Q.85 Chemistry ⭐ 360° Option Elimination Active

NEET 2025 Chemistry Question 85 Solution

Topic: Chemical Thermodynamics · Official NTA Booklet Code 45

Question

The standard heat of formation, in kcal/mol of Ba²⁺ is: [Given: standard heat of formation of SO₄²⁻ ion (aq) = −216 kcal/mol, standard heat of crystallisation of BaSO₄(s) = −4.5 kcal/mol, standard heat of formation of BaSO₄(s) = −349 kcal/mol]

✓

Verified Against NTA Final Answer Key

Option (1) : -128.5 kcal/mol

📚 Chemical Thermodynamics Medium

📋 Given Parameters & Problem Setup

Standard Heat of Formation of SO₄²⁻(aq) -216 kcal/mol

Standard Heat of Crystallisation of BaSO₄(s) -4.5 kcal/mol

Standard Heat of Formation of BaSO₄(s) -349 kcal/mol

Target Variable Standard heat of formation of Ba²⁺(aq) in kcal/mol

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The 30-Second Exam Hall Shortcut: ⚡ Hess’s Law Heat of Crystallisation Equation

⏱️ Save ~40 seconds

Crystallisation: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s), ΔH_cryst = -4.5. ΔH_cryst = ΔH_f(BaSO₄) - [ΔH_f(Ba²⁺) + ΔH_f(SO₄²⁻)] ⟹ -4.5 = -349 - ΔH_f(Ba²⁺) - (-216) ⟹ ΔH_f(Ba²⁺) = -349 + 216 + 4.5 = -128.5 kcal/mol!

1 Write the crystallisation reaction: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) with ΔH° = -4.5 kcal/mol.

2 Apply Hess’s Law: ΔH_rxn = Σ ΔH_f°(products) - Σ ΔH_f°(reactants).

3 Set up equation: -4.5 = ΔH_f°(BaSO₄(s)) - [ΔH_f°(Ba²⁺(aq)) + ΔH_f°(SO₄²⁻(aq))].

4 Substitute values: -4.5 = -349 - ΔH_f°(Ba²⁺(aq)) - (-216).

5 Simplify: -4.5 = -349 + 216 - ΔH_f°(Ba²⁺(aq)) = -133 - ΔH_f°(Ba²⁺(aq)).

6 Solve: ΔH_f°(Ba²⁺(aq)) = -133 - (-4.5) = -133 + 4.5 = -128.5 kcal/mol.

7 Select Option (1) in under 20 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Hess’s Law of Constant Heat Summation (NCERT Chemistry · Thermodynamics: Δ_r H° = Σ Δ_f H°(products) - Σ Δ_f H°(reactants)).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Define the Thermochemical Reaction of Crystallisation

The heat of crystallisation of barium sulfate represents the enthalpy change when aqueous barium and sulfate ions precipitate to form one mole of solid crystal:

\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \longrightarrow \text{BaSO}_4(s) \quad \Delta H_{cryst}^\circ = -4.5\text{ kcal/mol}

• This is the reverse of the enthalpy of solution.

Step 2

Apply Hess’s Law to Express Enthalpy of Reaction

The enthalpy of reaction equals the enthalpy of formation of products minus the enthalpies of formation of reactants:

\Delta H_{cryst}^\circ = \Delta H_f^\circ(\text{BaSO}_4, s) - \left[\Delta H_f^\circ(\text{Ba}^{2+}, aq) + \Delta H_f^\circ(\text{SO}_4^{2-}, aq)\right]

• Given values: ΔH_{cryst}° = -4.5 kcal/mol, ΔH_f°(BaSO₄, s) = -349 kcal/mol, ΔH_f°(SO₄²⁻, aq) = -216 kcal/mol.

Step 3

Solve for Standard Heat of Formation of Ba²⁺(aq)

Substitute numerical values and solve the linear equation:

-4.5 = -349 - \left[\Delta H_f^\circ(\text{Ba}^{2+}, aq) + (-216)\right] = -349 + 216 - \Delta H_f^\circ(\text{Ba}^{2+}, aq)

• -4.5 = -133 - \Delta H_f^\circ(\text{Ba}^{2+}, aq)
• \Delta H_f^\circ(\text{Ba}^{2+}, aq) = -133 + 4.5 = -128.5\text{ kcal/mol}.
• Matches Option (1) with absolute precision.

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. Direct algebraic resolution of Hess’s cycle: -133 + 4.5 = -128.5 kcal/mol.

💡

Why this is airtight: None; rigorous thermochemical calculation.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Omits the heat of crystallisation (-349 + 216 = -133).

⚠️

Student Fallacy (Why candidates pick this trap): Ignoring the heat of crystallisation term.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Sign inversion of the -133 value.

⚠️

Student Fallacy (Why candidates pick this trap): Sign error during Hess’s law subtraction.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Arithmetic scrambling of numbers.

⚠️

Student Fallacy (Why candidates pick this trap): Adding all terms indiscriminately.

⚠️

NTA Trap Alert: Omitting Heat of Crystallisation (-133 Trap)

Option 2 (-133.0) is obtained simply by subtracting: -349 - (-216) = -133. Candidates who forget to include the heat of crystallisation (-4.5 kcal/mol) fall directly into Option 2!

🛡️ How to avoid the -1 mark: Always write the balanced precipitation equation and apply ΔH_rxn = Σ ΔH_f(products) - Σ ΔH_f(reactants).

📖

Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Chemical Thermodynamics Standard Benchmark

Core Scientific Principle Tested:

"ΔH_reaction = Σ ΔH_f°(products) - Σ ΔH_f°(reactants)."

Syllabus Focus: Hess’s Law of Constant Heat Summation & Enthalpies of Formation

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

ΔH_f(Ba²⁺) = -349 + 216 + 4.5 = -128.5 kcal/mol (Option 1)!

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NEET 2025 Chemistry Question 85 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Hess’s Law Heat of Crystallisation Equation

Step-by-Step Rigorous Workflow & Mathematical Proof

Define the Thermochemical Reaction of Crystallisation

Apply Hess’s Law to Express Enthalpy of Reaction

Solve for Standard Heat of Formation of Ba²⁺(aq)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Omitting Heat of Crystallisation (-133 Trap)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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