Q.74 Chemistry ⭐ 360° Option Elimination Active

NEET 2026 Chemistry Question 74 Solution

Topic: Thermodynamics & Biomolecules · Official NTA Booklet Code 50

Question

A protein undergoes reversible thermal denaturation from its initial state N to denatured state D according to N ⇌ D. At 60 °C, the concentrations of both N and D are equal at equilibrium, and the standard enthalpy change of denaturation is 666 kJ mol⁻¹. The standard entropy change (ΔS° in kJ K⁻¹ mol⁻¹) of the protein upon denaturation at 60 °C is closest to

✓

Verified Against NTA Final Answer Key

Option (1) : 2.0

📚 Thermodynamics & Biomolecules Medium

📋 Given Parameters & Problem Setup

Equilibrium Process N ⇌ D (Native state to Denatured state)

Transition Temperature (T) 60 °C = 60 + 273.15 = 333.15 K ≈ 333 K

Equilibrium Condition [N] = [D] at 60 °C (K_eq = [D]/[N] = 1.0)

Enthalpy Change (ΔH°) 666 kJ mol⁻¹

Target Standard entropy change (ΔS° in kJ K⁻¹ mol⁻¹)

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The 30-Second Exam Hall Shortcut: ⚡ Equilibrium Denaturation Shortcut

⏱️ Save ~60 seconds

When [N] = [D], K = 1, so ΔG° = 0! Therefore ΔS° = ΔH° / T!

1 Equilibrium with [N] = [D] means equilibrium constant K_eq = 1.

2 Standard Gibbs free energy: ΔG° = -RT ln(K_eq) = -RT ln(1) = 0.

3 Since ΔG° = ΔH° - TΔS° = 0 ⇒ ΔH° = TΔS°.

4 Solve for ΔS°: ΔS° = ΔH° / T = 666 kJ mol⁻¹ / 333 K = 2.0 kJ K⁻¹ mol⁻¹!

5 Notice unit requested: kJ K⁻¹ mol⁻¹ (not J K⁻¹ mol⁻¹)!

6 Option (1) has 2.0! Done in 15 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Thermodynamic definition of reversible phase transition/conformational denaturation: at the melting temperature T_m where halves are equal, ΔG° = 0 and ΔS° = ΔH° / T_m.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Evaluate the Equilibrium Constant K

At the midpoint of thermal denaturation (T = 60 °C), the concentrations of native protein (N) and denatured protein (D) are equal: [N] = [D].

K = \frac{[\text{D}]}{[\text{N}]} = \frac{[\text{N}]}{[\text{N}]} = 1.0

• Equilibrium constant K = 1.0.

Step 2

Relate Equilibrium Constant to Gibbs Free Energy

The standard Gibbs free energy change of the reaction is related to the equilibrium constant by: ΔG° = -RT ln K

\Delta G^\circ = -RT \ln(1.0) = 0

• At the transition temperature, the native and denatured states are in thermodynamic equilibrium (ΔG° = 0).

Step 3

Calculate ΔS° in Required Units (kJ K⁻¹ mol⁻¹)

From the Gibbs-Helmholtz relation: ΔG° = ΔH° - T ΔS° = 0 ⇒ ΔS° = ΔH° / T Convert temperature to Kelvin: T = 60 + 273 = 333 K. Substitute ΔH° = 666 kJ mol⁻¹ and T = 333 K:

\Delta S^\circ = \frac{\Delta H^\circ}{T} = \frac{666\text{ kJ mol}^{-1}}{333\text{ K}} = 2.0\text{ kJ K}^{-1}\text{ mol}^{-1}

• ΔS° = 2.0 kJ K⁻¹ mol⁻¹.
• Matches Option (1).

⭐

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:

Exact thermodynamic calculation in kJ K⁻¹ mol⁻¹.

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Why this is airtight: None; flawless execution.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Unit confusion trap! 2000.0 is the value in J K⁻¹ mol⁻¹, but the question explicitly demanded kJ K⁻¹ mol⁻¹.

⚠️

Student Fallacy (Why candidates pick this trap): Multiplying by 1000 without reading the required unit in the prompt.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Simply reporting the temperature in Kelvin.

⚠️

Student Fallacy (Why candidates pick this trap): Mistaking the temperature value for entropy.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Dividing 666 by 60 °C (using Celsius instead of Kelvin: 666 / 60 = 11.1).

⚠️

Student Fallacy (Why candidates pick this trap): Classic error of substituting temperature in Celsius instead of Kelvin.

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NTA Trap Alert: Using Celsius (60 °C) or Overlooking the "kJ" in Units

Two massive traps in one question: (1) Dividing by 60 gives 11.1 (Option 4) — thermodynamics REQUIRES absolute temperature in Kelvin (333 K)! (2) Answering 2000 (Option 2) because entropy is usually in J, ignoring that the prompt explicitly asks for kJ K⁻¹ mol⁻¹.

🛡️ How to avoid the -1 mark: Always convert °C to K (60 + 273 = 333 K), and check the required unit in the parentheses (kJ K⁻¹ mol⁻¹).

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Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Thermodynamics & Biomolecules Standard Benchmark

Core Scientific Principle Tested:

"At phase/conformational equilibrium where K=1, ΔG° = 0 and ΔS° = ΔH°/T."

Syllabus Focus: Gibbs energy change and equilibrium constant; protein denaturation

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

666 divided by 333 is 2; in kJ units, Option 1 is true!

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NEET 2026 Chemistry Question 74 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Equilibrium Denaturation Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Evaluate the Equilibrium Constant K

Relate Equilibrium Constant to Gibbs Free Energy

Calculate ΔS° in Required Units (kJ K⁻¹ mol⁻¹)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Using Celsius (60 °C) or Overlooking the "kJ" in Units

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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