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 & Explanation

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

Reference Study Guides

Chemical Thermodynamics

More Chemical Thermodynamics Previous-Year Questions — Page 7

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.

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.

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

Q jee_main_2025_29_jan_morning First Law of Thermodynamics and Heat Capacity

500 ~J

of energy is transferred as heat to

0.5 ~mol

of Argon gas at

298 ~K

and

1.00 atm

. The final temperature and the change in internal energy respectively are : Given:

R = 8.3 JK⁻¹ mol⁻¹

A. $348K$ $348\mathrm{K}$ and $300J$ $300\mathrm{J}$
B. $378K$ $378\mathrm{K}$ and $300J$ $300\mathrm{J}$
C. $368K$ $368\mathrm{K}$ and $500J$ $500\mathrm{J}$
D. $378K$ $378\mathrm{K}$ and $500J$ $500\mathrm{J}$

Solution

Formulas Used

For an ideal gas undergoing a constant pressure process ( $1.00 atm$ $1.00\text{ atm}$):

qₚ = n \cdot Cₚ \cdot Δ T

Change in internal energy ( $Δ U$ $\Delta U$):

Δ U = n \cdot Cv \cdot Δ T

For a monoatomic gas like Argon:

$Cv = (3)/(2) R$ $C_v = \frac{3}{2} R$
$Cₚ = (5)/(2) R$ $C_p = \frac{5}{2} R$

Core Logic

Step 1: Calculate the final temperature ( $Tf$ $T_f$) Heat transferred at constant pressure ( $qₚ$ $q_p$) = $500 J$ $500\text{ J}$

500 = 0.5 \times ((5)/(2) \times 8.3) \times (Tf - 298)

500 = 0.5 \times 20.75 \times (Tf - 298)

500 = 10.375 \times (Tf - 298)

Tf - 298 = (500)/(10.375) \approx 48.2 K

Tf = 298 + 48.2 = 346.2 K \approx 348 K

---

Step 2: Calculate the change in internal energy ( $Δ U$ $\Delta U$)

Δ U = n \cdot Cv \cdot Δ T

Alternatively, using the ratio of heat capacities:

Δ U = ((Cv)/(Cₚ)) \times qₚ = (3)/(5) \times 500 J = 300 J

Thus, the final temperature is $348 K$ $348\text{ K}$ and the change in internal energy is $300 J$ $300\text{ J}$.

Pattern Recognition

For a monoatomic ideal gas under constant pressure, exactly $60%$ $60\%$ of the heat added ( $(Cv)/(Cₚ) = (3)/(5)$ $\frac{C_v}{C_p} = \frac{3}{5}$) goes into increasing the internal energy ( $Δ U$ $\Delta U$), while $40%$ $40\%$ is lost to expansion work ( $W$ $W$).

Correct Option: (A)

Q79 jee_main_2024_01_february_morning First Law of Thermodynamics

Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following:

A. $q = 0, Δ T \neq 0, w = 0$ $q = 0, \Delta T \neq 0, w = 0$
B. $q = 0, Δ T < 0, w \neq 0$ $q = 0, \Delta T < 0, w \neq 0$
C. $q \neq 0, Δ T = 0, w = 0$ $q \neq 0, \Delta T = 0, w = 0$
D. $q = 0, Δ T = 0, w = 0$ $q = 0, \Delta T = 0, w = 0$

Solution

Core Logic

Free expansion means expansion against a vacuum ( $Pₑₓₜ = 0$ $P_{ext} = 0$). Work done: $w = -Pₑₓₜ Δ V$ $w = -P_{ext} \Delta V$. Since $Pₑₓₜ = 0$ $P_{ext} = 0$, $w = 0$ $w = 0$.

Adiabatic condition means there is no heat exchange with the surroundings. Heat transfer: $q = 0$ $q = 0$.

According to the First Law of Thermodynamics, $Δ U = q + w$ $\Delta U = q + w$. Since $q = 0$ $q = 0$ and $w = 0$ $w = 0$, the change in internal energy $Δ U = 0$ $\Delta U = 0$.

For an ideal gas, internal energy is a function of temperature only ( $Δ U = nCvΔ T$ $\Delta U = nC_v\Delta T$). If $Δ U = 0$ $\Delta U = 0$, then $Δ T = 0$ $\Delta T = 0$.

Step 1: Final Parameter Check

Evaluating all parameters simultaneously: $q = 0$ $q = 0$ $w = 0$ $w = 0$ $Δ T = 0$ $\Delta T = 0$

Pattern Recognition

Adiabatic + Free Expansion of IDEAL gas $arrow$ $\rightarrow$ Nothing changes thermodynamically except volume and pressure. $q = 0, w = 0, Δ U = 0, Δ T = 0, Δ H = 0$ .

Chapter Mix

Class 11 Chemistry: Thermodynamics

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