Chemical Thermodynamics – NEET UG ChemistryPractice Questions & PYQs

When two separate chlorine atoms react to form a diatomic chlorine molecule, the creation of the stable covalent bond releases energy into the surroundings. Because this process releases heat rather than absorbing it, the reaction is exothermic, which dictates that $\Delta_rH < 0$. Regarding entropy, we must evaluate the change in the system's disorder by comparing the number of gas particles before and after the reaction. The transition from two independent, free-moving atoms to a single, more constrained diatomic molecule results in a decrease in the total number of particles and a corresponding reduction in spatial randomness, which implies $\Delta_rS < 0$.

In free expansion, gas expands against zero external pressure, meaning the work done by the system is calculated as $w = -P_{ex} \Delta V = 0$. Since the process is adiabatic, no heat is transferred, so $q = 0$, and the first law of thermodynamics, $\Delta U = q + w$, results in a total change in internal energy of zero. Because the internal energy of an ideal gas is strictly dependent on temperature, an unchanging internal energy necessarily implies that $\Delta T = 0$.

The work done on an ideal gas during expansion against a constant external pressure is defined by the formula $W = -P_{\text{ext}}(V_f - V_i)$. In this scenario, the gas expands from an initial volume of 0.001 m^3 to a final volume of 0.01 m^3 while working against a constant external pressure of 100,000 N/m^2. The difference in volume is calculated as 0.01 m^3 minus 0.001 m^3, resulting in 0.009 m^3.

Applying these values to the work formula gives $W = -100,000 \times 0.009$. Multiplying these figures results in -900 J. The negative sign confirms that energy is being transferred from the gas to its surroundings as it expands, representing the work done by the system.

Entropy reflects the degree of randomness or disorder within a system, where a decrease in the number of gaseous moles typically indicates a shift toward a more ordered state. When two moles of atomic hydrogen combine to form one mole of molecular hydrogen as represented by $2H(g) \rightarrow H_2(g)$, the total number of gaseous particles decreases. This reduction in the number of independent species signifies a loss of disorder, which leads to a negative change in entropy for the process.

During the adiabatic expansion of an ideal gas, the system is thermally isolated with $q = 0$, meaning the work done by the expanding gas comes directly at the expense of its internal energy, resulting in a decrease in temperature. Because the process follows the path from A to C, this energy loss dictates that the temperature at the final state C must be lower than at the initial state A, satisfying the condition $T_C < T_A$. Consequently, the statement $T_C > T_A$ contradicts the fundamental cooling effect inherent in adiabatic expansion and is the incorrect assertion.

For a gas expanding against a constant external pressure, the system performs work on its environment. This isothermal process, occurring at 300 K, uses the principle $W = -P_{ext} \Delta V$ to quantify the energy involved. Because the gas volume increases during the expansion, the work done by the gas is negative, indicating energy transfer out of the system.

The gas expands from an initial volume of 0.1 L to a final volume of 0.25 L, resulting in a volume change of 0.15 L. By applying the constant external pressure of 2 bar, we calculate the work as $W = -2 \times 0.15 = -0.3$ L bar. Multiplying this by the conversion factor where 1 L bar equals 100 J provides the result of -30 J.

The enthalpy change of a reaction is calculated by determining the difference between the total energy required to break the bonds of the reactants and the total energy released upon forming the bonds of the products. By assigning the variable $E$ to the bond dissociation energy of $\mathrm{X}_2$, the given ratio defines the bond dissociation energy of $\mathrm{Y}_2$ as $0.5E$ and that of $\mathrm{XY}$ as $E$. For the formation reaction $\frac{1}{2}\mathrm{X}_2 + \frac{1}{2}\mathrm{Y}_2 \rightarrow \mathrm{XY}$, the enthalpy change is given by the equation $\Delta H = \left(\frac{E}{2} + \frac{0.5E}{2}\right) - E$. Simplifying this expression results in $\Delta H = 0.5E + 0.25E - E$, which equates to $-0.25E$. Setting $-0.25E$ equal to the given value of $-200\text{ kJmol}^{-1}$ leads to the calculation of $E = \frac{200}{0.25}$, yielding a bond dissociation energy of $800\text{ kJmol}^{-1}$ for $\mathrm{X}_2$.

The first law of thermodynamics states that the change in internal energy, $\Delta U$, is the sum of the heat transferred to the system, $q$, and the work performed, $w$. Given that the container is well insulated, no heat enters or leaves the system, rendering the process adiabatic and setting $q = 0$, so the change in internal energy is equivalent to the work done. For a gas expanding against a constant external pressure, work is determined by the formula $w = -P_{\text{ext}} \Delta V$. Substituting the given values for pressure and volume, the work is $w = -2.5 \times (4.50 - 2.50) = -5.0$ L atm. Converting this energy value into joules by multiplying by the factor $101.3$ J per $1$ L atm gives a total change in internal energy of $\Delta U = -5.0 \times 101.3 = -505$ J.

The spontaneity of a reaction is governed by the Gibbs free energy change, where a process occurs spontaneously only when $\Delta G < 0$. Using the fundamental relationship $\Delta G = \Delta H - T\Delta S$, the condition for spontaneity becomes $\Delta H - T\Delta S < 0$, which rearranges to $T > \frac{\Delta H}{\Delta S}$. Converting the enthalpy change to 35500 J mol⁻¹ to maintain consistent units, we substitute this value and the entropy change of 83.6 J K⁻¹ mol⁻¹ into the inequality $T > \frac{35500}{83.6}$. This calculation demonstrates that the reaction is spontaneous at temperatures where $T > 425 \text{ K}$.

The spontaneity of a process is dictated by the Gibbs free energy equation $\Delta G = \Delta H - T\Delta S$, where a negative value of $\Delta G$ signifies a spontaneous reaction. For a reaction to remain spontaneous at all temperatures, the enthalpy must be exothermic, denoted as $\Delta H < 0$, while the entropy must increase, denoted as $\Delta S > 0$. Because the absolute temperature $T$ is always positive, these conditions ensure that the term $-T\Delta S$ is always negative, which keeps the total $\Delta G$ negative regardless of the specific temperature value.

The entropy change of a perfect gas is fundamentally dictated by its transition between states, defined by the general expression $\Delta S = nR \ln \left( \frac{p_i}{p_f} \right) + n C_p \ln \left( \frac{T_f}{T_i} \right)$. Because the process occurs isothermally, the temperature remains constant, which means the initial and final temperatures are equal and the temperature-dependent term becomes zero. This reduction isolates the influence of pressure alone, resulting in the final expression $\Delta S = nR \ln \left( \frac{p_i}{p_f} \right)$.

The relationship between the vapour pressure of a liquid and temperature in a liquid-vapour equilibrium is governed by the Clausius-Clapeyron equation, which is expressed as $\ln P = -\frac{\Delta H_v}{RT} + \text{constant}$. By differentiating this equation with respect to temperature $T$, we identify the derivative of the constant term as zero and the derivative of the term involving $-1/T$ as $1/T^2$. Consequently, this operation yields the expression $\frac{d\ln P}{dT} = \frac{\Delta H_v}{RT^2}$, which accurately captures the rate of change of the natural logarithm of vapour pressure with respect to temperature.

The enthalpy of combustion defines the energy released during the formation of one mole of a substance, serving as a constant ratio for any amount of that product. Because chemical energy scales linearly with the quantity of material, one must first determine the number of moles present in the provided mass and then scale the standard enthalpy value accordingly. Since carbon dioxide consists of one carbon atom and two oxygen atoms, its molar mass is $12 + (2 \times 16) = 44$ g/mol.

Dividing the given mass of 35.2 g by the molar mass of 44 g/mol reveals that 0.8 mol of carbon dioxide is formed. Multiplying this mole count by the energy released per mole, given as 393.5 kJ/mol, determines the total thermal energy output. Performing the calculation $0.8 \times 393.5 = 314.8$ results in 314.8 kJ, which effectively rounds to 315 kJ of heat released.

The spontaneity of any chemical or physical process is governed by the Gibbs free energy criterion, $\Delta G = \Delta H - T\Delta S$, which requires the total change in free energy to be negative. When a gas is adsorbed onto a solid surface, the particles transition from a disordered gaseous state to a restricted, ordered state, which necessarily leads to a decrease in entropy, represented as $\Delta S < 0$. Because adsorption is inherently an exothermic process that releases energy as molecules bind to the surface, the enthalpy change is also negative, or $\Delta H < 0$. To achieve a spontaneous process where $\Delta G < 0$, the magnitude of the negative enthalpy must be large enough to offset the unfavorable entropy term, requiring that the magnitude of $\Delta H$ exceeds that of $T\Delta S$.

The calculation of Gibbs free energy for this reaction relies on the relationship $\Delta G = \Delta H - T\Delta S$. We first determine the enthalpy change $\Delta H$ using $\Delta H = \Delta U + \Delta n_g RT$, where the change in gaseous moles is determined to be $\Delta n_g = 2 - 0 = 2$. With an internal energy change of 2.1 kcal and the gas constant taken as 2 cal $K^{-1}$ $mol^{-1}$, we compute $\Delta H = 2.1 + \frac{2 \times 2 \times 300}{1000} = 3.3$ kcal. Using this enthalpy change along with the entropy value of 20 cal $K^{-1}$ at 300 K, the Gibbs free energy is $\Delta G = 3.3 - \frac{300 \times 20}{1000} = -2.7$ kcal.

Hess's Law of constant heat summation states that the total enthalpy change for a chemical process remains the same regardless of whether the reaction occurs in a single step or a series of steps. By adding the two sequential reactions, $\mathrm{C}_{\;\text {(graphite) } \;}+\;\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}$ with enthalpy change $y$ and $\mathrm{CO}_{(g)}+\;\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}$ with enthalpy change $z$, the intermediate product $\mathrm{CO}_{(g)}$ cancels out to yield the overall reaction $\mathrm{C}_{\;\text {(graphite) } \;}+\mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}$. Given that the enthalpy change for this direct reaction is $x$, it follows that $x$ must be equal to the sum of the enthalpy changes of the individual steps, resulting in the relationship $x=y+z$.

At constant temperature and pressure, Avogadro's Law implies that gas volumes behave in proportion to molar quantities. The stoichiometric combustion of methane follows the equation $\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}$, while propane combusts according to $\mathrm{C}_{3}\mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}$.

Designating the volume of methane as $V$ litres, the propane volume becomes $5 - V$ litres. Since methane reacts with two volumes of oxygen and propane with five, the total oxygen volume consumed is described by the equation $2V + 5(5 - V) = 16$. Simplifying this yields $2V + 25 - 5V = 16$, which results in $3V = 9$, meaning there are 3 litres of methane and 2 litres of propane in the mixture.

Converting these volumes to moles using the molar volume of 22.4 litres per mole, the methane amount is $\frac{3}{22.4}$ moles and the propane amount is $\frac{2}{22.4}$ moles. Multiplying these by the respective enthalpies of combustion, 890 kJ per mole and 2220 kJ per mole, gives the total heat released as $\left(\frac{3}{22.4} \times 890\right) + \left(\frac{2}{22.4} \times 2220\right)$. This calculation produces approximately 317 kJ.

The enthalpy change of any chemical reaction, denoted by $\Delta H$, is fundamentally defined by the difference between the activation energy required for the forward reaction, $E_a(f)$, and the activation energy required for the reverse reaction, $E_a(r)$, as expressed by $\Delta H = E_a(f) - E_a(r)$. When the energy barriers for both the forward and reverse directions are identical, the reactants and products effectively sit at the same energy level, meaning no net energy is absorbed or released during the transformation. By substituting the given condition $E_a(f) = E_a(r)$ into the equation, we obtain $\Delta H = E_a(f) - E_a(f) = 0$, confirming that the reaction involves zero enthalpy change.

The electrolytic reduction of metal oxides is governed by the relationship $\Delta_rG = -nFE^\circ$, where the Gibbs free energy change dictates the minimum electrical potential required to drive a non-spontaneous reaction. In the decomposition of aluminum oxide, aluminum ions exist in a $+3$ oxidation state and must gain electrons to transform into metallic aluminum. Because the balanced equation involves $\frac{4}{3}$ moles of aluminum and each mole of aluminum requires $3$ moles of electrons for reduction, the total number of moles of electrons transferred in the process is $n = \frac{4}{3} \times 3 = 4$. By substituting the given energy value of $960,000 \text{ J/mol}$ and the Faraday constant of $96500 \text{ C/mol}$ into the equation, the required potential difference is calculated as follows:

$E^\circ = \frac{\Delta_rG}{nF} = \frac{960000}{4 \times 96500}$

Performing this calculation yields approximately $2.49 \text{ V}$, which establishes $2.5 \text{ V}$ as the minimum voltage necessary to drive the electrolytic reduction of aluminum oxide at 500°C.

Phase transitions such as the melting of ice occur under reversible equilibrium conditions, where the molar entropy change is directly related to the enthalpy of fusion and the absolute temperature by the expression $\Delta S = \frac{\Delta H_f}{T_f}$. Converting the enthalpy of fusion from 1.435 kcal/mol to 1435 cal/mol and using the absolute melting temperature of 273 K, the calculation is performed as:

$\Delta S = \frac{1435}{273} = 5.26 \text{ cal/(mol K)}$

This calculation provides the final value for the molar entropy change during the phase change of water from solid to liquid.