Chemical energetics II: entropy, Gibbs free energy, feasibility of reactions
Chemical Energetics II: Entropy, Gibbs Free Energy & Feasibility of Reactions
1️⃣ Entropy (S)
Entropy is a measure of how spread out the energy of a system is. Think of it like a messy room: the more disordered the room, the higher the entropy.
- 🔹 Disorder: More ways to arrange the same energy → higher entropy.
- 🔹 Temperature dependence: At higher temperatures, molecules move faster → entropy increases.
- 🔹 Units: J mol⁻¹ K⁻¹.
Key formula:
| Symbol | Definition |
|---|---|
| $S$ | Entropy of a system |
| $ΔS$ | Change in entropy |
2️⃣ Gibbs Free Energy (G)
Gibbs free energy tells us whether a reaction will happen spontaneously at constant temperature and pressure.
Formula:
$$G = H - TS$$
- 🔹 $H$ = enthalpy (heat content)
- 🔹 $T$ = absolute temperature (K)
- 🔹 $S$ = entropy
Spontaneity rule:
- 🔹 If $ΔG < 0$, the reaction is spontaneous.
- 🔹 If $ΔG > 0$, the reaction is non‑spontaneous.
- 🔹 If $ΔG = 0$, the system is at equilibrium.
3️⃣ Feasibility of Reactions
To decide if a reaction will occur, calculate $ΔG$ using:
$$ΔG = ΔH - TΔS$$
- 🔹 Exothermic reactions ($ΔH < 0$) often have $ΔG < 0$ at lower temperatures.
- 🔹 Endothermic reactions ($ΔH > 0$) can become spontaneous if $ΔS$ is large and positive.
- 🔹 Temperature can tip the balance: higher $T$ favours reactions with $ΔS > 0$.
🔬 Example: Combustion of Methane
$$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)}$$
Given:
- $ΔH° = -890$ kJ mol⁻¹
- $ΔS° = -200$ J mol⁻¹ K⁻¹
- $T = 298$ K
Calculate $ΔG°$:
$$ΔG° = (-890\,000) - 298(-200) = -890\,000 + 59\,600 = -830\,400\ \text{J mol}^{-1}$$
Since $ΔG° < 0$, combustion is spontaneous at room temperature. 🔥
Exam Tip: When you see $ΔH$ and $ΔS$, always plug them into $ΔG = ΔH - TΔS$. Remember that a negative $ΔG$ means the reaction will proceed on its own. 📚
4️⃣ Quick Reference Table
| Parameter | Symbol | Units | Significance |
|---|---|---|---|
| Enthalpy change | $ΔH$ | kJ mol⁻¹ | Heat released or absorbed |
| Entropy change | $ΔS$ | J mol⁻¹ K⁻¹ | Degree of disorder |
| Gibbs free energy change | $ΔG$ | kJ mol⁻¹ | Spontaneity indicator |
Exam Tip: For each reaction, write down $ΔH$ and $ΔS$ first. Then calculate $ΔG$ at the given temperature. If $ΔG$ is negative, write “spontaneous” in your answer. Remember to convert units if necessary! ??
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