1. Entropy and Spontaneity Section 15.2 (AHL)

2. Introduction Entropy can be regarded as a measure of the disorder or dispersal of energy in a system It measures the randomness or disorderness Is given the symbol “S” Compare a solid and a gas at the same temperature; the gas has greater entropy because its particles are moving rapidly in all directions

3. Effect of Temperature Change When the temperature is increased, disorder and hence entropy increases The reverse takes place when the temperature is lowered

4. Effect of Change of State The disorder of particles increases from solid to liquid to gas (of the same substances), increasing entropy

5. Effect of a Change of the Number of Particles If the number of particles increases, disorder and hence entropy increases Ex. N 2 O 4(g) → 2NO 2(g) (1 mole of gas to 2 moles, so entropy increases) Conversely, entropy decreases when the number of particles decreases

6. Effect of Mixing of Particles Mixing of particles increases entropy because it increases disorder This is easily seen when one substance is dissolved in another as the particles are free to move around randomly

7. Predicting the Sign of a Change in Entropy If the products are more disordered than the reactants, then the entropy change of the system is positive, + Δ S If the products are less disordered than the reactants, then the entropy change is negative, - Δ S

8. Some Examples Reaction or change Melting Boiling Condensing Freezing Crystallization from a solution Chemical reaction: solid or liquid from gas Entropy change Increase Large increase Large decrease Decrease Large decrease

9. Calculating Entropy Changes The units of entropy and entropy change are J K -1 mol -1 Entropy values are absolute values and can be measured experimentally Standard entropy change = sum of entropies of products – sum of entropies of reactants Δ S θ = Σ S θ [products] – Σ S θ [reactants] Entropy values are in the data booklet

10. Example Problem CH 4(g) + 2O 2(g) → CO 2(g) + 2H 2 O (l) S θ [methane] = 186 J K -1 mol -1 S θ [oxygen] = 205 J K -1 mol -1 S θ [carbon dioxide] = 214 J K -1 mol -1 S θ [water] = 70 J K -1 mol -1 Δ S θ = 214 + (2x 70) – (186 + 2 x 205) Δ S θ = -242 J K -1 mol -1

11. Spontaneity Spontaneous process: has a natural tendency to occur and involves an increase in the entropy of the universe (at the expense of energy available to do useful work). Non spontaneous processes result in a decrease in the entropy of the universe Standard Ambient conditions (SATP): 100 kPa of pressure and 298 K Spontaneous processes may occur very quickly or very slowly

12. Spontaneity Determination Gibbs free energy: the energy available to do work Spontaneity is determined by the sign of the Gibbs free energy change, ΔG θ (AKA free energy) Gibbs equation: ΔG θ = ΔH θ – TΔS θ T = standard temperature in K ΔH θ = enthalpy change

13. More on Spontaneity and Gibbs Free energy ΔS θ = entropy change ΔG θ is measured in kJ mol -1 - ΔG θ means the reaction or process is spontaneous +ΔG θ means the reaction or process is non- spontaneous ΔG θ = 0, the reaction or process is at equilibrium

14. Example Problem #1 Calculate the Gibbs free energy change for the following reaction under standard conditions N 2(g) + 3H 2(g) → 2NH 3(g) ΔH θ = -95.4 kJ ΔS θ = -198.3 J K -1 mol -1 (needs to be in kJ) T = 298 K

15. Problem #1 (Continued) ΔG θ = ΔH θ – TΔS θ ΔG θ = -95.4 – (298) (-0.1983) ΔG θ = -36.3 kJ mol -1 -ΔG θ means the reaction is spontaneous at this temperature Part 2: calculate the temperature at which the reaction ceases to occur spontaneously

16. Part 2 Need to find when ΔG θ = 0 0 =ΔH θ – TΔS θ TΔS θ = ΔH θ T = ΔH θ / ΔS θ T = -95.4 -0.1983 T = 481 K = 208 °C

17. Gibbs Free Energy Change of Formation This is the free energy change that occurs when 1 mol of a compound formed from its elements in their standard states under standard conditions Each compound has a Gibbs free energy change of formation, ΔG θ f Some values are in the data booklet (Table 12)

18. More The ΔG θ f of elements is zero Gibbs free energy change for a reaction = [sum of Gibbs free energy of formation of the products] – [sum of Gibbs free energy of formation of the reactants] ΔG θ = ΣG θ f(products) - ΣG θ f(reactants)

19. Problem # 2 Use the following Gibbs free energy changes of formation to calculate the free energy change for the decomposition of MgCO 3 MgCO 3(s) → MgO (s) + CO 2(g) ΔG θ (MgCO 3(s) ) = -1012 kJ mol -1 ΔG θ (MgO (s) ) = -569 kJ mol -1

20. Problem # 2 Continued ΔG θ (CO 2(g) ) = -394 kJ mol -1 ΔG θ = [-394 + (-569)] - (-1012) ΔG θ = -963 + 1012 ΔG θ = + 49 kJ mol -1 + ΔG θ indicates that the reaction is non- spontaneous under standard conditions

21. Spontaneity and Temperature Positive entropy changes are favorable and negative enthalpy changes are favorable and drive the reaction forward Negative entropy changes are unfavorable and positive enthalpy changes are unfavorable and drive the reaction backward See the following chart to summarize the effect of temperature on spontaneity

22. Think of the equation : ΔG θ = ΔH θ – TΔS θ, as this will guide you to determine the impact of each variable. Remember for a reaction to be spontaneous, : ΔG θ has to be NEGATIVE (-)

23. Problem # 3 Determine ΔG for the following reaction: CaCO 3(s) → CaO (s) + CO 2(g) (at 500K) The required data are: (BE CAREFUL OF UNITS!)

24. First Calculate ΔH ΔH = ΣΔH f(products) - ΣΔH f(reactants) ΔH = [(-636) + (-394)] - (-1207) ΔH = +177 kJ mol -1 = 177000 J mol -1

25. Calculate ΔS ΔS = ΣΔS (products) - ΣΔS (reactants) ΔS = (40 + 214) – 93 ΔS = 161 J K -1 mol -1

26. Calculate ΔG ΔG = ΔH θ – TΔS θ ΔG = 177000 – (500)(161) ΔG = 96500 J mol -1 or 96.5 kJ mol -1