## Entropy

A SPONTANEOUS PROCESS (e.g. diffusion) will proceed on its own without any external influence. Entropy, S˚ Entropy is a description of the number of ways atoms can share quanta of energy. If number of ways of arranging the energy (W) is high, then system is disordered and entropy (S) is high. Substances with more ways of arranging their atoms and energy (more disordered) have a higher entropy. Elements …tend to have lower entropies than… Compounds Simpler compounds Complex compounds Pure substances Mixtures solid Liquid gas Temperature Entropy Solids have lower entropies than liquids which are lower than gases. When a solid increases in Temperature its entropy increases as the particles vibrate more. There is a bigger jump in entropy with boiling than that with melting. Gases have large entropies as they are much more disordered Predicting Change in entropy ‘∆S’ Qualitatively Balanced chemical equations can often be used to predict if ∆S˚ is positive or negative. In general, a significant increase in the entropy will occur if: -there is a change of state from solid or liquid to gas – there is a significant increase in number of molecules between products and reactants. NH4Cl (s) HCl (g) + NH3 (g) ∆S˚ = +ve •change from solid reactant to gaseous products •increase in number of molecules both will increase disorder Na s + ½ Cl2 g NaCl s ∆S˚ = -ve •change from gaseous and solid reactant to solid •decrease in number of molecules both will decrease disorder An increase in disorder and entropy will lead to a positive entropy change ∆S˚ = +ve Calculating ∆S˚ quantitatively Data books lists standard entropies (S˚) per mole for a variety of substances. It is not possible for a substance to have a standard entropy of less than zero. Elements in their standard states do not have zero entropy. Only perfect crystals at absolute zero (T = 0 K) will have zero ∆S entropy: ˚ = Σ S˚products – ΣS˚reactants The unit of entropy is J K-1 mol-1 At 0K substances have zero entropy. There is no disorder as particles are stationary reactants Activation Energy: EA ∆H Progress of Reaction Energy 4 products melting boiling N Goalby chemrevise.org Example Calculate ∆S˚ for the following reaction at 25˚C: 2Fe2O3 (s) + 3C (s) 4Fe (s) + 3CO2 (g) ∆S˚ = S˚products – S˚reactants = (3 x 213.6 + 4 x 27.3) – (2 x 87.4 + 3 x 5.7) = + 558.1 J K-1 mol-1 = + 558 J K-1 mol-1 (3 S.F.) S [Fe2O3 ] = 87.4 J K-1 mol-1 S [C] = 5.7 J K-1 mol-1 S [Fe] = 27.3 J K-1 mol-1 S [CO2 ] = 213.6 J K-1 mol-1 Note: the entropy change is very positive as a large amount of gas is being created increasing disorder. A problem with ∆H A reaction that is exothermic will result in products that are more thermodynamically stable than the reactants. This is a driving force behind many reactions and causes them to be spontaneous (occur without any external influence). Some spontaneous reactions, however, are endothermic. How can this be explained? We need to consider something called entropy

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5.2.2 Enthalpy and entropy

Entropy (a) explanation that entropy is a measure of the dispersal of energy in a system which is greater, the more disordered a system HSW1 The model of entropy to explain thermodynamic stability. (b) explanation of the difference in magnitude of the entropy of a system: (i) of solids, liquids and gases (ii) for a reaction in which there is a change in the number of gaseous molecules (c) calculation of the entropy change of a system, ∆S, and related quantities for a reaction given the entropies of the reactants and products M2.2, M2.3, M2.4

## Gibbs’ free energy

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 ∆f H / 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∆ f H˚ [products] – Σ∆ f H˚ [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 Effect of Temperature on feasibility Changing Temperature will change the value of – T∆S in the above equation ∆G = ∆H – T∆S If the reaction involves an increase in entropy (∆S is +ve) then increasing Temperature will make it more likely that ∆G is negative and more likely that the reaction occurs e.g. NaCl + aq Na+ (aq) + Cl- (aq) If the reaction involves an decrease in entropy (∆S is – ve) then increasing Temperature will make it more less likely that ∆G is negative and less likely for the reaction to occur. E.g. HCl(g) + NH3 (g) ➝ NH4Cl(s) If the reaction has a ∆S close to zero then temperature will not have a large effect on the feasibility of the reaction as – T∆S will be small and ∆G won’t change much e.g. N2 (g) + O2 (g) 2NO (g) This graph shows how the free-energy change for formation of ammonia varies with temperature above 240 K. ½ N2 (g) + 3 /2 H2 (g) NH3 (g) Applying the equation of a straight line y= mx+c to the ∆G = ∆H – T∆S equation. c = ∆H The gradient of this graph is equal to -∆S The positive gradient means ∆S is negative which corresponds to the equation above showing increasing order. When ∆G <0 then the reaction is spontaneous. In this case at Temperatures below around 460K The slope of the line would change below 240K because ammonia would be a liquid and the entropy change would be different. Enthalpies of solution Using Born Haber cycles to determine enthalpy changes of solution MgCl2 (s) ∆LEH (MgCl2 ) + 2Cl- Mg (g) 2+ (g) + 2Cl- Mg (aq) 2+ (aq) ∆ hyd H Mg2+ ΔsolH When an ionic substance dissolves the lattice must be broken up. The enthalpy of lattice dissociation is equal to the energy needed to break up the lattice (to gaseous ions). This step is endothermic. The size of the lattice enthalpy depends on the size and charge on the ion. The smaller the ion and the higher its charge the stronger the lattice ΔsolH = – ∆ LEH + Σ∆ hyd H When an ionic lattice dissolves in water it involves breaking up the bonds in the lattice and forming new bonds between the metal ions and water molecules. For MgCl2 the ionic equation for the dissolving is MgCl2 (s) + aq Mg2+ (aq) + 2Cl- (aq) N Goalby chemrevise.org 6 -30 -20 -10 0 10 20 30 40 0 200 400 600 800 1000 ΔG kJ/ mol Temperature/ K 2 x ∆ hyd H ClMg2+ (aq) + 2Cl- (g) Example . Calculate the enthalpy of solution of NaCl given that the lattice enthalpy of formation of NaCl is -771 kJmol-1 and the enthalpies of hydration of sodium and chloride ions are -406 and -364 kJmol-1 respectively ΔsolH = – ∆LEH + Σ∆hydH = – (-771) + (-406-364) = + 1 kJmol-1 What does ΔsolH tell us? Generally ΔsolH is not very exo or endothermic so the hydration enthalpy is about the same as lattice enthalpy. In general the substance is more likely to be soluble if the ΔsolH is exothermic. If a substance is insoluble it is often because the lattice enthalpy is much larger than the hydration enthalpy and it is not energetically favourable to break up the lattice, making ΔsolH endothermic. BaSO4 (s) ∆LEH (BaSO4 ) + SO4 2- Ba (g) 2+ (g) + SO4 2- Ba (aq) 2+ (aq) ∆hydH Ba2+ ΔsolH INSOLUBLE ΔH solution endothermic. We must consider entropy, however, to give us the full picture about solubility. When a solid dissolves into ions the entropy increases as there is more disorder as solid changes to solution and number of particles increases. This positive ∆S can make ∆G negative even if ∆H solution is endothermic, especially at higher temperatures. Hydration enthalpies are exothermic as energy is given out as water molecules bond to the metal ions. The negative ions are attracted to the δ+ hydrogens on the polar water molecules and the positive ions are attracted to the δ – oxygen on the polar water molecules. The higher the charge density the greater the hydration enthalpy (e.g. smaller ions or ions with larger charges) as the ions attract the water molecules more strongly. e.g. Fluoride ions have more negative hydration enthalpies than chloride ions Magnesium ions have a more negative hydration enthalpy than barium ions For salts where ΔH solution is exothermic the salt will always dissolve at all Temperatures ∆S is positive due to the increased disorder as more particles so – T∆S always negative ∆G = ∆H – T∆S ∆H is negative ∆G is always negative For salts where ΔH solution is endothermic the salt may dissolve depending on whether the -T∆S value is more negative than ∆H is positive ∆S is positive due to the increased disorder as more particles so – T∆S always negative ∆G = ∆H – T∆S ∆H is positive Will dissolve if ∆G is negative Increasing the Temperature will make it more likely that ∆G will become negative, making the reaction feasible and the salt dissolve

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5.2.2 Enthalpy and entropy

Free energy (d) explanation that the feasibility of a process depends upon the entropy change and temperature in the system, T∆S, and the enthalpy change of the system, ∆H HSW1,5,6 Use of entropy, enthalpy and temperature for predicting feasibility. (e) explanation, and related calculations, of the free energy change, ∆G, as: ∆G = ∆H – T∆S (the Gibbs’ equation) and that a process is feasible when ∆G has a negative value M0.0, M2.2, M2.3, M2.4 HSW5 Link between ∆G and feasibility. (f) the limitations of predictions made by ∆G about feasibility, in terms of kinetics. M0.3 HSW6 The relative effects of entropy and rate in determining feasibility of processes.

Credits: Neil Goalby