Chemical Thermodynamics – NEET UG ChemistryPractice Questions & PYQs

At chemical equilibrium, the Gibbs free energy change for a reaction is zero. This principle is governed by the relation $\Delta G = \Delta H - T\Delta S$, where $T$ represents the temperature at which the forward and reverse processes occur at equal rates. By setting $\Delta G$ to zero, the expression rearranges to show that the equilibrium temperature is determined by the ratio of the enthalpy change to the entropy change:
$T = \frac{\Delta H}{\Delta S}$

Because the enthalpy change is provided as 30 kJ mol^-1 and entropy as 105 J K^-1 mol^-1, we must first convert the enthalpy to 30000 J mol^-1 to ensure consistent units. Dividing 30000 J mol^-1 by 105 J K^-1 mol^-1 results in a temperature of 285.7 K.

The change in Gibbs energy, $\Delta G_{\text{system}}$, provides a fundamental thermodynamic criterion for predicting the spontaneity of a process occurring at constant temperature and pressure. When a system reaches a state of equilibrium, the net change in Gibbs energy becomes $\Delta G_{\text{system}} = 0$, indicating that the system is stable and no longer driven toward reactants or products. In contrast, processes are spontaneous when $\Delta G_{\text{system}} < 0$ and are non-spontaneous in the forward direction when $\Delta G_{\text{system}} > 0$.

The enthalpy of hydrogenation of benzene is determined by calculating the hypothetical enthalpy for three localized double bonds and adjusting for the resonance energy stabilization. First, the hypothetical enthalpy for three double bonds is obtained by multiplying the enthalpy of hydrogenation of a single double bond, as represented by cyclohexene, by three: $3 \times -119.5 = -358.5$ kJ mol$^{-1}$. The actual enthalpy of hydrogenation is then found by subtracting the resonance energy from this hypothetical value, resulting in $-358.5 - (-150.4) = -208.1$ kJ mol$^{-1}$.

The relationship between enthalpy change and internal energy change for a gaseous reaction is governed by the equation
$\Delta H = \Delta E + \Delta n_g RT$
where $\Delta n_g$ represents the difference between the total moles of gaseous products and gaseous reactants. For the reaction $\text{H}_{2(g)} + \text{Br}_{2(g)} \rightarrow 2\text{HBr}_{(g)}$, the calculation yields $\Delta n_g = 2 - (1 + 1) = 0$. Because the change in gaseous moles is zero, the term $\Delta n_g RT$ vanishes, leaving the equality $\Delta H = \Delta E$.

Spontaneity in a chemical system is determined by the thermodynamic drive toward a state of minimum energy and maximum disorder. A reaction proceeds naturally when it is exothermic, releasing energy such that $\Delta H < 0$, and simultaneously increases the randomness of the system, represented by $\Delta S > 0$. Because this specific combination satisfies both fundamental requirements for a process to occur without external input, such a reaction is guaranteed to be spontaneous.

The standard enthalpy of neutralization is defined as $-57.33 \text{ kJ mol}^{-1}$ for the reaction of $H^+$ and $OH^-$ ions in water, but the reaction between $\mathrm{MgO}_{(s)}$ and $\mathrm{HCl}_{(aq)}$ includes an additional, highly exothermic step involving the hydration of $\mathrm{Mg}^{2+}$ ions. This supplemental energy release, which contributes to an overall measured reaction enthalpy of $-146 \text{ kJ}$, results in a total heat evolution that is significantly larger in magnitude than the standard $-57.33 \text{ kJ mol}^{-1}$ value.

A reaction proceeds spontaneously whenever the change in Gibbs free energy, represented by the equation $\Delta G = \Delta H - T \Delta S$, results in a negative value. When both the enthalpy change $\Delta H$ and entropy change $\Delta S$ are positive, the sign of $\Delta G$ depends entirely on the relative magnitudes of $\Delta H$ and the product $T \Delta S$. Therefore, the reaction becomes spontaneous only when the entropy-driven term $T \Delta S$ exceeds the enthalpy term $\Delta H$, ensuring that $\Delta G$ becomes negative.

The work done during the expansion of a gas against a constant external pressure is defined by the equation $W = -P_{\text{ext}} \Delta V$. Begin by determining the change in volume, which is 6 dm³ minus 4 dm³, resulting in 2 dm³. Because 1 dm³ is equal to 1 L, the volume change is 2 L. Using the given external pressure of 3 atm, the work becomes $W = -3 \,\text{atm} \times 2 \,\text{L} = -6 \,\text{L}\,\text{atm}$. Applying the conversion factor where $1 \,\text{L}\,\text{atm} = 101.32 \,\text{J}$ allows you to calculate the energy in Joules, yielding $W = -6 \times 101.32 \,\text{J} = -607.92 \,\text{J}$. Rounding this value gives a final result of -608 J.

The enthalpy change of a reaction is calculated by the difference between the total energy absorbed to break the bonds of the reactants and the total energy released when new bonds are formed in the products. Because breaking chemical bonds is an endothermic process and forming them is an exothermic process, the standard enthalpy change, $\Delta H^{\circ}$, is determined by subtracting the sum of the bond energies of the products from the sum of the bond energies of the reactants.

For the reaction $\mathrm{H}_{2} + \mathrm{Br}_{2} \rightarrow 2\mathrm{HBr}$, the energy required to break the reactants is the sum of 433 kJ for the $\mathrm{H}-\mathrm{H}$ bond and 192 kJ for the $\mathrm{Br}-\mathrm{Br}$ bond, which totals 625 kJ. The energy released when forming two moles of $\mathrm{H}-\mathrm{Br}$ bonds is calculated as $2 \times 364 = 728$ kJ. Subtracting the released energy from the required energy gives the final value:
$\Delta H^{\circ} = 625 - 728 = -103 \,\text{kJ}$
The negative sign indicates that the reaction releases energy, confirming that it is an exothermic process.

The fundamental relationship determining the Gibbs free energy change for a process at constant temperature is defined by the following expression:
$\Delta G = \Delta H - T \Delta S$
Calculating the standard Gibbs energy change for the oxidation of ammonia requires normalizing the units by converting the entropy value of -145.6 into kJ K⁻¹ mol⁻¹ by dividing by 1000, which is necessary to ensure consistency with the enthalpy units. Multiplying this converted entropy value by the temperature of 298 K yields an entropy term of -43.38 kJ mol⁻¹. Finally, applying the formula with the given enthalpy change of -382.64 kJ mol⁻¹, the $\Delta G$ is calculated as -382.64 - (-43.38), which equals -339.3 kJ mol⁻¹.

The Second Law of Thermodynamics dictates that any spontaneous or irreversible process inherently tends toward a state of increased total disorder across the universe. Because the universe consists of both the system and the surroundings, the net change in entropy for a spontaneous event must be positive to ensure this increase in disorder. This fundamental requirement means that for any process to occur spontaneously, the total change in entropy must satisfy $\Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$.

The entropy change during a phase transition, such as the melting of ice, occurring at a constant temperature is defined by the ratio of the enthalpy change to the absolute temperature. Because the melting of ice at 0 °C is a reversible process at equilibrium, the entropy change is determined by dividing the heat absorbed, which is equivalent to the enthalpy of fusion, by the temperature in Kelvin.

Converting the enthalpy of fusion from 6.0 kJ mol⁻¹ to 6000 J mol⁻¹ ensures unit consistency, while the temperature of 0 °C converts to 273 K on the absolute scale. Applying the relationship $\Delta S = \frac{6000}{273}$ yields approximately 21.978 J K⁻¹ mol⁻¹, which rounds to 21.98 J K⁻¹ mol⁻¹.

The enthalpy of solution represents the total energy change involved in dissolving a solute, which can be viewed as the net result of breaking intermolecular attractions in both the pure solvent and the pure solute, followed by the establishment of new solvent-solute interactions. According to the principle of conservation of energy applied through state functions, the overall enthalpy change for the process is simply the algebraic sum of the enthalpy changes for each individual step. Because the process involves overcoming initial attractions in the first two stages and forming new interactions in the final stage, the total heat of solution is mathematically expressed as $\Delta H_{\text{soln}} = \Delta H_{1} + \Delta H_{2} + \Delta H_{3}$.

The standard enthalpy of reaction is equivalent to the standard enthalpy of formation for a compound when that substance is produced directly from its constituent elements in their standard physical states. Because the enthalpy of formation for any element in its natural state is defined as zero, the heat of reaction simplifies to the formation enthalpy of the product whenever the reactants are exclusively such elements. In the chemical equation $Xe_{(g)}+2 {F}_{2(g)} \rightarrow {XeF}_{4(g)}$, both xenon and fluorine are elements \in their standard states, meaning their respective enthalpies of formation are zero. As a result, the total enthalpy of the reaction is equal to the enthalpy of formation for $XeF_4$.

The transformation of graphite into diamond is non-spontaneous at standard pressure and requires an input of work to overcome the standard free energy change $\Delta G^\circ$. The system reaches equilibrium when the work done by pressure balances this energy requirement, described by the thermodynamic equation $\Delta G = \Delta G^\circ + P \Delta V = 0$. First, we calculate the molar volumes for both allotropes using the molar mass of $12 \,\text{g} \,\text{mol}^{-1}$. The volume of graphite is approximately $\frac{12}{2.25} = 5.33 \,\text{cm}^3 \,\text{mol}^{-1}$ and the volume of diamond is approximately $\frac{12}{3.31} = 3.62 \,\text{cm}^3 \,\text{mol}^{-1}$.

The change in molar volume, $\Delta V = V_{\text{diamond}} - V_{\text{graphite}}$, is approximately $-1.71 \,\text{cm}^3 \,\text{mol}^{-1}$, which is equivalent to $-1.71 \times 10^{-6} \,\text{m}^3 \,\text{mol}^{-1}$. Substituting the given $\Delta G^\circ = 1895 \,\text{J} \,\text{mol}^{-1}$ into the equilibrium expression $P = -\frac{\Delta G^\circ}{\Delta V}$, we calculate the required pressure as $P = -\frac{1895}{-1.71 \times 10^{-6}} \,\text{Pa}$. This results in a value of approximately $11.08 \times 10^8 \,\text{Pa}$. Since this calculated value does not correspond to any of the specific numerical choices provided, the appropriate conclusion is that none of the options are correct.

The relationship between enthalpy change and internal energy change for a chemical reaction is given by the equation $\Delta H - \Delta E = \Delta n_g RT$, where $\Delta n_g$ represents the change in the number of moles of gaseous species between products and reactants. For the reaction $\mathrm{C}_{3} \mathrm{H}_{8(\mathrm{g})} + 5 \mathrm{O}_{2(\mathrm{g})} \rightarrow 3 \mathrm{CO}_{2(\mathrm{g})} + 4 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})}$, we must consider only the gaseous components, as liquid water does not contribute to the gaseous mole count. The product side contains 3 moles of gaseous carbon dioxide, while the reactant side consists of 1 mole of propane and 5 moles of oxygen, totaling 6 moles of gaseous reactants. Calculating the difference, we find $\Delta n_g = 3 - 6 = -3$, which leads to the final result of $\Delta H - \Delta E = -3 RT$.

The molar heat capacity represents the energy required to raise the temperature of one mole of a substance by one degree Kelvin. Because the problem provides the mass of water rather than the number of moles, the first step is to calculate the moles by dividing the 100 g of water by its molar mass of 18 g/mol. The heat added to the system is then distributed across this molar amount, causing a temperature rise that can be determined by the relationship between total heat, amount of substance, and molar heat capacity.

The calculation follows the formula $Q = n \cdot C \cdot \Delta T$ where substituting the given values leads to the equation:
$1000 = \left( \frac{100}{18} \right) \cdot 75 \cdot \Delta T$
Rearranging the equation to isolate the temperature change gives $\Delta T = \frac{1000 \cdot 18}{100 \cdot 75}$ which calculates to 2.4 K.

Entropy is defined by the fundamental thermodynamic relationship $\Delta S = \frac{q_{\text{rev}}}{T}$, where heat is measured in Joules and absolute temperature in Kelvin. Dividing the unit of energy by the unit of temperature gives the base entropy unit of Joules per Kelvin. Normalizing this value for one mole of substance leads to the standard molar entropy unit of J K^-1 mol^-1.

Hess's Law states that the total enthalpy change for a chemical process remains constant regardless of the path taken, allowing us to derive the enthalpy of formation for a compound by combining the known heats of combustion for its constituent elements and the compound itself. We represent the target formation reaction as:
$\mathrm{C}_{(s)} + 2 \mathrm{H}_{2(g)} \rightarrow \mathrm{CH}_{4(g)}$
The enthalpy change for this process is determined by summing the heat of combustion for carbon ($\Delta H = -94$ kcal/mol) with twice the heat of combustion for hydrogen ($\Delta H = -68$ kcal/mol), which gives a combined value of -94 - 136 = -230 kcal/mol. Since the target reaction produces methane rather than consuming it, we subtract the heat of combustion for methane ($\Delta H = -213$ kcal/mol) to effectively reverse its combustion equation, resulting in -230 - (-213), which equals -17 kcal/mol.

When dealing with the reversible expansion of an ideal gas at a constant temperature, we identify the process as isothermal. Because the temperature remains steady, the internal energy does not change, and the entropy variation is driven strictly by the expansion in volume, defined by the formula $\Delta S = n R \ln \frac{V_2}{V_1}$.

Since the gas is kept at 27 degrees Celsius, which is 300 K, the temperature component of the entropy formula simplifies to zero, leaving only the volume-dependent term. We substitute the given values, using 2 moles for the amount of gas and 2 cal/mol K for the gas constant, along with the volume ratio of 20 liters to 2 liters, which results in $\Delta S = 2 \times 2 \times \ln 10$. Applying the standard approximation $\ln 10 \approx 2.303$, the total entropy change calculates to 9.2 cal/K.