Q.87 Chemistry ⭐ 360° Option Elimination Active

NEET 2023 Chemistry Question 87 Solution

Topic: Chemical Thermodynamics · Official NTA Booklet Code G5

Question

The equilibrium concentrations of the species in the reaction A + B ⇌ C + D are 2, 3, 10 and 6 mol L⁻¹, respectively at 300 K. ΔG° for the reaction is (R = 2 cal/mol K)

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Verified Against NTA Final Answer Key

Option (4) : -1381.80 cal

📚 Chemical Thermodynamics Medium

📋 Given Parameters & Problem Setup

Reaction A + B ⇌ C + D

Equilibrium Concentrations [A]=2 M, [B]=3 M, [C]=10 M, [D]=6 M

Temperature (T) 300 K

Gas Constant (R) 2 cal / mol K

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The 30-Second Exam Hall Shortcut: ⚡ Kc to ΔG° Direct Multiplication: ΔG° = -2.303 RT log(Kc) = -1381.8 cal

⏱️ Save ~40 seconds

Equilibrium constant Kc = ([C][D])/([A][B]) = (10 × 6)/(2 × 3) = 60/6 = 10. Since log₁₀(10) = 1, ΔG° = -2.303 · R · T · log₁₀(Kc) = -2.303 × 2 × 300 × 1 = -2.303 × 600 = -1381.80 cal in 15 seconds!

1 Compute Kc: (10 × 6) / (2 × 3) = 60 / 6 = 10.

2 Evaluate log₁₀(Kc): log₁₀(10) = 1.

3 Formula: ΔG° = -2.303 × R × T × log₁₀(Kc).

4 Multiply: -2.303 × (2 cal/mol K) × (300 K) × 1 = -1381.80 cal.

5 Match Option (4) directly.

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Why You Can 100% Trust This Shortcut: Fundamental thermodynamic equation connecting Gibbs free energy and equilibrium constant.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculate the Equilibrium Constant (Kc)

For the reversible homogeneous reaction A + B ⇌ C + D, substitute given concentrations:

K_c = \frac{[\text{C}][\text{D}]}{[\text{A}][\text{B}]} = \frac{10 \times 6}{2 \times 3} = \frac{60}{6} = 10

Step 2

State the Relationship Between ΔG° and Kc

The standard Gibbs free energy change of reaction is related to the equilibrium constant by:

\Delta G^\circ = -RT \ln K_c = -2.303 R T \log_{10} K_c

Step 3

Substitute Numerical Values

Given R = 2 cal mol⁻¹ K⁻¹, T = 300 K, and log₁₀(10) = 1:

\Delta G^\circ = -2.303 \times (2\text{ cal mol}^{-1}\text{ K}^{-1}) \times (300\text{ K}) \times 1 = -1381.80\text{ cal}

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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 (1)

✗ Incorrect

Verdict & Scientific Reason:

Decimal factor error (divided by 100).

⚠️

Student Fallacy (Why candidates pick this trap): Candidate mismanaged decimal placement during multiplication.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Sign error: Omitted the negative sign in ΔG° = -RT ln K.

⚠️

Student Fallacy (Why candidates pick this trap): Candidate forgot that spontaneous forward equilibrium (Kc > 1) requires negative ΔG°.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Decimal factor error (divided by 10).

⚠️

Student Fallacy (Why candidates pick this trap): Candidate misplaced a power of ten in the 600 multiplier.

4 (4)

✓ 100% Correct

Verdict & Scientific Reason:

Accurate: ΔG° = -2.303 × 2 × 300 × 1 = -1381.80 cal.

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Why this is airtight: None. Candidate executed the formula and arithmetic flawlessly.

⚠️

NTA Trap Alert: Sign and Gas Constant Units Blunders

Two traps lurk here: (1) Forgetting the negative sign in ΔG° = -RT ln K; when Kc > 1, ΔG° MUST be negative. (2) Habitually plugging R = 8.314 J/mol K instead of using R = 2 cal/mol K as explicitly instructed in the prompt.

🛡️ How to avoid the -1 mark: Always verify units of R from the question statement and check that Kc > 1 ensures ΔG° < 0.

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Core NEET Syllabus Standard

National Eligibility cum Entrance Test (NEET UG) · Chemical Thermodynamics Standard Benchmark

Core Scientific Principle Tested:

"Standard Gibbs free energy change dictates the equilibrium position through the relation ΔG° = -2.303 RT log Kc."

Syllabus Focus: Gibbs Energy Change and Equilibrium Constant

Preserved in Rationalized Curriculum Syllabus & Academic Standards ↗

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NEET 2023 Chemistry Question 87 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Kc to ΔG° Direct Multiplication: ΔG° = -2.303 RT log(Kc) = -1381.8 cal

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculate the Equilibrium Constant (Kc)

State the Relationship Between ΔG° and Kc

Substitute Numerical Values

360° Option Elimination Engine Key Feature

NTA Trap Alert: Sign and Gas Constant Units Blunders

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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Entropy Changes in Isothermal Free Expansion of an Ideal Gas

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