Total enthalpy change for freezing of 1 ~mol of water at 10 C to ice at -10 C is (Given: DeltafusH = x kJ/mol…

Q75 jee_main_2026_22_january_morning Gibbs Energy and Equilibrium Constant

Dissociation of a gas

A₂

takes place according to the following chemical reactions. At equilibrium, the total pressure is 1 bar at 300K.

A₂(g) leftharpoons 2A(g)

The standard Gibbs energy of formation of the involved substances has been provided below:

Substance	$Δ Gf° / kJ mol⁻¹$
$A₂$ $A_{2}$	-100.00
A	-50.832

The degree of dissociation of

A₂(g)

is given by

(x \times 10⁻²)1/2

where

x =

____. (Nearest integer). [Given:

R = 8 J mol⁻¹K⁻¹

,

2 = 0.3010

,

3 = 0.48

]

Numerical Answer. Answer: 33 to 33

Solution

Related Formula

Δ G°reaction = Σ Δ G°f(products) - Σ Δ G°f(reactants)

Δ G° = -RT ln Kₚ

Kₚ = (4α² P₀)/(1 - α²)

Core Logic

First, find standard Gibbs free energy of the reaction:

Δ G°reaction = 2 \times Δ G°f(A) - Δ G°f(A₂)

Δ G° = 2(-50.832) - (-100.00) = -101.664 + 100.00 = -1.664 kJ mol⁻¹

Δ G° = -1664 J mol⁻¹

Use this to find $Kₚ$ $K_p$:

-1664 = -8 \times 300 \times ln Kₚ

1664 = 2400 ln Kₚ

ln Kₚ = (1664)/(2400) = 0.6933

Since $ln 2 \approx 0.693$ $\ln 2 \approx 0.693$, we have: $Kₚ = 2$ $K_p = 2$

Step 1: Calculate Degree of Dissociation

For the reaction $A₂ leftharpoons 2A$ $A_2 \rightleftharpoons 2A$, with initial moles 1 and degree of dissociation $α$ $\alpha$: Moles at equilibrium: $1-α$ $1-\alpha$ (for $A₂$ $A_2$) and $2α$ $2\alpha$ (for A). Total moles = $1+α$ $1+\alpha$.

Kₚ = (PA)²PA₂ = (((2α)/(1+α) P₀)²)/((1-α)/(1+α) P₀) = (4α² P₀)/(1 - α²)

Given total pressure $P₀ = 1 bar$ $P_0 = 1\text{ bar}$ and $Kₚ = 2$ $K_p = 2$:

2 = (4α² (1))/(1 - α²)

2 - 2α² = 4α²

6α² = 2 α² = (1)/(3) α = 1\sqrt(3)

Step 2: Match to Given Format

We are given $α = (x \times 10⁻²)1/2$ $\alpha = (x \times 10^{-2})^{1/2}$. Squaring both sides:

α² = x \times 10⁻²

(1)/(3) = x \times 10⁻² x = (100)/(3) = 33.33

Step 3: Rounding

Nearest integer is 33.

Pattern Recognition

Whenever $Δ G°$ $\Delta G^{\circ}$ yields an $RT ln Kₚ$ $RT \ln K_p$ around 0.693, $Kₚ$ $K_p$ is 2. The formula $Kₚ = 4α²/(1-α²)$ $K_p = 4\alpha^2/(1-\alpha^2)$ for $A₂ leftharpoons 2A$ $A_2 \rightleftharpoons 2A$ at $P=1$ $P=1$ is standard and should be memorized.

Chapter Mix

Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium

Q72 jee_main_2026_22_january_evening Born-Haber Cycle and Lattice Enthalpy

If the enthalpy of sublimation of Li is

155 kJ mol⁻¹

, enthalpy of dissociation of

F₂

is

150 kJ mol⁻¹

, ionization enthalpy of Li is

520 kJ mol⁻¹

, electron gain enthalpy of F is

-313 kJ mol⁻¹

, standard enthalpy of formation of LiF is

-594 kJ mol⁻¹

. The magnitude of lattice enthalpy of LiF is ____

kJ mol⁻¹

(Nearest integer).

Numerical Answer. Answer: 1031 to 1031

Solution

Related Formula

Δf H⁰ = Δsub H + IE + (1)/(2)Δbond H + Δeg H + L.E.

Core Logic

Step 1: Substitute given thermodynamic cycle values into Born-Haber equation:

-594 = 155 + 520 + (150)/(2) + (-313) + L.E.

-594 = 155 + 520 + 75 - 313 + L.E.

-594 = 437 + L.E.

L.E. = -594 - 437 = -1031 kJ mol⁻¹

Step 2: Magnitude of lattice enthalpy is $1031 kJ mol⁻¹$ $1031\text{ kJ mol}^{-1}$.

Born-Haber cycle diagram for LiF for Q72 - JEE Main 2026 Evening

Pattern Recognition

Sees: Born-Haber cycle parameters for ionic solid. Shortcut: Add sublimation, ionization, half-dissociation, and electron gain enthalpies, then subtract from formation enthalpy.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q58 jee_main_2026_23_january_morning First Law of Thermodynamics and Sign Conventions

A cup of water at

5°C

(system) is placed in a microwave oven and the oven is turned on for one minute during which, the water begins to boil. Which of the following option is true?

A. $q = +ve, w = 0, Δ U = -ve$ $q = +ve, w = 0, \Delta U = -ve$
B. $q = +ve, w = -ve, Δ U = +ve$ $q = +ve, w = -ve, \Delta U = +ve$
C. $q = -ve, w = -ve, Δ U = -ve$ $q = -ve, w = -ve, \Delta U = -ve$
D. $q = +ve, w = -ve, Δ U = -ve$ $q = +ve, w = -ve, \Delta U = -ve$

Solution

Related Formula

Δ U = q + w

where: $Δ U$ $\Delta U$ = change in internal energy $q$ $q$ = heat added to system $w$ $w$ = work done on the system

Core Logic

Analyze the state changes step-by-step applying IUPAC sign conventions for thermodynamics.

First Law of Thermodynamics and Sign Conventions diagram for Q58 - JEE Main 2026 Morning

Step 1: Heat Transfer

Since heat is supplied by the microwave oven to the water (system), the system absorbs heat. Thus, $q = +ve$ $q = +ve$.

Step 2: Work Done

As water boils, it converts from liquid to vapor, which implies a massive volume expansion. Work is done by the system against the atmosphere. Thus, work done on the system is negative: $w = -ve$ $w = -ve$.

Step 3: Internal Energy Change

The temperature of water increases from $5^° C$ $5^\circ C$ to $100^° C$ $100^\circ C$, and liquid converts to gas. The internal energy of steam at $100^° C$ $100^\circ C$ is much greater than that of liquid water at $5^° C$ $5^\circ C$. Hence, $Δ U = +ve$ $\Delta U = +ve$.

Pattern Recognition

Heating + Boiling inherently means heat is entering ( $q > 0$ $q > 0$), expanding volume pushes outward doing work ( $w < 0$ $w < 0$), and increasing thermal energy raises the internal state function ( $Δ U > 0$ $\Delta U > 0$).

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q63 jee_main_2026_24_january_morning Work Done in Isothermal Processes

Match the List-I with List-II

List-I (Isothermal process for ideal gas system)	List-II Work done ( $Vf > Vᵢ$ $V_f > V_i$)
A. Reversible expansion	I. $w = 0$ $w = 0$
B. Free expansion	II. $w = -nRT ln (Vf)/(Vᵢ)$ $w = -nRT \ln \frac{V_f}{V_i}$
C. Irreversible expansion	III. $w = -pₑₓ(Vf - Vᵢ)$ $w = -p_{ex}(V_f - V_i)$
D. Irreversible compression	IV. $w = -pₑₓ(Vᵢ - Vf)$ $w = -p_{ex}(V_i - V_f)$

Choose the correct answer from the options given below :

A. $A-IV, B-I, C-III, D-II$ $\text{A-IV, B-I, C-III, D-II}$
B. $A-IV, B-II, C-III, D-I$ $\text{A-IV, B-II, C-III, D-I}$
C. $A-I, B-III, C-II, D-IV$ $\text{A-I, B-III, C-II, D-IV}$
D. $A-II, B-I, C-III, D-IV$ $\text{A-II, B-I, C-III, D-IV}$

Solution

Related Formula

WRev = -\int Pgas dV

WIrrev = -Pₑₓₜ Δ V

Core Logic

(A) Reversible isothermal expansion:

WRev = -nRT ln [ (Vf)/(Vᵢ) ]

Matches with II.

(B) Free expansion (expansion into vacuum):

Pₑₓₜ = 0

$W = 0$ $W = 0$ Matches with I.

(C) Irreversible expansion against constant external pressure:

Wirrev = -Pₑₓₜ Δ V

Wirrev = -Pₑₓₜ (Vf - Vᵢ)

Matches with III.

(D) Irreversible compression: Same core formula but since it is a compression from $Vf$ $V_f$ back to $Vᵢ$ $V_i$, the volume change term flips to $Δ V = (Vᵢ - Vf)$ $\Delta V = (V_i - V_f)$, leading to:

Wirrev = -Pₑₓₜ (Vᵢ - Vf)

Matches with IV.

Step 1: Final Conclusion

The correct matches are A-II, B-I, C-III, D-IV.

Pattern Recognition

Free expansion always implies zero work because $Pₑₓₜ = 0$ $P_{\text{ext}} = 0$. Reversible implies a continuous integration over volume resulting in the natural log formula.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q67 jee_main_2026_24_january_evening Enthalpy of Atomization and Bond Enthalpy

The heat of atomisation of methane and ethane are 'x' kJ mol

⁻¹

and 'y' kJ mol

⁻¹

respectively. The longest wavelength (

λ

) of light capable of breaking the C-C bond can be expressed in SI unit as:

A. $hc1000( y-6x4)⁻¹$
B. $NAhc250(4y-6x)$ $\frac{N_{A}hc}{250(4y-6x)}$
C. $NAhc250(y-6x)$ $\frac{N_{A}hc}{250(y-6x)}$
D. $NAhc(y-(6x)/(4))⁻¹$

Solution

Related Formula

E = (hc)/(λ) λ = (hc)/(E)

Where E is the energy required to break one bond.

Core Logic

For Methane ( $CH₄$ $\mathrm{CH}_4$): $CH₄(g) arrow C(g) + 4H(g) ΔᵣH = x kJ / mole$ The energy is used to break 4 C-H bonds: $4 \times εC-H = 1000x J / mole$

For Ethane ( $C₂H₆$ $\mathrm{C}_2\mathrm{H}_6$): $C₂H₆(g) arrow 2C(g) + 6H(g) ΔᵣH = y kJ / mole$ The energy is used to break 1 C-C bond and 6 C-H bonds: $εC-C + 6 \times εC-H = 1000y J / mole$

Step 1: Extract C-C Bond Energy

From methane: $εC-H = (1000x)/(4) = 250x J / mole$ Substitute into ethane equation: $εC-C + 6 \times (250x) = 1000y$ $εC-C = 1000y - 1500x = [y - (3x)/(2)] \times 1000 J / mole$

Step 2: Wavelength Calculation

The energy calculated above is per mole. Energy required to break one C-C bond is: $E = εC-CNA$

Now, equating this to photon energy: $εC-CNA = (hc)/(λ)$ $λ = hc \cdot NAεC-C$

Substitute $εC-C$ $\varepsilon_{\mathrm{C-C}}$: $λ = hc \cdot NA[y - (3x)/(2)] \times 1000$ $λ = hc \cdot NA(2y - 3x)/(2) \times 1000$ $λ = 2 \cdot hc \cdot NA1000(2y - 3x) = hc \cdot NA500(2y - 3x) = hc \cdot NA250(4y - 6x)$

Pattern Recognition

Always convert molar quantities to per-bond (atomic scale) quantities by dividing by Avogadro's number ( $NA$ $N_A$) when equating macroscopic enthalpy data to single photon quantum limits ( $hc/λ$ $hc/\lambda$). Don't forget to multiply $kJ$ $kJ$ to $J$ $J$ by 1000 for standard SI unit consistency.

Chapter Mix

Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Structure of Atom

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Chemical Thermodynamics Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Calculate Degree of Dissociation

Step 2: Match to Given Format

Step 3: Rounding

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Heat Transfer

Step 2: Work Done

Step 3: Internal Energy Change

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Extract C-C Bond Energy

Step 2: Wavelength Calculation

Pattern Recognition

Chapter Mix

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