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…

Q41 jee_main_2025_24_jan_evening Hess's Law of Constant Heat Summation

S(g) + (3)/(2) O₂(g) arrow SO₃(g) + 2x kcal

SO2(g) + (1)/(2)O2(g) arrow SO3(g) + y kcal

The heat of formation of

SO₂(g)

is given by:

A. \frac{2x}{y}\mathrm{\ kcal}
B. y - 2x\mathrm{\ kcal}
C. 2x + y\mathrm{\ kcal}
D. x + y\mathrm{\ kcal}

Solution

Related Formula

Using Hess's Law, the enthalpy change of a net reaction can be determined by linearly combining the steps:

Δ Hnet = Σ Δ Hproducts - Σ Δ Hreactants

Core Logic

The heat of formation of $SO₂(g)$ $\mathrm{SO}_2(g)$ corresponds to the target thermochemical equation:

Target: S(g) + O2(g) arrow SO2(g) Δ Hf = ?

Let's write out the given equations along with their enthalpy changes (remembering that exothermic reactions release heat, so $Δ H = -Q$ $\Delta H = -Q$): 1. $S(g) + (3)/(2)O₂(g) arrow SO₃(g) Δ H₁ = -2x kcal$ 2. $SO₂(g) + (1)/(2)O₂(g) arrow SO₃(g) Δ H₂ = -y kcal$

To isolate $SO₂(g)$ $\mathrm{SO}_{2}(g)$ on the product side, subtract Equation (2) from Equation (1):

[S(g) + (3)/(2)O2(g)] - [SO2(g) + (1)/(2)O2(g)] arrow SO3(g) - SO3(g)

S(g) + O2(g) arrow SO2(g)

Now apply the same operation to the enthalpy values:

Δ Hf = Δ H1 - Δ H₂ = -2x - (-y) = y - 2x kcal

This matches Option (2).

Pattern Recognition

To isolate your target species on the desired side of the equation, use Hess's Law to add or subtract the given elemental equations. Make sure to invert the sign of the enthalpy change if you reverse a reaction.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q33 jee_main_2025_24_jan_morning Spontaneity and Gibbs Energy Change

Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.

A. Both $Δ H$ $\Delta H$ and $Δ S$ $\Delta S$ are (+ve)
B. $Δ H$ $\Delta H$ is (-ve) but $Δ S$ $\Delta S$ is (+ve)
C. $Δ H$ $\Delta H$ is (+ve) but $Δ S$ $\Delta S$ is (-ve)
D. Both $Δ H$ $\Delta H$ and $Δ S$ $\Delta S$ are (-ve)

Solution

Related Formula

Δ G = Δ H - TΔ S

Core Logic

An endothermic profile specifies that $Δ H > 0$ $\Delta H > 0$. For the system to become spontaneous ( $Δ G < 0$ $\Delta G < 0$) specifically when shifting to higher temperatures ( $T$ $T$), the temperature-dependent entropic subtraction term ( $-TΔ S$ $-T\Delta S$) must outweigh the enthalpic barrier. This transition demands a positive structural entropy step, i.e., $Δ S > 0$ $\Delta S > 0$.

Hence, both $Δ H$ $\Delta H$ and $Δ S$ $\Delta S$ are positive.

Pattern Recognition

Spontaneity driven purely by elevated thermal thresholds mandates matching positive signs for enthalpy and entropy.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q47 jee_main_2025_24_jan_morning Gibbs Free Energy and Equilibrium Temperature

Standard entropies of

X₂, Y₂

and

XY₅

are 70, 50 and

110 ~J ~K⁻¹ ~mol⁻¹

respectively. The temperature in Kelvin at which the reaction

(1)/(2) X _ 2 + (5)/(2) Y _ 2 arrow X Y _ 5 Δ H ^ ° = - 3 5 k J m o l ^ - 1

will be at equilibrium is (Nearest integer)

Numerical Answer. Answer: 700 to 700

Solution

Related Formula

Δ Sᵣₓₙ⁰ = Σ Sproducts⁰ - Σ Sreactants⁰ and T = (Δ H⁰)/(Δ S⁰) at equilibrium (Δ G⁰ = 0)

Core Logic

First, calculate the standard entropy change for the reaction system ( $Δ Sᵣₓₙ⁰$ $$\Delta S_{\text{rxn}}^{0}$$):

Δ Sᵣₓₙ⁰ = S⁰(XY₅) - [ (1)/(2)S⁰(X₂) + (5)/(2)S⁰(Y₂) ]

Δ Sᵣₓₙ⁰ = 110 - [ ((1)/(2) \times 70) + ((5)/(2) \times 50) ] = 110 - [35 + 125]

Δ Sᵣₓₙ⁰ = 110 - 160 = -50 J K⁻¹ mol⁻¹

At thermodynamic equilibrium, the change in Gibbs free energy drops to zero ( $Δ G⁰ = 0$ $\Delta G^0 = 0$):

0 = Δ H⁰ - TΔ S⁰ T = (Δ H⁰)/(Δ S⁰)

Convert the enthalpy value into Joules ( $Δ H⁰ = -35 \times 10³ J/mol$ ) and substitute the parameters:

T = -35000 J mol⁻¹-50 J K⁻¹ mol⁻¹ = 700 Kelvin

Pattern Recognition

Ensure all variables use matching energy units (Joules vs. Kilojoules) before setting up your final division step.

Chapter Mix

Class 11 Chemistry: Chemical Thermodynamics

Q jee_main_2025_28_jan_evening First Law of Thermodynamics and State Functions

An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path

A\to B\to C arrow Darrow A

as shown in the three cases below. Choose the correct option regarding

Δ U

:

Thermodynamic cyclic path diagrams for Q33 - JEE Main 2025 — The diagram displays three distinct cyclic paths (Case I, Case II, Case III) on volume vs pressure graphs.

A. $Δ U (Case-III) > Δ U (Case-II) > Δ U (Case-I)$
B. $Δ U (Case-I) > Δ U (Case-II) > Δ U (Case-III)$
C. $Δ U (Case-I) > Δ U (Case-III) > Δ U (Case-II)$
D. $Δ U (Case-I) = Δ U (Case-II) = Δ U (Case-III)$

Solution

Related Formula

For any state function like Internal Energy ( $U$ $U$), the cyclic integral over a complete closed loop is identically zero:

\oint dU = 0 Δ Ucyclic = 0

Core Logic

Internal energy ( $U$ $U$) depends only on the initial and final states of the thermodynamic system, not on the path followed.

In all three listed cases, the ideal gas undergoes a complete cyclic path that returns to its original configuration state $A$ $A$.

Step 1: Final Evaluation

Since every transformation begins and ends at point $A$ $A$:

Δ UCase-I = 0

Δ UCase-II = 0

Δ UCase-III = 0

Therefore, $Δ U (Case-I) = Δ U (Case-II) = Δ U (Case-III)$ .

Pattern Recognition

Do not waste time calculating path areas or values if the question asks for a state function change ( $Δ U$ $\Delta U$, $Δ H$ $\Delta H$, $Δ S$ $\Delta S$, $Δ G$ $\Delta G$) over a cyclic loop. The answer is instantly zero for all cases!

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q47 jee_main_2025_28_jan_evening Hess's Law / Enthalpy of Formation

Consider the following data: Heat of formation of

CO₂(g) = -393.5 ~kJ~mol⁻¹

Heat of formation of

H₂O(l) = -286.0 ~kJ~mol⁻¹

Heat of combustion of benzene

= -3267.0 ~kJ~mol⁻¹

The heat of formation of benzene is ______

kJ~mol⁻¹

(Nearest integer).

Numerical Answer. Answer: 48 to 48

Solution

Related Formula

Enthalpy of reaction from enthalpy of formation data:

Δ Hreaction = Σ Δ Hf(Products) - Σ Δ Hf(Reactants)

Core Logic

Write out the balanced thermochemical equation for the combustion of benzene ( $C₆H₆$ $C_6H_6$):

C₆H₆(l) + (15)/(2)O₂(g) arrow 6CO₂(g) + 3H₂O(l)

Given parameters:

$Δ Hc = -3267.0 kJ/mol$ $\Delta H_c = -3267.0\mathrm{\ kJ/mol}$
$Δ Hf[CO₂] = -393.5 kJ/mol$
$Δ Hf[H₂O] = -286.0 kJ/mol$
$Δ Hf[O₂] = 0 kJ/mol$ $\Delta H_f[O_2] = 0\mathrm{\ kJ/mol}$

Step 1: Applying Hess's Law

Substitute these values into the reaction expression:

-3267 = [6(-393.5) + 3(-286.0)] - Δ Hf[C₆H₆]

-3267 = [-2361.0 - 858.0] - Δ Hf[C₆H₆]

-3267 = -3219.0 - Δ Hf[C₆H₆]

Δ Hf[C₆H₆] = -3219.0 + 3267.0 = 48 kJ/mol

Pattern Recognition

Always set up products minus reactants when using heat of formation data. Pay close attention to stoichiometric coefficients (multiply $CO₂$ $CO_2$ by $6$ $6$ and $H₂O$ $H_2O$ by $3$ $3$) to ensure accurate bookkeeping.

Chapter Mix

Class 11 Chemistry: 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 7

Solution

Related Formula

Core Logic

Pattern Recognition

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Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying Hess's Law

Pattern Recognition

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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: