Thermodynamic Terms
System: The part of the universe under study. Surroundings: Everything else outside the system. Boundary: Separates system from surroundings.
Types of Systems
- Open: Exchanges both matter and energy.
- Closed: Exchanges only energy.
- Isolated: Exchanges neither.
State Functions and Path Functions
State functions: Depend only on initial and final states (P, V, T, U, H, S, G). Path functions: Depend on the path taken (q, w).
Internal Energy (U) and Enthalpy (H)
Enthalpy
H = U + PV
Change in Enthalpy
At constant pressure: Delta H = Delta U + P Delta V = q_p
Relation Between Delta H and Delta U
Delta H = Delta U + Delta n_g RT
Where Delta n_g = moles of gaseous products - moles of gaseous reactants.
Enthalpy Changes
Standard Enthalpy of Formation (Delta_f H^theta)
Enthalpy change when 1 mole of compound is formed from its elements in their standard states.
Example: C(s) + O2(g) -> CO2(g), Delta_f H^theta = -393.5 kJ/mol.
Standard Enthalpy of Combustion (Delta_c H^theta)
Enthalpy change when 1 mole of substance is completely burnt in excess oxygen.
Standard Enthalpy of Reaction
Delta_r H^theta = sum Delta_f H^theta (products) - sum Delta_f H^theta (reactants)
Enthalpy of Atomisation
Energy required to break all bonds in 1 mole of substance into gaseous atoms.
Bond Enthalpy
Average energy required to break a specific type of bond. Used to estimate reaction enthalpies:
Delta_r H = sum(text(Bond enthalpies of reactants)) - sum(text(Bond enthalpies of products))
Enthalpy of Solution
Enthalpy change when 1 mole of solute dissolves in a solvent.
Enthalpy of Neutralisation
Enthalpy change when 1 mole of H+ reacts with 1 mole of OH-.
For strong acid-strong base: Delta_neut H = -57.1 kJ/mol.
Hess's Law of Constant Heat Summation
The total enthalpy change for a reaction is independent of the path taken.
DeltaH = DeltaH_1 + DeltaH_2 + DeltaH_3 + ...
Applications
- Enthalpy changes for reactions that cannot be measured directly.
- Using known enthalpies to find unknown ones.
Example: Find enthalpy of formation of CO from:
C + O2 -> CO2, Delta H = -393.5 kJ
CO + 1/2 O2 -> CO2, Delta H = -283 kJ
By Hess's law: Delta_f H(CO) = -393.5 - (-283) = -110.5 kJ/mol.
Entropy (S)
A measure of randomness or disorder.
Second Law of Thermodynamics
The entropy of the universe always increases for spontaneous processes.
Entropy Change
Delta S = q_rev/T
Factors Increasing Entropy
- Increase in temperature.
- Increase in volume.
- Solid to liquid to gas phase changes.
- Increase in number of gaseous molecules.
- Mixing of substances.
Gibbs Free Energy (G)
G = H - TS
Gibbs Free Energy Change
Delta G = Delta H - T Delta S
Spontaneity Criteria
Delta G < 0: Spontaneous process.Delta G > 0: Non-spontaneous.Delta G = 0: Equilibrium.
Effect of Temperature on Spontaneity
| Delta H | Delta S | Delta G | Spontaneity |
|---|---|---|---|
| - | + | Always - | Spontaneous at all T |
| + | - | Always + | Non-spontaneous at all T |
| - | - | Depends on T | Spontaneous at low T |
| + | + | Depends on T | Spontaneous at high T |
Standard Gibbs Free Energy of Formation
Delta_r G^theta = sum Delta_f G^theta (products) - sum Delta_f G^theta (reactants)
Relation with Equilibrium Constant
Delta G^theta = -RT ln K
Delta G = Delta G^theta + RT ln Q (where Q is reaction quotient)
Worked Examples
Example 1: For a reaction, Delta H = +100 kJ and Delta S = +250 J/K at 300 K. Is it spontaneous?
Solution: Delta G = 100 - 300*0.25 = 100 - 75 = +25 kJ. Delta G > 0, so non-spontaneous at 300 K.
At what temperature does it become spontaneous? T > Delta H/Delta S = 100/0.25 = 400 K.
Example 2: Calculate Delta H for 2H2 + O2 -> 2H2O given bond energies: H-H = 436, O=O = 498, O-H = 464 kJ/mol.
Solution: Bonds broken: 2 H-H + 1 O=O = 2*436 + 498 = 1370 kJ.
Bonds formed: 4 O-H = 4*464 = 1856 kJ.
Delta H = 1370 - 1856 = -486 kJ.
Common Mistakes
- Sign convention: In Hess's law, reversing a reaction flips the sign of Delta H.
- Bond enthalpy vs formation enthalpy: Bond enthalpy is always positive (energy required to break bonds).
- Units of entropy: J/K mol (not kJ/K mol). Convert to kJ when using
Delta G = Delta H - T Delta S. - Standard states: Standard state is pure substance at 1 bar pressure.
ISC Exam Focus
- Theory (70%): First law, enthalpy definitions, Hess's law, entropy, Gibbs free energy.
- Application (30%): Numerical problems on Delta H, Delta G, spontaneity, bond energies.
- ISC frequently asks: "Calculate Delta H using bond energies/Hesss law" and "Determine spontaneity using Gibbs free energy."
- Standard enthalpy of formation and combustion problems are common.
Self-Test Questions
Q1: Define enthalpy of formation with an example.
Answer: Enthalpy change when 1 mole of compound forms from elements. Example: C + O2 -> CO2, Delta_f H = -393.5 kJ/mol.
Q2: State Hess's law. Answer: Total enthalpy change is independent of path, depending only on initial and final states.
Q3: Calculate Delta H for C + 1/2 O2 -> CO given:
C + O2 -> CO2, Delta H = -393.5 kJ
CO + 1/2 O2 -> CO2, Delta H = -283 kJ
Answer: Delta H = -393.5 - (-283) = -110.5 kJ/mol.
Q4: Define entropy. In which direction does entropy increase? Answer: Measure of randomness. Entropy increases with temperature, volume, gas formation, melting, evaporation.
Q5: For a reaction, Delta H = -200 kJ and Delta S = -100 J/K. Find Delta G at 500 K and predict spontaneity.
Answer: Delta G = -200 - 500*(-0.1) = -200 + 50 = -150 kJ. Delta G < 0, so spontaneous.
Q6: Derive the relation between Delta G and equilibrium constant.
Answer: Delta G^theta = -RT ln K. At equilibrium, Delta G = 0.
