Chemical Thermodynamics for NEET UG
Chemical thermodynamics tells you how much heat a reaction gives out or takes in, and whether it will happen on its own. The key quantities are ΔH, ΔS and ΔG.
ChemistryClass 1171 MCQs2 lessons13 flashcards3 memory tricks
Practise Chemical Thermodynamics free
What you'll learn
- Enthalpy and Hess's law
Use enthalpy changes and Hess's law to find ΔH of a reaction. - Entropy, Gibbs energy and spontaneity
Use ΔG = ΔH − TΔS to predict whether a reaction is spontaneous.
Key formulas
| From enthalpies of formation | ΔH°_rxn = ΣΔH°f(products) − ΣΔH°f(reactants) |
|---|---|
| From bond enthalpies | ΔH = Σ(bonds broken) − Σ(bonds formed) |
| Link between ΔH and ΔU | ΔH = ΔU + Δn_g RT |
| Gibbs energy change | ΔG = ΔH − TΔS |
| Spontaneous | ΔG < 0 |
| At equilibrium | ΔG = 0 |
| Link to equilibrium constant | ΔG° = −2.303 RT log K |
Memory tricks
Intensive properties don't depend on amount (temperature, density, pressure). Extensive properties do (mass, volume, enthalpy, entropy).
State functions: U, H, S, G, P, V, T. Path functions: heat q and work w.
ΔG = ΔH − TΔS. (−, +): always spontaneous. (+, −): never. (−, −): spontaneous at low T. (+, +): spontaneous at high T.
Try 3 NEET-style questions
Q1. For the reaction N₂(g) + 3H₂(g) → 2NH₃(g) at temperature T, the value of ΔH − ΔU is
- +2RT
- −2RT
- −RT
- zero
Show answer
Answer: (2) −2RT
ΔH − ΔU = Δn_g RT, where Δn_g = moles of gaseous products − moles of gaseous reactants = 2 − 4 = −2. So ΔH − ΔU = −2RT.
Q2. The enthalpy of fusion of ice at 273 K is 6.0 kJ mol⁻¹. The entropy of fusion of ice is about
- 0.022 J K⁻¹ mol⁻¹
- 1638 J K⁻¹ mol⁻¹
- 6.0 J K⁻¹ mol⁻¹
- 22.0 J K⁻¹ mol⁻¹
Show answer
Answer: (4) 22.0 J K⁻¹ mol⁻¹
ΔfusS = ΔfusH / T = 6000 J mol⁻¹ / 273 K ≈ 22.0 J K⁻¹ mol⁻¹.
Q3. For a reaction, ΔH = +30 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. Assuming both are constant, the reaction becomes spontaneous
- at any temperature
- above 3000 K
- above 300 K
- above 0.3 K
Show answer
Answer: (3) above 300 K
At equilibrium ΔG = 0, so T = ΔH/ΔS = 30 000 J mol⁻¹ / 100 J K⁻¹ mol⁻¹ = 300 K. Above this TΔS > ΔH and ΔG < 0.
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