NEET asked ΔH vs ΔU Calculation in 2023, 2024 — Expect it in 2025

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1. TITLE

NEET asked ΔH vs ΔU Calculation in 2023, 2024 — Expect it in 2025

2. THE ACTUAL NEET QUESTION

NEET 2024 Re — Question
For the following reaction at 300 K
$A_2(g) + 3B_2(g) \rightarrow 2AB_3(g)$
the enthalpy change is +15 kJ, then the internal energy change is :
A) 19988.4 J
B) 200 J
C) 1999 J
D) 1.9988 kJ

3. OFFICIAL ANSWER

NEET Answer Key: (A)

4. COMPLETE SOLUTION

For Physical Chemistry:
- Step-by-step calculation:
1. Identify the reaction and states: $A_2(g) + 3B_2(g) \rightarrow 2AB_3(g)$.
2. Calculate $\Delta n_g$ (change in gaseous moles): $\Delta n_g = n_{g(products)} - n_{g(reactants)}$.
$\Delta n_g = 2 - (1 + 3) = -2$.
3. Ensure units match. Since $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ and the options are mostly in Joules, convert $\Delta H$ to Joules: $+15 \text{ kJ} = 15000 \text{ J}$.
4. Substitute into the equation: $15000 = \Delta U + (-2) \times 8.314 \times 300$.
5. Solve for $\Delta U$: $15000 = \Delta U - 4988.4$.
6. $\Delta U = 15000 + 4988.4 = 19988.4 \text{ J}$.
- Formula used: $\Delta H = \Delta U + \Delta n_g RT$.
- Why this approach: The question specifically provides the enthalpy change ($\Delta H$) and asks for internal energy change ($\Delta U$) at a constant temperature. The ideal gas relation equating the two via pressure-volume work ($p\Delta V = \Delta n_g RT$) is the only direct path.

5. WHY NEET ASKED THIS

Explain the examiner's intent:
- What concept were they testing? They were testing your understanding of the relationship between enthalpy and internal energy, specifically your ability to accurately calculate $\Delta n_g$ (the difference in moles of only gaseous substances).
- What's the common mistake they expected students to make? The examiner expected students to forget to convert the enthalpy value from kilojoules (kJ) to Joules (J) before subtracting the $RT$ term (which is inherently in Joules because of $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$).
- Which option was the "trap"? Option D (1.9988 kJ) and C (1999 J) act as traps for students who mess up the decimal places or forget unit conversions during calculation.

6.  TREND ANALYSIS

PATTERN ALERT: 

Prediction for NEET 2025:
Expect a conceptual variation where you are given a combustion reaction containing liquids and solids (e.g., combustion of benzene or benzoic acid) and asked to calculate either the sign or the exact value of $\Delta H - \Delta U$. The trap will be counting the moles of liquids/solids into the $\Delta n_g$ calculation.

7. SIMILAR QUESTIONS TO PRACTICE

"If you understood this, you can solve:"
- NEET 2023 Q9: Identify the correct relation between $\Delta H$ and $\Delta U$.
- NEET 2003 Q67: Find $\Delta H - \Delta E$ for the reaction $C_3H_8 + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$.
- NEET 2000 Q84: Relation for $C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$.

8. QUICK REVISION FOR THIS TYPE

For $\Delta H$ vs $\Delta U$ calculation questions, always:
- [First check] the physical states of all reactants and products in the balanced equation.
- [Then apply] the formula $\Delta n_g = n_{products} - n_{reactants}$ strictly counting only the gaseous ($g$) moles. Ignore solids ($s$) and liquids ($l$).
- [Watch out for] unit discrepancies! Enthalpy ($\Delta H$) is usually given in kJ, while the gas constant $R$ ($8.314$) is in Joules. Always multiply your $\Delta H$ by 1000 before plugging it into $\Delta H = \Delta U + \Delta n_g RT$.

9. NCERT BASE

This question tests: NCERT Class 11, Part 1, Chapter 5 (Thermodynamics), Page 143-144.

10. CTA

All PYQs from this chapter → Chemical Thermodynamics Guide
Practice similar questions → Mock exams