Chemistry Physical - How to Solve: Thermodynamics & Thermochemistry (ΔH, ΔS, ΔG, Hess’s Law, Spontaneity) – NEET UG Guide

How to Solve: Thermodynamics & Thermochemistry (ΔH, ΔS, ΔG, Hess’s Law, Spontaneity) – NEET UG Guide

Introduction Mastering thermodynamics unlocks 10-12 marks in NEET Chemistry—enough to push you from a 600 to a 650+ score. These concepts explain why ice melts, why batteries work, and how your body converts food into energy. If you can solve ΔG, ΔH, and Hess’s Law problems, you’ll ace every thermochemistry question on exam day.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand:
1. First Law of Thermodynamics – Energy cannot be created or destroyed, only transferred (ΔU = q + w).
2. Enthalpy (H) – Heat content of a system at constant pressure (ΔH = qₚ).
3. Standard States – Pure substances at 1 atm pressure and 25°C (298 K).

If any of these are unclear, review them now—thermodynamics builds on these basics.

KEY TERMS & FORMULAS

1. Enthalpy Change (ΔH)

• Definition: Heat absorbed or released in a reaction at constant pressure.
• Formula: ΔH = ΣΔH_products – ΣΔH_reactants
• ΔH = Enthalpy change (kJ/mol)
• MEMORISE THIS: ΔH is negative for exothermic reactions (heat released), positive for endothermic (heat absorbed).

2. Entropy Change (ΔS)

• Definition: Measure of disorder or randomness in a system.
• Formula: ΔS = ΣS_products – ΣS_reactants
• ΔS = Entropy change (J/mol·K)
• MEMORISE THIS: ΔS is positive when disorder increases (e.g., solid → liquid → gas), negative when disorder decreases.

3. Gibbs Free Energy (ΔG)

• Definition: Determines spontaneity of a reaction.
• Formula: ΔG = ΔH – TΔS
• ΔG = Gibbs free energy (kJ/mol)
• T = Temperature (K)
• MEMORISE THIS:
  • ΔG < 0 → Spontaneous (reaction occurs on its own)
  • ΔG = 0 → Equilibrium
  • ΔG > 0 → Non-spontaneous (requires energy input)

4. Hess’s Law

• Definition: The total enthalpy change for a reaction is the same, regardless of the pathway taken.
• Formula: ΔH_total = ΣΔH_steps
• MEMORISE THIS: If you reverse a reaction, flip the sign of ΔH. If you multiply a reaction by a factor, multiply ΔH by the same factor.

5. Standard Enthalpy of Formation (ΔH°f)

• Definition: Enthalpy change when 1 mole of a compound forms from its elements in their standard states.
• Formula: ΔH°_reaction = ΣΔH°f(products) – ΣΔH°f(reactants)
• Given on exam sheet (but know how to use it).

STEP-BY-STEP METHOD

Step 1: Identify What’s Being Asked

• Is the question about ΔH, ΔS, ΔG, or spontaneity?
• Is it a Hess’s Law problem (multiple reactions)?
• Is it a formation enthalpy problem (ΔH°f)?

Step 2: Write Down All Given Data

• List ΔH, ΔS, T, or ΔG values.
• If Hess’s Law, write all given reactions.
• If ΔH°f, list standard enthalpies of formation for all compounds.

Step 3: Choose the Right Formula

• ΔH problem? Use ΔH = ΣΔH_products – ΣΔH_reactants or Hess’s Law.
• ΔS problem? Use ΔS = ΣS_products – ΣS_reactants.
• ΔG problem? Use ΔG = ΔH – TΔS.
• Spontaneity? Check ΔG sign.

Step 4: Manipulate Reactions (Hess’s Law Only)

• Reverse a reaction? Flip the ΔH sign.
• Multiply/divide a reaction? Multiply/divide ΔH by the same factor.
• Add reactions? Add their ΔH values.

Step 5: Plug in Values & Solve

• Substitute numbers into the formula.
• Check units! (kJ vs. J, °C vs. K).
• Convert temperature to Kelvin if needed (T(K) = T(°C) + 273).

Step 6: Interpret the Answer

• ΔG < 0? Reaction is spontaneous.
• ΔG > 0? Reaction is non-spontaneous.
• ΔH negative? Exothermic (heat released).
• ΔS positive? Disorder increases.

WORKED EXAMPLES

Example 1 – Basic: Calculating ΔH from Formation Enthalpies

Question: Calculate ΔH for the reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) Given: - ΔH°f(CH₄) = -74.8 kJ/mol - ΔH°f(CO₂) = -393.5 kJ/mol - ΔH°f(H₂O) = -285.8 kJ/mol

Solution:
1. Write the formula: ΔH°_reaction = ΣΔH°f(products) – ΣΔH°f(reactants)
2. List ΔH°f values: - Products: CO₂ = -393.5 kJ/mol, H₂O = -285.8 kJ/mol (×2) - Reactants: CH₄ = -74.8 kJ/mol, O₂ = 0 kJ/mol (elements in standard state)
3. Plug in values: ΔH° = [(-393.5) + 2(-285.8)] – [(-74.8) + 2(0)] ΔH° = [-393.5 – 571.6] – [-74.8] ΔH° = [-965.1] – [-74.8] ΔH° = -890.3 kJ/mol
4. Interpretation: The reaction is exothermic (ΔH is negative).

What we did and why: We used standard enthalpies of formation to calculate the overall enthalpy change of the reaction. Since O₂ is an element in its standard state, its ΔH°f = 0.

Example 2 – Medium: Hess’s Law

Question: Given:
1. C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ
2. CO(g) + ½O₂(g) → CO₂(g) ΔH = -283.0 kJ Find ΔH for: C(s) + ½O₂(g) → CO(g)

Solution:
1. Goal: Get the target reaction from the given reactions.
2. Reverse Reaction 2 (to get CO₂ → CO + ½O₂): CO₂(g) → CO(g) + ½O₂(g) ΔH = +283.0 kJ (sign flipped)
3. Add Reaction 1 and the reversed Reaction 2: C(s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ CO₂(g) → CO(g) + ½O₂(g) ΔH = +283.0 kJ Net reaction: C(s) + ½O₂(g) → CO(g)
4. Add ΔH values: ΔH = -393.5 + 283.0 = -110.5 kJ
5. Interpretation: The formation of CO from C and O₂ is exothermic.

What we did and why: We used Hess’s Law to combine reactions and find the unknown ΔH. Reversing a reaction flips the ΔH sign, and adding reactions adds their ΔH values.

Example 3 – Exam-Style: ΔG and Spontaneity

Question: For the reaction: N₂(g) + 3H₂(g) → 2NH₃(g) ΔH = -92.2 kJ/mol, ΔS = -198.7 J/mol·K At what temperature (in °C) will the reaction become non-spontaneous?

Solution:
1. Understand spontaneity: ΔG must be > 0 for non-spontaneity.
2. Use ΔG = ΔH – TΔS, set ΔG = 0 (transition point): 0 = ΔH – TΔS
3. Rearrange to solve for T: T = ΔH / ΔS
4. Convert units: - ΔH = -92.2 kJ/mol = -92,200 J/mol (to match ΔS units) - ΔS = -198.7 J/mol·K
5. Plug in values: T = (-92,200) / (-198.7) = 464.0 K
6. Convert to °C: T(°C) = 464.0 – 273 = 191°C
7. Interpretation: - Below 191°C: ΔG < 0 (spontaneous) - Above 191°C: ΔG > 0 (non-spontaneous)

What we did and why: We found the temperature at which ΔG = 0 (equilibrium). Above this temperature, the reaction becomes non-spontaneous because the entropy term (TΔS) dominates.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 1-minute thermodynamics survival guide for NEET:
1. ΔH = heat change (negative = exothermic, positive = endothermic).
2. ΔS = disorder (positive = more disorder, negative = less).
3. ΔG = ΔH – TΔS (negative = spontaneous, positive = non-spontaneous).
4. Hess’s Law: Flip reactions → flip ΔH, multiply reactions → multiply ΔH.
5. Elements in standard state (O₂, N₂, C) have ΔH°f = 0.
6. Always convert T to Kelvin (T(K) = T(°C) + 273).
7. If ΔG < 0, the reaction happens on its own—no extra energy needed!

Now go crush those thermodynamics questions!