1. Entropy and Free Energy

2. Why do reactions take place? Feasible reactions take place spontaneously, although the rate may be slow. Generally the more negative ΔH the more likely the reaction. So why are endothermic reactions possible?

3. Randomness When a gas evaporates it spreads out. This increases its degree of randomness.

4. The same is true on mixing gases; +

5. And dissolving solids; +

6. Entropy Endothermic reactions are feasible if they increase disorder. IE Product particles are more randomly arranged than reactant particles. Randomness is expressed mathematically as entropy (S). A feasible endothermic reaction will have a positive entropy change (ΔS).

7. All entropies are +ve. Entropies of elements in their standard states are not zero. Entropies increase with temperature as the particles spread out. So entropies are quoted at 298K and 101kPa. NB Entropies of solids < liquids < gases SubstanceS (JK -1 mol -1 ) Iron27 Iron Oxide88 Calcium carbonate 93 Ice48 Water70 Steam189 Carbon dioxide 214

8. Calculating entropy changes 1) Add the entropies of the products. 2) Add the entropies of the reactants. 3) Subtract the entropies of the reactants from that of the products. 4) Then if ΔS is positive the reaction will be feasible.

9. Eg; Calcium carbonate decomposes when heated to form calcium oxide. CaCO 3 → CaO + CO 2 ΔH = +178 kjmol -1

10. CaCO 3 → CaO + CO 2 Entropies of products; CaO = 40 CO 2 = 214 40 + 214 = 254 Entropy of reactants; CaCO 3 = 93 Entropy change ΔS = 254 - 93 = +161 JK -1 mol - 1

11. Gibbs Free Energy The feasibility of a reaction is determined by; 1) ΔH 2) ΔS These two factors are combined to calculate Gibbs Free Energy (G). ΔG = ΔH – TΔS If ΔG is negative a reaction is feasible.

12. Influence of temperature on reactions ΔG is temperature dependent. This means that reactions can become feasible as temperature is raised. Eg; CaCO 3 → CaO + CO 2 At 298K ΔH = 178 kjmol -1 ΔS = 161 jK -1 mol -1 = 0.161 kjK -1 mol -1 T ΔS = 298 x 0.161 = 47.98 kjmol -1 ΔG = 178 – 47.98 = 130 kjmol -1 So the reaction is not feasible.

13. Instead the reverse reaction occurs; CaO + CO 2 → CaCO 3 ΔG = -130 kjmol -1 But at 1500K ΔH = 178 kjmol -1 ΔS = 161 jK -1 mol -1 = 0.161 kjK -1 mol -1 T ΔS = 1500 x 0.161 = 241.5 kjmol -1 ΔG = 178 – 241.5 = -63.5 kjmol -1 The reaction now has a negative ΔG so it has become feasible.

14. Zero values of ΔG When ΔG is 0 the reaction is just feasible. The temperature at which this occurs can be calculated. Eg; CaCO 3 → CaO + CO 2 ΔG = ΔH – TΔS 0 = 178 – T0.161 T = 178 / 0.161 = 1106K

15. Kinetic factors Neither ΔH nor ΔS gives any indication of the rate of a reaction. Some reactions are predicted to be feasible on the basis of ΔH and ΔS but in practice are so slow that they unfeasible. This is because of a kinetic barrier. IE There is a high activation energy.

16. reactants products energyenergy exergonic reaction Reaction profile a activation energy, E a transition state (or activated complex) bonds breaking bonds forming Course of reaction ReplayReplay Close windowClose window

17. Eg; Stability of graphite C + O 2 → CO 2 ΔH = -394 kjmol -1 S CO 2 = 213.6 jK -1 mol -1 S O 2 = 205 jK -1 mol -1 S C = 5.7 jK -1 mol -1 ΔS = +2.91 jK -1 mol -1 ΔG = -394 – (298 x 0.00291) = -394.86 kjmol -1 So the reaction is feasible at 298K. But in practise it is too slow.