Gibbs Free Energy Change, ∆G Gibbs free energy is a term that combines the effect of enthalpy and entropy into one number The balance between entropy and enthalpy determines the feasibility of a reaction. This is given by the relationship : ∆G = ∆H – T∆S For any spontaneous change, ∆G will be negative. A reaction that has increasing entropy (+ve ∆S) and is exothermic (-ve ∆H ) will make ∆G be negative and will always be feasible ∆G = ∆H – T∆S Units: KJ mol-1 Unit of S= J K-1 mol-1 Need to convert to KJ K-1 mol-1 ( ÷1000) Units: KJ mol-1 Convert from ˚C to K (+ 273) Example : Data for the following reaction, which represents the reduction of aluminium oxide by carbon, are shown in the table. Al2O3 (s) + 3C(s) → 2Al(s) + 3CO(g) Calculate the values of ∆H , ∆S and ∆G for the above reaction at 298 K Substance ∆fH / kJmol–1 ∆S / JK–1mol–1 Al2O3 (s) -1669 51 C(s) 0 6 Al(s) 0 28 CO(g) -111 198 1. Calculate ∆S ∆S˚ = Σ S˚products – Σ S˚reactants = (2 x 28 + 3×198) – (51 + 3 x 6) = +581J K-1 mol-1 (3 S.F.) 2. Calculate ∆H˚ ∆H˚ = Σ∆fH˚ [products] – Σ∆fH˚ [reactants] = (3 x -111) – -1669 = +1336 kJ mol-1 3. Calculate ∆G ∆G = ∆H – T∆S = +1336 – 298x 0.581 = +1163kJ mol-1 ∆G is positive. The reaction is not feasible Calculating the temperature a reaction will become feasible Calculate the temperature range that this reaction will be feasible N2 (g) + O2 (g) 2 NO(g) ∆ H = 180 kJ mol-1 ∆S = 25 J K-1 mol-1 The reaction will be feasible when ∆ G ≤0 Make ∆G = 0 in the following equation ∆G = ∆H – T∆S 0 = ∆H – T∆S So T= ∆H / ∆S T = 180/ (25/1000) = 7200K The T must be >7200K which is a high Temp! ∆G during phase changes As physical phase changes like melting and boiling are equilibria, the ∆G for such changes is zero. What temperature would methane melt at? CH4(s) CH4 (l) ∆H = 0.94 kJmol-1 ∆S = 10.3 Jmol-1K-1 Make ∆G = 0 in the following equation ∆G = ∆H – T∆S 0 = ∆H – T∆S So T= ∆H / ∆S T= 0.94 / (10.3÷1000) T= 91K If ∆G is negative there is still a possibility, however, that the reaction will not occur or will occur so slowly that effectively it doesn’t happen. If the reaction has a high activation energy the reaction will not occur. 6

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3.1.8.2 Gibbs free-energy change, ∆G, and entropy change, ∆S (A-level only)

The balance between entropy and enthalpy determines the feasibility of a reaction given by the relationship: ∆G = ∆H – T∆S (derivation not required).

For a reaction to be feasible, the value of ∆G must be zero or negative.