NEET 2023 Q12: ΔG ∘ and K: The Spontaneity Link Trend — Full solution + Why they asked this

PYQ + Trends

1. TITLE

NEET 2023 Q12: ΔG ∘ and K: The Spontaneity Link Trend — Full solution + Why they asked this

2. THE ACTUAL NEET QUESTION

NEET 2023 — Question
The equilibrium concentrations of the species in the reaction $A+B\rightleftharpoons C+D$ are 2, 3, 10 and $6~mol~L^{-1}$, respectively at $300~K$. $\Delta G^{\circ}$ for the reaction is $(R=2~cal/mol~K)$
A) -137.26 cal
B) -1381.80 cal
C) -13.73 cal
D) 1372.60 cal

3. OFFICIAL ANSWER

NEET Answer Key: (B)

4. COMPLETE SOLUTION

• Step-by-step calculation
• First, calculate the equilibrium constant ($K_{eq}$) using the given concentrations:
  $K_{eq} = \frac{[C][D]}{[A][B]} = \frac{10 \times 6}{2 \times 3} = \frac{60}{6} = 10$
• Use the relation between standard Gibbs energy change and the equilibrium constant:
  $\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$

• Substitute the values into the formula:
  $R = 2 \text{ cal mol}^{-1}\text{K}^{-1}$
  $T = 300 \text{ K}$
  $K = 10$
  $\Delta G^{\circ} = -2.303 \times 2 \times 300 \times \log(10)$
  $\Delta G^{\circ} = -2.303 \times 600 \times 1$
  $\Delta G^{\circ} = -1381.8 \text{ cal}$

• Formula used
  $K_{eq} = \frac{[Products]}{[Reactants]}$
  $\Delta G^{\circ} = -2.303 RT \log K$

• Why this approach
  The standard Gibbs energy change ($\Delta G^{\circ}$) is mathematically linked directly to the equilibrium constant ($K$). By calculating $K_{eq}$ from the provided equilibrium concentrations, you can immediately find $\Delta G^{\circ}$ using the standard thermodynamic equation.

5. WHY NEET ASKED THIS

• What concept were they testing?
  The examiner was testing the student's ability to bridge chemical equilibrium with thermodynamics by evaluating $K_{eq}$ and substituting it into the Gibbs free energy equation.
• What's the common mistake they expected students to make?
  Students often forget to multiply by the $2.303$ factor when converting from natural log ($\ln$) to base-10 log ($\log$), or they might mistakenly use $R = 8.314$ (Joules) instead of the provided $R = 2$ (Calories).
• Which option was the "trap"?
  Option D ($1372.60 \text{ cal}$) is a trap for students who forget the negative sign in the formula. Remember, a $K$ value greater than $1$ indicates a spontaneous forward reaction, meaning $\Delta G^{\circ}$ must be negative.

6. COMPARISION TABLE

Prediction for NEET 2026:
Expect a reverse calculation where $\Delta G^{\circ}$ is given, and you must calculate the equilibrium constant $K$ using antilogs. Alternatively, expect a conceptual question linking the sign of $\Delta G^{\circ}$ to $K$ (e.g., confirming that if $\Delta G^{\circ} > 0$, then $K < 1$).

7. SIMILAR QUESTIONS TO PRACTICE

"If you understood this, you can solve:"
- NEET 2020 Q20: Hydrolysis of sucrose with a given $K_c$.
- NEET 2015 Cancelled Q43: Identifying the correct thermodynamic statement for a reversible process in equilibrium ($\Delta G^{\circ}=-2.30 RT \log K$).
- Finding $K_c$ from a given $\Delta G^{\circ}$ using standard state conditions.

8. QUICK REVISION FOR THIS TYPE

For ΔG ∘ and K questions, always:
- First check the units of the universal gas constant $R$ ($8.314 \text{ J K}^{-1}\text{mol}^{-1}$ vs. $2 \text{ cal K}^{-1}\text{mol}^{-1}$).
- Then apply the thermodynamic relationship: $\Delta G^{\circ} = -RT \ln K = -2.303 RT \log K$.
- Watch out for the negative sign! If $K > 1$, $\Delta G^{\circ}$ is negative (products are favored). If $K < 1$, $\Delta G^{\circ}$ is positive (reactants are favored).

9. NCERT BASE

This question tests: NCERT Class 11, Unit 6, Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G.

10. CTA

All PYQs from this chapter → Equilibrium Guide
Practice similar questions → [PYQ Test: 6-Equilibrium]