Entropy Entropy, S, is the quantitative measure of the degree of disorder in a system. Entropy is ‘a measure of disorder’ or ‘the amount of randomness’ in a system. All systems possess some degree of randomness because particles are always in constant motion. There is a general tendency towards greater entropy – nature tends to move from order to disorder in isolated systems. For example, gas molecules spread out over time to fill a space, increasing their entropy. Energy here is changing, from being concentrated to being more spread out: particles are becoming more disordered and so entropy increases. Gas molecules spread out Entropy increases This graph shows how entropy gradually changes with temperature. At 0K, perfect crystals have 0 entropy. The largest jumps in entropy values occur at the melting and boiling points. The greatest change occurs at boiling point because a gas is significantly more disordered than a solid or liquid.

Entropy increases when a solid dissolves: CuSO 4.5H 2 O is a solid lattice. When it dissolves in water: CuSO 4.5H 2 O  Cu 2+ (aq) + SO 4 2- (aq) + 5H 2 O (l) The copper and sulphate ions are spreading out – they are becoming more disordered so entropy increases. Entropy increases when a solid dissolves: CuSO 4.5H 2 O is a solid lattice. When it dissolves in water: CuSO 4.5H 2 O  Cu 2+ (aq) + SO 4 2- (aq) + 5H 2 O (l) The copper and sulphate ions are spreading out – they are becoming more disordered so entropy increases. Entropy increases when there is an increase in the number of gas moles: Mg (s) + 2HCl (aq)  MgCl 2 (aq) + H 2 (g) In a reaction where a gas evolved, entropy increases. Entropy increases when there is an increase in the number of gas moles: Mg (s) + 2HCl (aq)  MgCl 2 (aq) + H 2 (g) In a reaction where a gas evolved, entropy increases. ΔS = ΣS θ (products) - ΣS θ (reactants) The standard entropy, S θ of a substance is the entropy content of one mole of the substance under standard conditions. It has units J -1 K -1 mol -1. The entropy change of a reaction can be calculated using the standard entropies of the products and reactants: If a change makes a system more random, ΔS is positive. If a change makes a system more ordered, ΔS is negative. N 2 (g) + 3H 2 (g)  2NH 3 (g) N 2 (g)H 2 (g)NH 3 (g) SθSθ Use the data in the table to calculate ΔS for this reaction. ΔS = ΣS θ (products) - ΣS θ (reactants) ΔS = (2x193) – [(192) + (3 x 131)] ΔS = -199 JK -1 mol -1 We would expect the entropy value to be negative because more ordered system with fewer gaseous moles is produced.

Gibbs Free Energy A spontaneous process proceeds on its own. Spontaneous processes lead to lower energy and increased stability. Exothermic reactions generally are spontaneous at room temperature. The enthalpy content decreases and excess energy is released to the surroundings. This increases stability. Some endothermic reactions can take place spontaneously at room temperature. The enthalpy content of the system increases: this must also increase stability. The reason that such a reaction can take place is entropy. A process is spontaneous if a chemical system becomes more stable and its overall energy decreases. The overall energy decrease results from contributions form entropy and enthalpy. The entropy contribution to the overall energy depends on temperature. Energy derived from entropy = TΔS so as temperature increases, the energy derived from entropy becomes more significant. ΔG = ΔH - T ΔS The feasibility of a reaction depends on the balance between enthalpy and entropy. Spontaneous processes happen when ΔG is negative. The free energy change ΔG is the balance between enthalpy, entropy and temperature for a process:

Use the data in the table to: (a)determine the feasibility of the reaction at 25° temperature. (b)determine the minimum temperature for the reaction to take place (c)explain why the reaction becomes feasible at high temperatures. ZnCO 3 (s) ZnO (s) CO 2 (g) ΔH θ f kJmol S θ JK -1 mol ZnCO 3 (s)  ZnO (s) + CO 2 (g) (a) ΔH θ = ΣΔH θ f (products) - ΣΔH θ f (reactants) = [(-83) + (-393.5)] – (-547.5) = +71 kJmol -1 ΔS = ΣS θ (products) - ΣS θ (reactants) ΔS = (44+214) – (82) ΔS = +176 JK -1 mol -1 ΔG = ΔH - T ΔS ΔH = +71 kJmol -1 ΔS = +176 JK -1 mol -1 = kJK -1 mol -1 T= = 298K ΔG = ΔH - T ΔS ΔG = 71 – (298 x 0.176) ΔG = kJmol -1 ΔG is positive therefore the reaction is not feasible. (c) As temperature increases, TΔS gets bigger. Eventually, when TΔS = H, ΔG will be 0 and the reaction will then be feasible.

Changing temperature and ΔG ΔG = ΔH - T ΔS ΔHΔHΔSΔS Feasibility? -ve+veAlways -veAlways feasible In these circumstances, ΔG will always be negative so the reaction will always be feasible. ΔHΔHΔSΔSΔG = ΔH - T ΔSFeasibility? +ve-veAlways +veNever feasible In these circumstances, ΔG will always be positive; two negatives give a positive, so ΔG will never be less than 0, hence the reaction will never be feasible. ΔHΔHΔSΔSΔG = ΔH - T ΔSFeasibility? -ve -ve at low temperature Feasible at low temperature If ΔS is negative, T ΔS becomes a positive term in the equation. Provided that the T ΔS term is smaller than the value of ΔH, the reaction will be feasible because ΔG will be less than 0. For example, if ΔH is -110, and ΔS is -20, at a temperature of 5K, the TΔS value will be Putting this into the expression for ΔG : -110 – (-100) = -10. The reaction is feasible at this low temperature. ΔHΔHΔSΔSΔG = ΔH - T ΔSFeasibility? +ve -ve at high temperature Feasible at high temperature For the reaction to be feasible, ΔG must be negative. For this to be so, if ΔS is positive, the TΔS term in the equation must be larger than the ΔH value. This will be true at high temperatures. Hence, reactions with positive ΔH and ΔS values are only feasible at high temperatures.