Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

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, the melting point decreases, causing the entire solid phase (ice) to disappear completely. 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. Class 11 Chemistry: Chemical Thermodynamics

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct obse

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, the meltin

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

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, the melting point decreases, causing the entire solid phase (ice) to disappear completely. 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. Class 11 Chemistry: Chemical Thermodynamics

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct obse

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, the meltin

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K . If pressure of the…

More Chemical Thermodynamics Previous-Year Questions — Page 5

Q50 jee_main_2025_28_jan_morning Bond Enthalpy Calculation

The formation enthalpies,

Δ Hf

for

H(g)

and

O(g)

are 220.0 and

250.0~kJ~mol⁻¹

, respectively, at

298.15K

, and

Δ Hf⁻

for

H₂O(g)

-242.0kJ mol⁻¹

at the same temperature. The average bond enthalpy of the O-H bond in water at

298.15K

___________________________________________________ (nearest integer).

Numerical Answer. Answer: 466 to 466

Solution

Related Formula

Reaction enthalpy based on atomization processes:

Δᵣ H = Σ Δf H(products) - Σ Δf H(reactants)

Step 1: Map the Dissociation Reaction

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

H₂O(g) arrow 2H(g) + O(g)

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

Δᵣ H = 2 \times B.E.(O-H)

Step 2: Calculate $Δᵣ H$ $\Delta_r H$

Using the enthalpies of formation:

Δᵣ H = [2 \times Δf H(H(g)) + Δf H(O(g))] - Δf H(H₂O(g))

Δᵣ H = [2 \times 220.0 + 250.0] - (-242.0)

Δᵣ H = [440.0 + 250.0] + 242.0 = 690.0 + 242.0 = 932.0 kJ mol⁻¹

Step 3: Solve for Single Bond Enthalpy

2 \times B.E.(O-H) = 932.0

B.E.(O-H) = (932.0)/(2) = 466 kJ mol⁻¹

Pattern Recognition

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

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q jee_main_2025_03_april_morning Intensive and Extensive Properties

Which of the following properties will change when system containing solution 1 will become solution 2?

A. Molar heat capacity
B. Density
C. Concentration
D. Gibbs free energy

Solution

Core Logic

Let us compute the concentration of both solutions:

Concentration of Solution 1 = 10 mol10 L = 1 mol/L

Concentration of Solution 2 = 1 mol1 L = 1 mol/L

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

Q jee_main_2025_03_april_morning Bond Enthalpy and Enthalpy of Formation

Given:

Δ Hsub[C(graphite)] = 710 kJ mol⁻¹

ΔC-HH = 414 kJ mol⁻¹

ΔH-HH = 436 kJ mol⁻¹

ΔC=CH = 611 kJ mol⁻¹

The

Δ Hf

for

CH₂=CH₂

is ________

kJ mol⁻¹

(nearest integer value)

Numerical Answer. Answer: 25 to 25

Solution

Related Formula

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

Δ Hf° = Σ Δ Hatomization (reactants) - Σ B.E.(products)

Core Logic

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

2C(graphite) + 2H₂(g) arrow CH₂=CH₂(g)

To construct this enthalpy pathway:

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

Step 1: Arithmetic Calculation

Δ Hf° = [2 \times 710] + [2 \times 436] - 611 - [4 \times 414]

Δ Hf° = 1420 + 872 - 611 - 1656 = 2292 - 2267 = 25 kJ mol⁻¹

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

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q37 jee_main_2025_04_april_evening Thermochemistry

Consider the given data : (a)

HCl(g) + 10H₂O(l)arrow HCl.10H₂O Δ H = - 6 9. 0 1 k J m o l ^ - 1

(b)

HCl(g) + 40H₂O(l)arrow HCl.40H₂O Δ H = - 7 2. 7 9 k J m o l ^ - 1

Choose the correct statement :

A. Dissolution of gas in water is an endothermic process
B. The heat of solution depends on the amount of solvent.
C. The heat of dilution for the HCl $(HCl.10H₂O$ to $HCl.40H₂O)$ is $3.78kJ mol⁻¹$ $3.78\mathrm{kJ mol}^{-1}$.
D. The heat of formation of HCl solution is represented by both (a) and (b)

Solution

Related Formula

Δ Hdilution = Δ H₂ - Δ H₁

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):

HCl\cdot10H₂O + 30H₂O arrow HCl\cdot40H₂O

Δ H = -72.79 - (-69.01) = -3.78 ~kJ\cdot mol⁻¹

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

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15K$273.15\mathrm{K}$ . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

More Chemical Thermodynamics Previous-Year Questions — Page 5

Solution

Related Formula

Step 1: Map the Dissociation Reaction

Step 2: Calculate Δᵣ H$\Delta_r H$

Step 3: Solve for Single Bond Enthalpy

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Identifying the Variable

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Arithmetic Calculation

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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Ice and water are placed in a closed container at a pressure of 1 atm and temperature $273.15K$ $273.15\mathrm{K}$ . If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

Step 2: Calculate $Δᵣ H$ $\Delta_r H$