A Level Chemistry 3.1.8 Thermodynamics - Lattice Enthalpy (AQA)

A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS LATTICE ENTHALPY This section is all about salts (ionic substances) and the enthalpy of the forces of attraction between the ions. The structure of an ionic substance can be described a...

A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS LATTICE ENTHALPY This section is all about salts (ionic substances) and the enthalpy of the forces of attraction between the ions. The structure of an ionic substance can be described as a lattice as the ions are in a regular, alternating arrangement. Enthalpy of Lattice Dissociation (ΔHLATT DISS) The enthalpy change when 1 mole of an ionic substance dissociates fully into gaseous ions e.g. NaCl(s) → Na+(g) + Cl-(g) Essentially, what is happening here is that the forces of attraction between the ions are broken, creating individual ions. Breaking any form of electrostatic attraction requires energy. This is why the Enthalpy of Lattice Dissociation is always positive (+ΔH) Enthalpy of Lattice Association (ΔHLATT ASS) The enthalpy change when 1 mole of an ionic substance is formed from its gaseous ions e.g. Na+(g) + Cl-(g) → NaCl(s) Essentially, what is happening here is that individual gaseous ions come together (associate) to form an ionic lattice. Forming any form of electrostatic attraction releases energy. This is why the Enthalpy of Lattice Association is always negative (-ΔH). AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS HINTS | TIPS | HACKS As with any enthalpy change, the enthalpy values are equal and opposite. e.g. for NaCl ΔHLATT DISS = +787 kJ.mol-1 ΔHLATT ASS = -787 kJ.mol-1 The stronger the attractions between the ions, the greater the lattice Enthalpy. i.e. ΔHLATT DISS is more positive and ΔHLATT ASS is more negative. Factors that influence the strength of attraction between the ions (and therefore the size of ΔHLATT) are: - Charge on the ion. The greater the charge the stronger the attraction - Ionic radius. The smaller the ionic radius, the stronger the attraction AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS BORN-HABER CYCLES It isn’t always possible to measure the ΔHformation of a salt directly (experimentally), so we need a work around using other experimental data to calculate it. This is where Born-Haber cycles come in. Think of them as a giant Hess’ Energy Cycle (Year 12) where we use an alternative route to calculate a missing enthalpy value. The Born-Haber cycle looks at each stage of the process of converting the constituent elements in their standard states into the solid salt. Both the metal and the non-metal have to go through THREE basic stages: 1. Atomisation - Both elements are converted into individual gaseous atoms (ΔHa) 2. Ionisation - The metal atoms become positive gaseous ions (ΔHI.E.) and the non-metal atoms become negative gaseous ions (ΔHE.A) 3. Lattice Association - The gaseous ions then come together to form the lattice (ΔHlatt ass) ΔHI.E. & ΔHE.A Na(g) + Cl(g) Na+(g) + Cl-(g) ΔHa Na(s) + ½Cl2(g) ΔHLATT ASS ΔHf NaCl(s) So, if we can calculate the sum of the enthalpy changes of the alternative route, we can calculate the ΔHf for the ionic substance. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS ΔH DEFINITIONS YOU NEED TO KNOW Building Born-Haber cycles can be tricky, so it’s important that you understand the definitions of the changes that occur at each of the stages of the alternative route. The skill here lies in linking the definition with the equation. Remember, the definition of any ΔH is essentially the equation in words. Also, pay attention to the signs of these! SYMBOL NAME DEFINITION SIGN The enthalpy change when 1 mole of a compound is formed from its constituent ΔHf FORMATION + or - elements in their standard states e.g. C(s) + 2H2(g) → CH4(g) The average enthalpy change when 1 mole of an element in its standard state is atomised to ΔHa ATOMISATION + produce 1 mole of gaseous atoms e.g. ½Cl2(g) → Cl(g) The average enthalpy change when 1 mole of BOND *ΔHB.E. covalent bonds is broken in the gaseous state + ENTHALPY e.g. Cl2(g) → 2Cl(g) IONISATION The enthalpy change when 1 mole of electrons ΔHI.E. ENERGY is removed from 1 mole of gaseous atoms/ions + (metal ions) e.g. Mg(g) → Mg+ (g) + e- ELECTRON The enthalpy change when 1 mole of electrons AFFINITY ΔHE.A. are added to 1 mole of gaseous atoms/ions + or - (non-metal e.g. Cl(g) + e- → Cl-(g) ions) The enthalpy change when 1 mole of an ionic LATTICE ΔHlatt ass substance is formed from its gaseous ions - ASSOCIATION e.g. Mg2+(g) + 2Br-(g) → MgBr2(s) * Watch out for “bond enthalpy”! It can be used instead of atomisation. They basically both do the same thing (produce individual atoms) but the number of individual atoms produced is different. Compare the two in the table above. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.2.4 THERMODYNAMICS There are 4 different combinations of ions that you can get when it comes to Born- Haber Cycles: 1. Group 1 + Group 7 2. Group 1 + Group 6 3. Group 2 + Group 7 4. Group 2 + Group 6 Each these contains the same essential enthalpy changes, but have slightly different features. When creating Born-Haber Cycles, you will be working with metals from Groups 1 & 2 and non-metals from Groups 6 & 7. Group 1 undergo 1st I.E. only to create 1+ ions Group 2 undergo both 1st & 2nd I.E. in two stages to create 2+ ions Group 7 undergo 1st electron affinity only to create 1- ions Group 6 undergo both 1st & 2nd electron affinity in two stages to create 2- ions The other important factor here is the actual formula of the salt as you must take the number of ions into account. This comes into play with the following combinations: Group 1 + Group 6 e.g. K2O, Na2S Group 2 + Group 7 e.g. MgCl2, CaF2 This means you will need to account for two ions in both the equations you write and the calculations that you do using the ΔH values. How To Build Born- Haber Cycles AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS To build a Born-Haber Cycle, you will need to both identify these changes and write equations. The changes are highlighted in bold. Let’s play “spot the difference”: between one side of each equation and the other… e.g. KCl ΔHa K K(s) + ½Cl2(g) → K(g) + ½Cl2(g) K is atomised ΔH1st I.E. K K(g) + ½Cl2(g) → K+(g) + ½Cl2(g) + e- K is ionised (loses e-) ΔHa Cl2 K+(g) + ½Cl2(g) → K+(g) + Cl(g) + e- Cl2 is atomised ΔH1st E.A. Cl K+(g) + Cl(g) → K+(g) + Cl-(g) Cl is ionised (gains e-) Now we have K+(g) + Cl-(g) which can come together to form the lattice ΔHLATT. ASS. KCl K+(g) + Cl-(g) → KCl(s) Think of each stage the Born-Haber Cycle as one of these equations. K+(g) + Cl(g) + e- ΔHa ½Cl2 ΔHE.A. Cl K+(g) + ½Cl2(g) + e- K+(g) + Cl-(g) ΔHI.E. K K(g) + ½Cl2(g) ΔHa K ΔHlatt ass. KCl K(s) + ½Cl2(g) ΔHf KCl KCl(s) AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS Now we have a Born-Haber Cycle built, we can insert the values for the individual ΔH’s ands use them to calculate an unknown value (e.g. ΔHf KCl) K+(g) + Cl(g) + e- ΔHa ½Cl2 = +122 kJ.mol-1 ΔHE.A. Cl = -349 kJ.mol-1 K+(g) + ½Cl2(g) + e- K+(g) + Cl-(g) -1 ΔHI.E. K = +419 kJ.mol K(g) + ½Cl2(g) ΔHa K = +90 kJ.mol-1 ΔHlatt ass. KCl K(s) + ½Cl2(g) = -719 kJ.mol-1 ΔHf KCl KCl(s) ΔHf KCl = +90 + 419 + 122 - 349 - 719 = -437 kJ.mol-1 You could be asked to calculate any one of How To Calculate Values these values given the data, so be prepared Using Born-Haber Cycles to rearrange the calculation to do so. Don’t forget to take the number of ions in the formula into account. e.g. you double the ΔH value for the changes involving the ion that there are two of (K in K2O) AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS THE PERFECT IONIC MODEL When we think of an ionic solid, we imagine a + - diagram like this one; perfectly spherical ions arranged in a regular lattice. - + However, the truth is that many ionic solids have some “covalent character”. What this means is - + that the electrons in the negative ions can still be shared a little with the positive ions. + - + ((((((( - + - The Perfect Ionic Model: In Reality: - Ions are spherical - Ions are distorted - Electrons are not shared - Electrons are partially shared - Purely electrostatic attractions - Partial covalent bonding The amount of covalent character an ionic substance has increases when: 1. The + ion has a smaller radius and has a greater charge 2. The - ion is larger radius and has a greater charge e.g. Look for one difference between the two salts you are comparing. LiCl will have greater covalent + - + - character than KCl as the Li+ ion has a smaller radius than the K+ LiCl KCl ion. (They have the same charge) CaO will have greater covalent character than CaF2 as the O2- ion 2+ 2- - 2+ has a greater charge than the F- ion. (They have the same radius) CaO CaF2 AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS PERFECT IONIC MODEL Vs BORN-HABER The lattice enthalpy (ΔHLATT) for a salt can be calculated using the theoretical values based on the perfect ionic model. You do not have to be able to calculate this. Just know that it is possible! We have also seen that we can calculate lattice enthalpy using the experimental values in a Born-Haber Cycle. (You do have to be able to do this!) The difference here is that the perfect ionic model calculation is theoretical and assumes that all ions are spherical and that electrons are not shared. Now we know that this isn’t true! However, the data we use in Born-Haber Cycle calculations is experimental, it is real world data that reflects the covalent character of ionic solids. The greater the covalent character an ionic solid has, the greater the difference between the calculated values for ΔHLATT e.g. Theoretical Experimental Calculation of ΔHLATT Calculation of ΔHLATT NaBr +732 +733 AgBr +890 +758 You can see from the data that there is a greater difference between the two values for AgBr than there is for NaBr. This means that AgBr shows greater covalent character than NaBr. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS ENTHALPY OF SOLUTION Enthalpy of solution (ΔHSOL) is the enthalpy change when 1 mole of a solid ionic substance is dissolved in water. e.g. NaCl(s) + (aq) → Na+(aq) + Cl-(aq) ΔHSOL can be exothermic (negative) or endothermic (positive) depending on the ionic substance. In order to understand why, like in Born-Haber Cycles, we need to look at the individual processes that happen when a solid ionic substance dissolves in water: 1. The ions must be separated from one another. We’ve already seen this! It is known as ΔHLATTICE DISSOCIATION. e.g. NaCl(s) → Na+(g) + Cl-(g) 2. The ions must then form electrostatic attractions with the water molecules to “dissolve”. This is a new ΔH known as Enthalpy of Hydration or ΔHHYD. Each ion in the lattice has it’s own equation / value for this: e.g. Na+(g) + (aq)→ Na+(aq) Cl-(g) + (aq)→ Cl-(aq) This two-step process can be shown as a Hess’ Energy Cycle: △Hsol Na(s) Na+(aq) + Cl-(aq) △Hlattice △Hhyd Na+ dissociation △Hhyd Cl- Na+(g) + Cl-(g) AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS As with the enthalpy changes involved in Born-Haber Cycles, it’s really important for you to know the definitions and equations for the changes involved in dissolving. SYMBOL NAME DEFINITION SIGN The enthalpy change when 1 mole of a solid ΔHsol SOLUTION ionic substance is dissolved in water + or - e.g. NaCl(s) + (aq) → Na+(aq) + Cl-(aq) The enthalpy change when 1 mole of an ionic LATTICE ΔHlatt diss substance dissociates fully into gaseous ions + DISSOCIATION e.g. NaCl(s) → Na+(g) + Cl-(g) The enthalpy change when 1 mole of gaseous ΔHhyd HYDRATION ions is associated with water molecules - e.g. Na+(g) + (aq) → Na+(aq) Notice that: ΔHlatt diss is always positive! Energy must be input to break the electrostatic attractions in the solid ΔHhyd is always negative! Energy is released as the water molecules arrange themselves around the ions and form new electrostatic attractions with them. e.g. 𝜹+ 𝜹+ H H O H ))))))))) Cl- H O 𝜹+ 𝜹- ))))))))) Na+ 𝜹+ 𝜹- 𝜹- )))) )))) )))) )))) ) H ) O 𝜹+ O H 𝜹+ 𝜹- 𝜹+ H H 𝜹+ So, ΔHsol is a balance between the energy input to dissociate the lattice and the energy released when the ions form attractions with water molecules. The is why ΔHsol is different for every salt and can be either positive or negative. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 A LEVEL CHEMISTRY 3.1.8 THERMODYNAMICS CALCULATING ΔHsol ϴ ϴ ϴ ΔH solution = ΔH latt diss + ∑ΔH hydration -1 (kJ.mol ) (always positive) (always negative) Enthalpy of solution (ΔHsol) = lattice dissociation enthalpy plus the sum of the enthalpies of hydrations for the ions How To Calculate Enthalpy Of Solution HINTS | TIPS | HACKS The greater the charge and the smaller the ion, the more exothermic ΔH hydration is. e.g. Mg2+ has a more negative ΔH hydration than Ca2+ as it has a smaller ionic radius. Ca2+ has a more negative ΔH hydration than K+ as it has a greater charge. The exam will only ask you to compare two ions in the same group OR the same period! If the △H solution for a salt is negative, it is more likely to be soluble in water. If it is positive is less likely to be soluble in water. Be prepared to rearrange the expression to find values for ΔH lattice dissociation or ΔH hydration in exam questions Also be prepared to use the expression to find ΔHLATT ASS! Don’t forget this is the same as ΔHLATT ASS, just a negative value. AQA www.chemistrycoach.co.uk © scidekick ltd 2024