Chemical Thermodynamics – JEE Main ChemistryPractice Questions & PYQs

Bond dissociation energies allow for the calculation of reaction enthalpy by subtracting the energy released during bond formation from the energy absorbed during bond cleavage. For the reaction $X_2 + Y_2 \rightarrow 2XY$, one mole of $X_2$ and one mole of $Y_2$ break apart to form two moles of $XY$. Defining the bond dissociation energy of $X_2$ as $E$, the given ratio implies the energy for $XY$ is $E$ and for $Y_2$ is $0.5E$. The enthalpy change for the reaction is determined by the total energy to break the bonds minus the energy released from forming them, leading to the expression $\Delta H = E + 0.5E - 2E = -0.5E$. Given that the formation of one mole of $XY$ releases 200 kJ/mol, the total enthalpy change for the production of two moles is 400 kJ/mol released. Since the reaction enthalpy is -400 kJ/mol, the relationship $-0.5E = -400$ kJ/mol allows for solving for the energy $E$, which results in 800 kJ/mol.

The relationship between enthalpy change and internal energy change for a reaction involving gases is defined by the equation $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ represents the difference between the sum of the stoichiometric coefficients of gaseous products and gaseous reactants. For the reaction $\mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3$, there are 2 moles of gaseous products and $1 + 3 = 4$ moles of gaseous reactants, leading to a change of $\Delta n_g = 2 - 4 = -2$. Substituting this into the thermodynamic expression gives $\Delta H = \Delta U - 2RT$, which demonstrates that $\Delta H$ is strictly less than $\Delta U$ by an amount of $2RT$.

Hess's Law dictates that the total enthalpy change of a reaction remains constant regardless of the path taken, allowing us to manipulate known thermochemical equations algebraically to determine the enthalpy of an unknown reaction. The enthalpy of formation for carbon monoxide corresponds to the reaction $C + 0.5 O_2 \rightarrow CO$. We are provided with the combustion equations for carbon and carbon monoxide: $C + O_2 \rightarrow CO_2$ with $\Delta H = -393.5\ \mathrm{kJ\ mol^{-1}}$ and $CO + 0.5 O_2 \rightarrow CO_2$ with $\Delta H = -283.0\ \mathrm{kJ\ mol^{-1}}$. By subtracting the second equation from the first, the carbon dioxide terms on both sides cancel out, effectively leaving the desired formation reaction. Consequently, we perform the same operation on the enthalpy values, calculating $-393.5 - (-283.0)$, which yields an enthalpy of formation of $-110.5\ \mathrm{kJ\ mol^{-1}}$.

Work done by an ideal gas during expansion against a constant external pressure is defined by the formula $w = -P\Delta V$. First, calculate the change in volume by subtracting the initial volume of 1 x 10^-3 cubic meters from the final volume of 1 x 10^-2 cubic meters, which results in an expansion of 9 x 10^-3 cubic meters. Multiplying this change in volume by the constant external pressure of 1 x 10^5 Newtons per square meter provides the energy involved. As the gas pushes against the surroundings, the work done is calculated by $w = -(1 \times 10^5) \times (9 \times 10^{-3})$, resulting in a final value of -900 Joules.

Enthalpy is classified as a state function, which means its value depends strictly on the initial and final states of a system rather than the path taken to get from one to the other. According to Hess's Law, the total $\Delta H$ for a reaction remains constant regardless of whether the transformation happens directly in a single step or through a series of intermediate stages. Consequently, the specific sequence of intermediate reactions does not influence the overall energy change of the system.

For any spontaneous or irreversible process, the universal evolution tends toward a state of higher disorder, which is defined by the Second Law of Thermodynamics as an increase in entropy for an isolated system, represented as $(dS)_{V,E} > 0$. When the process is restricted to constant temperature and pressure, the direction of spontaneous change is instead governed by the Gibbs free energy, where the system naturally seeks a lower energy state to reach equilibrium, resulting in the condition $(dG)_{T,P} < 0$. These two relations function as the fundamental thermodynamic markers indicating that a process will proceed naturally without external intervention.

Thermodynamics dictates that the standard Gibbs free energy change $\Delta G^\circ$ is fundamentally linked to the equilibrium state of a reversible reaction, quantified by the equilibrium constant $K_c$. This relationship is defined by the equation $\Delta G^\circ = -RT \ln K_c$, which describes how the energy difference between products and reactants at standard conditions influences the position of equilibrium. By simple algebraic rearrangement, this relationship is also represented as $-\Delta G^\circ = RT \ln K_c$.

Internal energy is fundamentally a state function, meaning its value depends solely on the current state of a system rather than the specific sequence or type of path taken to reach that state. When a system undergoes a cyclic process by traveling from state $A$ to $B$ and eventually returning to $A$, the initial and final states are identical. Because the net change in a state function over any complete cycle is always zero, regardless of whether the individual path segments are reversible or irreversible, the total change in internal energy is zero.

The enthalpy change of a reaction is determined by the energy required to break the bonds in the reactant molecules minus the energy released when new bonds are formed in the product molecules. This energy balance reflects the net heat absorbed or evolved during the transformation.

For the hydrogenation reaction $\mathrm{H}_2\mathrm{C}=\mathrm{CH}_2(\mathrm{g}) + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{H}_3\mathrm{C}-\mathrm{CH}_3(\mathrm{g})$, we account for the specific bonds involved. The reactants possess one $\mathrm{C}=\mathrm{C}$ double bond, four $\mathrm{C}-\mathrm{H}$ single bonds, and one $\mathrm{H}-\mathrm{H}$ single bond. The product contains one $\mathrm{C}-\mathrm{C}$ single bond and six $\mathrm{C}-\mathrm{H}$ single bonds. By setting up the energy balance, we see that four $\mathrm{C}-\mathrm{H}$ bonds appear on both sides of the equation and can be canceled out to simplify the calculation.

The formula for the enthalpy change becomes:
$\Delta H = (\mathrm{C}=\mathrm{C}) + (\mathrm{H}-\mathrm{H}) - (\mathrm{C}-\mathrm{C}) - 2(\mathrm{C}-\mathrm{H})$
Substituting the known bond energies into this expression, we get:
$\Delta H = 615 + 435 - 347 - 2(414)$
Performing the arithmetic gives 1050 - 1175, which results in an enthalpy change of -125 kJ.

Thermal capacity, also known as heat capacity, represents the total quantity of heat energy required to increase the temperature of a specific body by 1 K. Unlike specific heat, which is an intrinsic property defined for a unit mass of a material, thermal capacity depends on the total mass and composition of the actual object in question. This relationship is defined by the formula $C = \frac{Q}{\Delta T}$, which quantifies the heat energy $Q$ required to produce a temperature change $\Delta T$.

The spontaneity of a chemical reaction is governed by the Gibbs free energy equation $\Delta G = \Delta H - T \Delta S$, where a process is feasible only when $\Delta G$ is negative. Since the reaction is endothermic, $\Delta H$ is positive, meaning the only way to achieve a negative $\Delta G$ at higher temperatures is for the term $T \Delta S$ to be positive and sufficiently large to outweigh the enthalpy change. This requires $\Delta S$ to be positive, allowing the $T \Delta S$ term to increase with temperature until it exceeds $\Delta H$, thereby driving the reaction toward spontaneity at the boiling point.

The First Law of Thermodynamics is rooted in the principle of the conservation of energy, which stipulates that energy cannot be generated out of nothing but must instead be transformed from one state to another. For a heat engine to perform work, the energy output is strictly governed by the net heat exchanged, requiring a balance between the heat absorbed, the heat rejected, and the mechanical work produced. By asserting that the work done is $J(Q_1+Q_2)$, the statement overlooks the physical reality that a portion of the input thermal energy is inevitably converted into other forms of energy rather than entirely into useful mechanical work. Consequently, this representation creates a mathematical contradiction where the engine seemingly creates energy or ignores the fundamental efficiency constraints required by the law of conservation of energy.

Thermodynamic feasibility in chemical reactions is determined by the enthalpy change ($\Delta H$), where a negative value indicates a spontaneous process and a more negative value implies greater stability and feasibility. Because the formation of $2 \mathrm{ZnO}$ releases $-412 \text{ J}$, which is a higher negative value than the $-393 \text{ J}$ released during the formation of $2 \mathrm{CO}_2$, the oxidation of zinc is thermodynamically more favored than that of carbon. This enhanced spontaneity for zinc indicates that it has a greater affinity for oxygen, allowing zinc to act as the agent capable of oxidizing carbon.