7 Enthalpy change of solution and solubility of salt

Enthalpies of solution Using Hess’s law to determine enthalpy changes of solution MgCl2 (s) H lattice dissociation (MgCl2 ) + 2Cl- Mg (g) 2+ (g) + 2Cl- Mg (aq) 2+ (aq)  hyd H Mg2+ + 2 x  hyd H Cl- ΔHsolution In general Hsolution =  HL dissociation +  hydH 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 Sometimes in questions  HLatt formation is given instead of  HLatt dissociation in order to catch you out. Remember the difference between the two. OR H solution = –  HL formation +  hyd Hhyd 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 7 -30 -20 -10 0 10 20 30 40 0 200 400 600 800 1000 ΔG kJ/ mol Temperature/ K 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  sol H = –  HLatt formation +  hyd H = – (-771) + (-406-364) = + 1 kJmol-1 What does ΔHSolution tell us? Generally ΔH solution 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 ΔH solution 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 ΔH solution endothermic. BaSO4 (s) H lattice dissociation (BaSO4 ) + SO4 2- Ba (g) 2+ (g) + SO4 2- Ba (aq) 2+ (aq)  hydH Ba2+ +  hydH SO4 2- 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