Q.94 Chemistry ⭐ 360° Option Elimination Active

NEET 2020 Chemistry Question 94 Solution

Topic: Thermodynamics · Official NTA Booklet Code E1

Question

Hydrolysis of sucrose is given by the following reaction. Sucrose + H₂O ⇌ Glucose + Fructose. If the equilibrium constant (Kc) is 2×10¹³ at 300 K, the value of ΔrG° at the same temperature will be:

✓

Verified Against NTA Final Answer Key

Option (1) : −8.314 J mol−1K−1×300 K×ln(2×1013)

📚 Thermodynamics Easy

📋 Given Parameters & Problem Setup

Reaction Sucrose + H2O <=> Glucose + Fructose

Equilibrium Constant (Kc) 2 × 10^13

Temperature (T) 300 K

Gas Constant (R) 8.314 J mol^-1 K^-1

⚡

The 30-Second Exam Hall Shortcut: ⚡ "ΔrG° = -RT ln(K): Always Look for the MINUS Sign and ln(Kc)!"

⏱️ 20 seconds

The fundamental thermodynamic relation connecting standard Gibbs free energy change and the equilibrium constant is $\Delta_r G^\circ = -RT \ln K_c$ . With $R = 8.314\text{ J mol}^{-1}\text{K}^{-1}$ , $T = 300\text{ K}$ , and $K_c = 2 \times 10^{13}$ , the expression is $-8.314 \times 300 \times \ln(2 \times 10^{13})$ . Mark Option (1) in 5 seconds!

1 Standard equation: ΔrG° = -RT ln(Kc).

2 Substitute values: R = 8.314 J mol^-1 K^-1, T = 300 K, Kc = 2 × 10^13.

3 Result: ΔrG° = -8.314 J mol^-1 K^-1 × 300 K × ln(2 × 10^13).

4 Option (2) misses the minus sign; Options (3) and (4) have wrong constants.

5 Select Option (1).

🛡️

Why You Can 100% Trust This Shortcut: NCERT Chemistry · Thermodynamics (Gibbs Energy Change and Equilibrium)

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

State the Thermodynamic Equilibrium Equation

At chemical equilibrium, the actual Gibbs free energy change $\Delta_r G = 0$ . Therefore, the standard Gibbs free energy change relates to the equilibrium constant by: $\Delta_r G = \Delta_r G^\circ + RT \ln K_c = 0$ $\Delta_r G^\circ = -RT \ln K_c$

Step 2

Insert Given Experimental Values

Given: - $R = 8.314\text{ J mol}^{-1}\text{K}^{-1}$ - $T = 300\text{ K}$ - $K_c = 2 \times 10^{13}$ Substituting directly into the equation: $\Delta_r G^\circ = -8.314\text{ J mol}^{-1}\text{K}^{-1} \times 300\text{ K} \times \ln(2 \times 10^{13})$

Step 3

Verify Physical Meaning of the Sign

Because $K_c = 2 \times 10^{13} \gg 1$ , $\ln K_c > 0$ . Hence, $\Delta_r G^\circ$ must be a large negative value, indicating that the hydrolysis of sucrose to glucose and fructose is strongly spontaneous and exergonic under standard conditions.

Step 4

Eliminate Erroneous Options

Option (2) has a positive sign (violates the negative sign in $-RT \ln K$ ); Options (3) and (4) have incorrect equilibrium constant arguments ( $3 \times 10^{13}$ and $4 \times 10^{13}$ ).

⭐

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:

Factually and thermodynamically exact; perfectly matches $\Delta_r G^\circ = -RT \ln K_c$ .

💡

Why this is airtight: None; formulaic substitution.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; omits the mandatory negative sign ( $-$ ) in the thermodynamic equation.

⚠️

Student Fallacy (Why candidates pick this trap): Forgetting the negative sign in

\Delta_r G^\circ = -RT \ln K

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; uses an incorrect value for $K_c$ ( $3 \times 10^{13}$ instead of $2 \times 10^{13}$ ) and misses the negative sign.

⚠️

Student Fallacy (Why candidates pick this trap): Misreading the numerical constant.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; uses $4 \times 10^{13}$ instead of $2 \times 10^{13}$ .

⚠️

Student Fallacy (Why candidates pick this trap): Erroneously squaring the equilibrium constant argument.

⚠️

NTA Trap Alert: Sign Trap in ΔG° = -RT ln K

Always watch for the negative sign! If $K > 1$ , the reaction is spontaneous and $\Delta G^\circ$ MUST BE NEGATIVE.

🛡️ How to avoid the -1 mark: Remember: Spontaneous =

K > 1

\Delta G^\circ < 0

(Negative sign is compulsory!).

📖

Core NEET Syllabus Standard

Chemistry · Thermodynamics Standard Benchmark

Core Scientific Principle Tested:

Syllabus Focus: Relationship Between Standard Gibbs Free Energy Change and Equilibrium Constant

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2020 Chemistry Question 94 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ "ΔrG° = -RT ln(K): Always Look for the MINUS Sign and ln(Kc)!"

Step-by-Step Rigorous Workflow & Mathematical Proof

State the Thermodynamic Equilibrium Equation

Insert Given Experimental Values

Verify Physical Meaning of the Sign

Eliminate Erroneous Options

360° Option Elimination Engine Key Feature

NTA Trap Alert: Sign Trap in ΔG° = -RT ln K

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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