})+\mathrm{3CO}_2(\mathrm{g})$, which has an enthalpy change of -26.8 kJ, and the reaction $\mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g})\rightarrow\mathrm{Fe}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g})$, which has an enthalpy change of -16.5 kJ. Subtracting twice the second reaction from the first aligns the reactants and products, meaning the overall enthalpy change is calculated as -26.8 - 2(-16.5). Simplifying this expression gives -26.8 + 33.0, which equals +6.2 kJ.

Thermodynamic work is determined by the resistance encountered during a change in volume, described by the relation $W = -P_{ext} \Delta V$. Since the gas expands into a vacuum, there is no opposing external force, meaning $P_{ext} = 0$. Consequently, regardless of the three moles of ideal gas involved, the lack of external pressure results in zero work done.

Chemical equilibrium is defined by the state where the Gibbs free energy change of the reaction is zero, governed by the equation
$\Delta G = \Delta H - T\Delta S = 0$
The equilibrium temperature is determined by rearranging this relationship to $T = \frac{\Delta H}{\Delta S}$. For the reaction $\frac{1}{2}\mathrm{X}_2+\frac{3}{2}\mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_3$, the standard entropy change is calculated by subtracting the total entropy of the reactants from the product, while accounting for the stoichiometric coefficients. This calculation results in
$\Delta S^\circ = 50 - (\frac{1}{2} \times 60 + \frac{3}{2} \times 40) = -40 \text{ J K}^{-1} \text{ mol}^{-1}$
Converting the enthalpy change of $-30 \text{ kJ}$ into $-30,000 \text{ J}$ and substituting these values into the temperature equation yields
$T = \frac{-30,000}{-40} = 750 \text{ K}$

Phase transitions occurring at a constant pressure, such as the vaporization of water, reach equilibrium when the Gibbs free energy change is zero. The relationship governing this state is expressed by the thermodynamic equation $\Delta G = \Delta H - T\Delta S$, where $\Delta G$ is the Gibbs free energy change, $\Delta H$ is the enthalpy change, $T$ is the temperature, and $\Delta S$ is the entropy change. Because the system is at equilibrium when $\Delta G$ equals zero, the expression rearranges to identify the temperature as follows:

$T = \frac{\Delta H}{\Delta S}$

To maintain consistency in units, the enthalpy change must be converted from 40.63 kJ mol⁻¹ to 40630 J mol⁻¹ to align with the entropy value of 108.8 J K⁻¹ mol⁻¹. Substituting these quantities into the equation gives $T = \frac{40630}{108.8}$, which results in a temperature of 373.4 K.

The spontaneity of a reaction is fundamentally governed by the relationship between the equilibrium constant and the reaction quotient, where $K_p > Q$ indicates that the forward reaction rate exceeds the backward rate, resulting in a spontaneous process. For the thermodynamic condition $\Delta G^\circ < RT \ln Q$, the analysis dictates that the reaction is non-spontaneous. Equilibrium is reached when $K_p = Q$, signifying that the rates of the forward and backward reactions are perfectly balanced. Finally, the inequality $T > \frac{\Delta H}{\Delta S}$ rearranges to $T \Delta S > \Delta H$, which implies spontaneity only when the enthalpy change is positive, characterizing the reaction as spontaneous and endothermic.

The total enthalpy change of a chemical reaction is defined by the difference between the energy invested to break the bonds of the reactants and the energy released upon the formation of the product bonds. This principle reflects the net energy balance where an exothermic process occurs if the energy released during product formation exceeds the energy required for reactant bond cleavage. For the hydrogenation of ethene, the reactant side consists of one carbon-carbon double bond, four carbon-hydrogen bonds, and one hydrogen-hydrogen single bond, while the ethane product comprises one carbon-carbon single bond and six carbon-hydrogen bonds.

Summing the bond energies for the reactants, $606.10 + 431.37 + 4 \times 410.50 = 2679.47$, provides the total energy input required for dissociation. For the product, $336.9 + 6 \times 410.50 = 2799.9$, represents the total energy liberated. Subtracting the product bond energy from the reactant bond energy, $\Delta H = 2679.47 - 2799.9$, results in an enthalpy change of $-120.43$ kJ mol^-1, which aligns with the provided experimental value of $-120.0$ kJ mol^-1.

A chemical reaction proceeds spontaneously when the change in Gibbs free energy, $\Delta G$, is negative, which is governed by the relationship $\Delta G = \Delta H - T \Delta S$. Using the provided values of enthalpy change, $\Delta H = 170 \text{ kJ} = 170,000 \text{ J}$, and entropy change, $\Delta S = 170 \text{ J K}^{-1}$, the condition for spontaneity requires $\Delta H - T \Delta S < 0$. Rearranging this inequality to isolate temperature, $T \Delta S > \Delta H$, leads to the calculation $T(170) > 170,000$. This implies that the temperature must be greater than $1000 \text{ K}$ for the reaction to occur spontaneously, making $1110 \text{ K}$ the temperature at which the reaction is spontaneous.

Thermodynamic properties are categorized as either state functions, which rely solely on the system's current condition, or path functions, which depend on the specific sequence of steps taken to reach that condition. Since heat $q$ and work $w$ describe energy transfer through a process, they are classified as path functions and therefore fail to be state functions. In contrast, quantities such as the sum $q + w$, representing the change in internal energy, and the term $H - TS$, representing Gibbs free energy, rely exclusively on the system's endpoints, making them state functions.

Decomposition reactions such as the breakdown of $\mathrm{PCl}_{5(\mathrm{g})}$ require an absorption of thermal energy from the surroundings to break chemical bonds, which classifies the process as endothermic and ensures that $\Delta H > 0$. Because the reaction generates two moles of gas from a single mole of gaseous reactant, the inherent spatial disorder of the system increases, leading to a positive entropy change, $\Delta S > 0$.

The enthalpy change of a reaction is calculated using bond dissociation enthalpies by finding the difference between the total energy consumed to break bonds in the reactants and the total energy released during bond formation in the products. For the balanced chemical equation representing the formation of two moles of hydrogen chloride, $\mathrm{H}_2 + \mathrm{Cl}_2 \rightarrow 2\mathrm{HCl}$, we determine the total energy required to break the bonds in the reactants and the total energy released when forming the bonds in the products.

Summing the bond dissociation enthalpies for the reactant molecules results in $434 + 242 = 676$ kJ, while the energy liberated when forming two moles of hydrogen chloride bonds is $2 \times 431 = 862$ kJ. Subtracting the energy released from the energy required, we find the total enthalpy change for this specific reaction is $676 - 862 = -186$ kJ. Because the enthalpy of formation refers to the heat change for the production of exactly one mole of a substance from its constituent elements, we calculate the final value as $-186 / 2 = -93$ kJ mol$^{-1}$.

The standard enthalpy of formation is defined as the heat change accompanying the formation of exactly one mole of a substance from its constituent elements in their standard physical states. Evaluating the provided reactions, equation (ii) represents the synthesis of one mole of $\mathrm{H}_2\mathrm{O}_{(l)}$ from its pure elemental components, hydrogen gas and oxygen gas, which perfectly fulfills this specific definition. Since this equation describes the direct formation of the compound from its basic elements, the associated enthalpy change of $-\mathrm{X}_2\mathrm{kJmol}^{-1}$ corresponds directly to the enthalpy of formation for liquid water.

Enthalpy of formation for a substance is determined by the energy required to break the bonds of the reactants and the energy released upon the formation of new bonds in the products. For the formation of one mole of $HCl$ gas, the reaction is $\frac{1}{2} \text{H}_2 + \frac{1}{2} \text{Cl}_2 \rightarrow \text{HCl}$, and the enthalpy change is expressed by the following equation:

$\Delta H_f = \left(\frac{1}{2} \text{B.E.}_{\text{H-H}} + \frac{1}{2} \text{B.E.}_{\text{Cl-Cl}}\right) - \text{B.E.}_{\text{HCl}}$

Substituting the provided values of 430 kJ/mol for the $\text{H-H}$ bond energy, 240 kJ/mol for the $\text{Cl-Cl}$ bond energy, and -90 kJ/mol for the enthalpy of formation, the calculation is structured as:

$-90 = \left(\frac{1}{2} \times 430 + \frac{1}{2} \times 240\right) - \text{B.E.}_{\text{HCl}}$

Evaluating the reactant side of the equation results in 335 kJ/mol. This simplifies the expression to $-90 = 335 - \text{B.E.}_{\text{HCl}}$. By isolating the bond enthalpy of $HCl$, the calculation confirms a value of 425 kJ/mol.