Consider the given data : (a) HCl(g) + 10H2O(l)arrow HCl.10H2O Delta H = - 6 9. 0 1 k J m o l ^ - 1 (b) HCl(g) + 40H2O(l)arrow HCl.40H2O Delta H = - 7 2. 7 9 k J m o l ^ - 1 Choose the correct statement :

Analyzing the thermodynamic statements: - Delta H values are negative, so the dissolution of 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 vs -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 makes option (3) incorrect. 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. Class 11 Chemistry: Chemical Thermodynamics

Consider the given data : (a) HCl(g) + 10H2O(l)arrow HCl.10H2O Delta H = - 6 9. 0 1 k J m o l ^ - 1 (b) HCl(g) + 40H2O(l)arrow HCl.40H2O Delta H = - 7 2. 7 9 k J m o l ^ - 1 Choose the correct stateme

Analyzing the thermodynamic statements: - Delta H values are negative, so the dissolution of 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 vs -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 makes opti

Consider the given data : (a) HCl(g) + 10H2O(l)arrow HCl.10H2O Delta H = - 6 9. 0 1 k J m o l ^ - 1 (b) HCl(g) + 40H2O(l)arrow HCl.40H2O Delta H = - 7 2. 7 9 k J m o l ^ - 1 Choose the correct statement :

Analyzing the thermodynamic statements: - Delta H values are negative, so the dissolution of 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 vs -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 makes option (3) incorrect. 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. Class 11 Chemistry: Chemical Thermodynamics

Consider the given data : (a) HCl(g) + 10H2O(l)arrow HCl.10H2O Delta H = - 6 9. 0 1 k J m o l ^ - 1 (b) HCl(g) + 40H2O(l)arrow HCl.40H2O Delta H = - 7 2. 7 9 k J m o l ^ - 1 Choose the correct stateme

Analyzing the thermodynamic statements: - Delta H values are negative, so the dissolution of 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 vs -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 makes opti

Consider the given data : (a) HCl(g) + 10H2O(l)arrow HCl.10H2O Delta H = - 6 9. 0 1 k J m o l ^

More Chemical Thermodynamics Previous-Year Questions — Page 9

Q83 jee_main_2024_30_january_evening Hess's Law of Constant Heat Summation

Two reactions are given below:

2Fe(s) + (3)/(2)O2(g) arrow Fe₂O3(s), Δ H° = -822 kJ/mol

C(s) + (1)/(2)O2(g) arrow CO(g), Δ H° = -110 kJ/mol

Then enthalpy change for following reaction

3C(s) + Fe₂O3(s) arrow 2Fe(s) + 3CO(g)

Numerical Answer. Answer: 492 to 492

Solution

Related Formula

According to Hess's Law, the net enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps into which it can be divided.

Core Logic

Let the given reactions be: (1) $2Fe(s) + (3)/(2)O2(g) arrow Fe₂O3(s), Δ H₁ = -822 kJ/mol$ (2) $C(s) + (1)/(2)O2(g) arrow CO(g), Δ H₂ = -110 kJ/mol$

Target Reaction (3):

3C(s) + Fe₂O3(s) arrow 2Fe(s) + 3CO(g), Δ H₃ = ?

To construct the target reaction:

We need $3 CO(g)$ $3 \mathrm{CO}_{(g)}$ on the product side, so we multiply reaction (2) by 3.
We need $Fe₂O3(s)$ $\mathrm{Fe}_2\mathrm{O}_{3(s)}$ on the reactant side and $2 Fe(s)$ $2 \mathrm{Fe}_{(s)}$ on the product side, so we reverse reaction (1).

Step 1: Calculate Net Enthalpy

Target Reaction (3) = $3 \times (2) - (1)$ $3 \times (2) - (1)$

Δ H₃ = 3 \times Δ H₂ - Δ H₁

Δ H₃ = 3(-110) - (-822)

Δ H₃ = -330 + 822 = 492 kJ/mol

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q jee_main_2024_30_jan_morning Work Done in Cyclic Process

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

Aarrow Barrow Carrow A

as shown in the diagram. The total work done in the process is ________ J.

Work Done in Cyclic Process diagram for Q83 - JEE Main 2024 Morning — The image is a graph of Volume (dm3) vs Pressure (kPa) showing a triangular cyclic process starting from A(10,10) to B(10,30) to C(30,10) and back to A.

Numerical Answer. Answer: 200 to 200

Solution

Related Formula

Wcyclic = Area enclosed in P-V graph

Core Logic

The work done in a cyclic process is equal to the magnitude of the area enclosed by the cycle on a Pressure-Volume graph. Note that the provided graph is Volume ( $V$ $V$) on the y-axis versus Pressure ( $P$ $P$) on the x-axis. The path $A arrow B arrow C arrow A$ is traced in a clockwise direction on the $V$ $V$- $P$ $P$ graph. Clockwise on a $V$ $V$- $P$ $P$ graph corresponds to anti-clockwise on a standard $P$ $P$- $V$ $V$ graph, meaning net expansion work is done by the gas, making it positive conventionally (or negative depending on chemistry sign convention, but magnitude is asked for).

Step 1: Calculating Area

The enclosed region is a right-angled triangle. Base of triangle on P-axis = $30 - 10 = 20 kPa$ $30 - 10 = 20 \text{ kPa}$ Height of triangle on V-axis = $30 - 10 = 20 dm³$ $30 - 10 = 20 \text{ dm}^3$

Area = (1)/(2) \times base \times height

Area = (1)/(2) \times 20 \times 20 = 200 kPa ³

Step 2: Unit conversion

$1 kPa = 10³ Pa$ $1 \text{ kPa} = 10^3 \text{ Pa}$ $1 dm³ = 1 Litre = 10⁻³ m³$

W = 200 \times 10³ Pa \times 10⁻³ m³

W = 200 J

Pattern Recognition

$1 kPa \cdot 1 L = 1 Joule$ . This direct conversion saves time without converting explicitly to standard SI units ( $Pa$ $Pa$ and $m³$ $m^3$).

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q90 jee_main_2024_31_jan_evening Work Done in Isothermal Reversible Expansion

5 moles

of an ideal gas expands from

10 L

to a volume of

100 L

300 K

under isothermal and reversible condition then work

w

, is

-x J

. The value of

x

is ________ (Given

R = 8.314 J K⁻¹mol⁻¹

)

Numerical Answer. Answer: 28720 to 28721

Solution

Related Formula

W = -2.303 nRT ( (V₂)/(V₁) )

Core Logic

For an isothermal and reversible expansion of an ideal gas, work is done by the system on the surroundings, hence it is negative by IUPAC convention. Given: $n = 5 moles$ $n = 5\text{ moles}$ $R = 8.314 J K⁻¹mol⁻¹$ $T = 300 K$ $T = 300\text{ K}$ $V₁ = 10 L$ $V_1 = 10\text{ L}$ $V₂ = 100 L$ $V_2 = 100\text{ L}$

Step 1: Calculating Work Done

W = -2.303 \times 5 \times 8.314 \times 300 \times ( (100)/(10) )

W = -2.303 \times 5 \times 8.314 \times 300 \times (10)

W = -2.303 \times 12471 \times 1

W = -28720.713 J

Step 2: Final Formatting

The question asks for work $w = -x J$ $w = -x\text{ J}$. So $x = 28720.713$ $x = 28720.713$, which rounds to $28721$ $28721$.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Q88 jee_main_2024_31_jan_morning Gibbs Free Energy and Equilibrium

Consider the following reaction at 298 K.

(3)/(2)O2(g) leftharpoons O3(g). Kₚ = 2.47 \times 10⁻²⁹

ΔᵣG

for the reaction is ________ kJ. (Given

R = 8.314 J K⁻¹ mol⁻¹

)

Numerical Answer. Answer: 163 to 164

Solution

Related Formula

ΔᵣG = -RT ln Kₚ

Step 1: Calculation

ΔᵣG = -8.314 \times 10⁻³ kJ K⁻¹ mol⁻¹ \times 298 K \times ln(2.47 \times 10⁻²⁹)

= -8.314 \times 10⁻³ \times 298 \times (-65.87)

= 163.19 kJ

Step 2: Nearest Integer

Rounding $163.19$ $163.19$ to the nearest integer gives $163$ $163$.

Chapter Mix

Class 11 Chemistry: Thermodynamics

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

More Chemical Thermodynamics Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Step 1: Calculate Net Enthalpy

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Area

Step 2: Unit conversion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Work Done

Step 2: Final Formatting

Chapter Mix

Solution

Related Formula

Step 1: Calculation

Step 2: Nearest Integer

Chapter Mix

More Chemical Thermodynamics Questions — jee_main_2025_04_april_evening

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