Total enthalpy change for freezing of 1 ~mol$1 \mathrm{~mol}$ of water at 10° C$10^{\circ} \mathrm{C}$ to ice at -10° C$-10^{\circ} \mathrm{C}$ is (Given: ΔfusH = x$\Delta_{\mathrm{fus}}\mathrm{H} = \mathrm{x}$ kJ/mol, Cₚ[H₂O( )] = y J mol⁻¹ K⁻¹$\mathrm{C}_{\mathrm{p}}[\mathrm{H}_2\mathrm{O}(\ell)] = \mathrm{y}\text{ J mol}^{-1}\text{ K}^{-1}$, Cₚ[H₂O(s)] = z J mol⁻¹ K⁻¹$\mathrm{C}_{\mathrm{p}}[\mathrm{H}_2\mathrm{O}(\text{s})] = \mathrm{z}\text{ J mol}^{-1}\text{ K}^{-1}$)

Core Logic

Water has a unique property where the density of the liquid phase is greater than the density of the solid phase (ice). Consequently, the molar volume of ice is larger than that of liquid water:

According to Le Chatelier's Principle, increasing the pressure favors the phase that occupies a smaller volume to alleviate the applied stress. Thus, shifting the system forward converts ice into liquid water:

If the pressure is increased considerably (such as doubling it to 2 atm) at 273.15K$273.15\mathrm{K}$, the melting point decreases, causing the entire solid phase (ice) to disappear completely.

Pattern Recognition

Sees: Ice-water system under pressure change. Trap: Assuming that an increase in pressure always favors the solid phase. Water has an anomalous phase curve with a negative slope.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Related Formula

Reaction enthalpy based on atomization processes:

Step 1: Map the Dissociation Reaction

Consider the dissociation of gas phase water molecules into constituent gaseous atoms:

The total energy required corresponds to breaking exactly two O-H$\mathrm{O}-\mathrm{H}$ bonds:

Step 2: Calculate Δᵣ H$\Delta_r H$

Using the enthalpies of formation:

Step 3: Solve for Single Bond Enthalpy

Pattern Recognition

Sees: Atomization state values used to evaluate single bond metrics. Shortcut: Remember Total Dissociation Energy = Σ Δf H(atoms) - Δf H(molecule)$\text{Total Dissociation Energy} = \sum \Delta_f H(\text{atoms}) - \Delta_f H(\text{molecule})$. Halving the result gives the average bond enthalpy.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Core Logic

Let us compute the concentration of both solutions:

Since concentration is identical, both systems share matching compositions. Consequently, all intensive properties (independent of mass/size) like concentration, density, and molar heat capacity remain exactly equal.

Step 1: Identifying the Variable

Gibbs free energy (G$G$) is an extensive property that scales directly with the amount of matter in the system. Because Solution 1 contains a larger total mass and volume than Solution 2, its overall Gibbs free energy value will change.

Pattern Recognition

Shortcut: Look for the only extensive property in the options. Density, concentration, and molar parameters are always intensive. Gibbs free energy (G$G$) scales with total matter quantity.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Related Formula

The standard enthalpy of formation can be evaluated using atomization and bond dissociation enthalpies:

Core Logic

The target formation reaction for ethylene (CH₂=CH₂$\text{CH}_2=\text{CH}_2$) from standard elemental states is:

To construct this enthalpy pathway:

• Sublime 2 mol$2\text{ mol}$ of solid graphite: 2 × Δ Hsub°[C]$2 \times \Delta H_{\text{sub}}^{\circ}[\text{C}]$
• Dissociate 2 mol$2\text{ mol}$ of gaseous H-H$\text{H}-\text{H}$ bonds: 2 × ΔH-HH°$2 \times \Delta_{\text{H}-\text{H}}H^{\circ}$
• Form 1 mol$1\text{ mol}$ of C=C$\text{C}=\text{C}$ double bonds: -1 × ΔC=CH°$-1 \times \Delta_{\text{C}=\text{C}}H^{\circ}$
• Form 4 mol$4\text{ mol}$ of C-H$\text{C}-\text{H}$ single bonds: -4 × ΔC-HH°$-4 \times \Delta_{\text{C}-\text{H}}H^{\circ}$

Step 1: Arithmetic Calculation

Rounding to the nearest integer gives 25.

Pattern Recognition

Shortcut: Group energy components systematically. Reactant state atomization costs +2292 kJ$+2292\text{ kJ}$. Exothermic structural bond formation releases -2267 kJ$-2267\text{ kJ}$. The net difference yields a small endothermic value of +25 kJ mol⁻¹$+25\text{ kJ mol}^{-1}$.

Evaluation Rubric / Model Answer

25

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Related Formula

Core Logic

Analyzing the thermodynamic statements:

• Δ H$\Delta H$ values are negative, so the dissolution of HCl(g)$HCl(g)$ is clearly exothermic, eliminating option (1).
• Since the enthalpy release changes when the moles of water solvent shift from 10 to 40 (-69.01$-69.01$ vs -72.79$-72.79$), the heat of solution depends explicitly on the amount of solvent (Statement 2 is true).
• Let's check Statement 3: By subtracting equation (a) from (b):

The value is negative, indicating an exothermic process, so calling it +3.78$+3.78$ makes option (3) incorrect.

Pattern Recognition

The standard integral enthalpy of solution varies with solvent concentration until infinite dilution is achieved. Thus, concentration dependence is a core property of partial molar solution variables.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics