OCR A2 F Lattice enthalpy

OCR A2 F325 5.2.1 Lattice enthalpy

Contents 5.2.1 Lattice enthalpy Background theory
Ionisation and electron affinity Drawing the cycles

Born-Haber cycles Lattice, solution and hydration enthalpy
We will be covering: Lattice, solution and hydration enthalpy Using energy cycles to calculate enthalpy changes Grade Success Criteria B/C Recall the definitions lattice, solvation and hydration enthalpy A/B Explain the factors that determine the size of the lattice and solution enthalpy A/A* Calculate enthalpy changes using energy cycles

Ionic Bonding An ionic bond is an electrostatic attraction between oppositely charged ions Ionic bonding occurs between a metal and a non- metal and involves the transfer of electrons from the metal to the non-metal The bigger the difference in electronegativity between the two atoms, the more ionic the bond Ionic bonds are strong, consequently the melting and boiling points of ionic compounds are very high

Forming ionic compounds
Ionic compounds are represented by dot-and-cross diagrams: Ca 2 electrons transferred 2+ Ca 2.8.8 O + 2- O 2.6 2.8 CaO, calcium oxide

Lattice enthalpy Lattice enthalpy (the energy binding a crystal lattice together) is used to measure the strength of bonding in ionic compounds It is equivalent to the bond enthalpy of a covalent substance It is defined as follows: The standard lattice enthalpy, HLEʅ, is the enthalpy change when 1 mole of an ionic crystal lattice is formed from its constituent ions in gaseous form under standard conditions. An example is: Na+(g) + Cl-(g)  NaCl(s) HLEʅ = -787 kJ mol-1

Calculating lattice enthalpy
Lattice enthalpy cannot be measured directly – why?* To determine lattice enthalpy, an energy cycle based on Hess’s Law is used This cycle, called a Born-Haber cycle, uses enthalpy values that have been measured and recorded in data books to calculate the lattice enthalpy of a compound A Born-Haber cycle is actually several Hess’s Law cycles joined together, so is quite complicated *you can’t separate gaseous ions and then measure the enthalpy change when they combine

Calculating lattice enthalpy
To work out how to construct a cycle to calculate the lattice enthalpy of NaCl, start with the definition of lattice enthalpy: The standard lattice enthalpy, HLEʅ, is the enthalpy change when 1 mole of an ionic crystal lattice is formed from its constituent ions in gaseous form under standard conditions. The lattice enthalpy is therefore the enthalpy change for a process such as: Na+(g) + Cl-(g)  NaCl(s) ∆H= HLEʅ (NaCl(s))

The lattice enthalpy of NaCl
Na+(g) + Cl-(g)  NaCl(s) ∆H= Hlattʅ (NaCl(s)) This enthalpy change cannot be determined experimentally We need to therefore find another route to make NaCl that has an enthalpy change we can measure, then use Hess’s Law to calculate the lattice enthalpy The route we use is the formation of NaCl from its elements in their standard state This allows us to now build a Hess’s Law cycle

Na+(g) + Cl-(g) NaCl(s) HLEʅ (NaCl(s)) enthalpy of formation of the gaseous ions Na+ and Cl- Hfʅ (NaCl(s)) Na(s) + ½ Cl2(g) this has to be further broken down look this up in a data book

Forming gaseous ions is a two-part process The element must be first atomised (turned into gaseous atoms) and the energy required is the enthalpy of atomisation, Hat The gaseous atoms are then turned into ions For a metal, this process is called ionisation (electrons are removed) and the energy required is the ionisation enthalpy, Hie1 For a non-metal, this means adding electrons and the energy required is the electron affinity, Hea1

More definitions to learn: Standard enthalpy of atomisation, Hatʅ, of a compound is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state. First ionisation enthalpy, Hie1, is the energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous ions with a +1 charge The first electron affinity, Hea1, is the enthalpy change that occurs when 1 mole of gaseous atoms gain 1 mole of electrons to form 1 mole of gaseous ions with a –1 charge.

Na+(g) + Cl-(g) NaCl(s) HLEʅ (NaCl(s)) Hfʅ (Na+(g)) + Hfʅ (Cl- (g)) Hfʅ (NaCl(s)) Na(s) + ½ Cl2(g)

HLEʅ (NaCl(s)) = Hfʅ (NaCl(s)) - Hfʅ (Na+(g)) - Hfʅ (Cl-(g))
Applying Hess’s Law 1 Na+(g) + Cl-(g) NaCl(s) HLEʅ (NaCl(s)) Hfʅ (Na+(g)) + Hfʅ (Cl- (g)) Hfʅ (NaCl(s)) Na(s) + ½ Cl2(g) 3 2 By Hess’s Law: 1 = = 3 - 2 HLEʅ (NaCl(s)) = Hfʅ (NaCl(s)) - Hfʅ (Na+(g)) - Hfʅ (Cl-(g))

The two-part process of forming gaseous ions can also be represented by a Hess’s Law cycle: enthalpy of formation of gaseous ions Elements Gaseous ions enthalpy of atomisation of metal + enthalpy of atomisation of non-metal enthalpy of ionisation (metal) + electron affinity (non-metal) Gaseous atoms

For Na+ and Cl-, this looks like: Hfʅ (Na+(g)) + Hfʅ (Cl-(g)) Na(s) + ½ Cl2(g) Na+(g) + Cl-(g) Hatʅ (Na(s)) + Hatʅ (½Cl2(g)) Hie1 (Na(g)) + Hea1 (Cl(g)) Na(g) + Cl(g)

This is actually two cycles on top of each other, one for Na+, one for Cl- Na(s) + ½ Cl2(g) Na+(g) + Cl-(g) Na(g) + Cl(g) Hfʅ (Na+(g)) + Hfʅ (Cl-(g)) Hie1 (Na(g)) + Hea1 (Cl(g)) Hatʅ (Na(s)) + Hatʅ (½Cl2(g))

It is best to separate this into two cycles One deals with forming Na+: Na(s) Hfʅ (Na+(g)) Na+(g) Hatʅ (Na(s)) Hie1 (Na(g)) Na(g)

Hfʅ (Na+(g)) = Hatʅ (Na(s)) + Hie1 (Na(g))
2 Na(g) 3 By Hess’s Law: 1 = 2 + 3 Hfʅ (Na+(g)) = Hatʅ (Na(s)) + Hie1 (Na(g))

The other deals with forming Cl-: ½ Cl2(g) Hfʅ (Cl-(g)) Cl-(g) Hatʅ (½Cl2(g)) Hea1 (Cl(g)) Cl(g)

Hfʅ (Cl-(g)) = Hatʅ (½Cl2(g)) + Hea1 (Cl(g))
3 By Hess’s Law: 1 = 2 + 3 Hfʅ (Cl-(g)) = Hatʅ (½Cl2(g)) + Hea1 (Cl(g))

We now have three equations that can be combined to give an overall equation for the lattice enthalpy The first one is from the overall cycle we saw at the start: The second one relates to forming gaseous Na+: HLEʅ (NaCl(s)) = Hfʅ (NaCl(s)) - Hfʅ (Na+(g)) - Hfʅ (Cl-(g)) Hfʅ (Na+(g)) = Hatʅ (Na(s)) + Hie1 (Na(g))

The third one relates to forming gaseous Cl-: Combining these (by substituting equations 2 and 3 into 1) gives an expression that allows us to calculate the lattice enthalpy: Hfʅ (Cl-(g)) = Hatʅ (½Cl2(g)) + Hea1 (Cl(g)) HLEʅ (NaCl(s)) = Hfʅ (NaCl(s)) - Hatʅ (Na(s)) - Hie1 (Na(g)) Hatʅ (½Cl2(g)) - Hea1 (Cl(g))

HLEʅ (salt) = Hfʅ (salt) – Σ(all other enthalpy changes)
More simply… The lattice enthalpy of a salt is equal to its enthalpy of formation minus the sum of all the other enthalpy changes in the cycle HLEʅ (salt) = Hfʅ (salt) – Σ(all other enthalpy changes)

Complete Hess’s Law cycle for lattice enthalpy
Na(s) + ½ Cl2(g) Na(g) + Cl(g) 1 atomisation formation enthalpy 4 2 ionisation NaCl(s) Na+(g) + Cl-(g) 3 lattice enthalpy By Hess’s Law: = 4 ∴ 3 = lattice enthalpy = enthalpy of formation – Σeverything else

Born-Haber cycle When the processes used to calculate lattice enthalpy are represented on an energy diagram, the result is a Born-Haber cycle In a Born-Haber cycle, enthalpy changes have to be placed relative to each other depending on whether they are endothermic or exothermic You need to remember some essential periodic trends to understand this

Effective nuclear charge and shielding
Two important phenomena that help to explain many periodic trends (and the reactivity of most elements) are effective nuclear charge and the shielding effect

Effective nuclear charge
+ 7 protons = +7 6 electrons = -6 Effective nuclear charge = 7/6 = 1.17 N 2+ 7 protons = +7 5 electrons = -5 Effective nuclear charge = 7/5 = 1.4 N 3+ 7 protons = +7 4 electrons = -4 Effective nuclear charge = 7/4 = 1.75 N 7 protons = +7 7 electrons = -7 Effective nuclear charge = 7/7 = 1.0 Effective nuclear charge is how much of the nuclear charge is experienced by each electron The bigger its value, the greater the electrostatic attraction between each electron and the nucleus This pulls them closer and makes them harder to remove

Shielding and repulsion
The outer electron is shielded from the full attractive force of the nucleus by the inner electrons (orbitals coloured in blue) K Or: the outer electron is attracted by the nucleus but repelled by the inner electrons, so is further from the nucleus and at higher energy

Ionisation energies across a period
General trend: ionisation energies increase across a period Reason: Nuclear charge increases, electrons are added to the same principal energy level, no significant shielding, effective nuclear charge increases, electrons attracted more strongly, more energy required to remove them

Ionisation energies down a group
Li Na General trend: ionisation energies decrease down a group Reason: Nuclear charge increases, electrons are added to higher energy shells, further from the nucleus, significant shielding from inner electrons, effective nuclear charge decreases, electrons attracted less strongly, less energy required to remove them

Successive ionisation energies
+ 2+ N N N Trend: Ionisation energies rise steadily as electrons are removed from the same principal energy level. Reason: Ionic charge increases, no significant increase in shielding, less electron-electron repulsion so electrons pulled in more tightly, effective nuclear charge increases, more energy needed

Ionisation energy trends
First ionisation enthalpy is the energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of gaseous ions with a +1 charge This is always endothermic because it takes energy to do this Second ionisation enthalpy is the energy required to remove 1 mole of electrons from one mole of gaseous 1+ ions to produces one mole of gaseous +2 ions. This is even more endothermic because more energy is required to overcome stronger electrostatic attraction

Ionisation energy trends
Mg Mg + 2+ Mg Hie1 (Mg(g)) = kJ mol-1 Hie2 (Mg+(g)) = kJ mol-1 Hie3 (Mg2+(g)) = kJ mol-1 – why is this so large?

Electron affinity trends
The first electron affinity is the enthalpy change that occurs when 1 mole of gaseous atoms gain 1 mole of electrons to form 1 mole of gaseous ions with a –1 charge The first electron affinity is exothermic for atoms that normally form negative ions This can be explained by considering the more precise definition of electron affinity: The enthalpy change when one infinitely distant electron is added to each atom in one mole of gaseous atoms under standard conditions It is a favourable process to bring an atom from far way to the outer shell of an atom, where it feels the attractive nuclear force, hence exothermic

- O O + infinitely distant electron ion stabilised due to electrostatic attraction between electron and nucleus Hea1 (O(g)) = -147 kJ mol-1

Second electron affinity is the enthalpy change when one mole of gaseous 1- ions gains one electron per ion to produce gaseous 2- ions. The second electron affinity is endothermic because it takes energy to overcome the repulsive force between the negative ion and the electron. O - O 2- + Hea2 (O(g)) = kJ mol-1

Enthalpy of atomisation
Standard enthalpy of atomisation of an element is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state. It corresponds to breaking covalent bonds for molecular compounds, or disrupting metallic bonding in metals. These processes require energy so it is always endothermic For solid elements (metals and non-metals such as carbon), it is equivalent to the enthalpy of sublimation.

Drawing Born-Haber Cycles
First draw a vertical line to represent the y-axis, which will be the enthalpy scale Then draw a horizontal line to represent the x-axis and put the ionic solid on this line A line is then drawn at the top to represent the highest energy point (optional) Next, the elements needed to form the ionic solid are placed in their standard states on a line above the ionic solid This line corresponds to zero energy Continue to add the other steps until you end up with gaseous ions needed to form the ionic solid

Born-Haber Cycle For Sodium Chloride
Route 1 = Route 2 (Hess’ Law) ∆Hʅf(NaCl) = ∆Hʅat(Na) + ∆Hʅie1(Na) + ∆Hʅat(Cl) + ∆Hʅea1(Cl) + ∆Hʅlatt(NaCl) Na+(g) + e- + Cl(g) enthalpy ∆Hʅat(Cl) ∆Hʅea1(Cl) Na+(g) + ½ Cl2(g) + e- Na+(g) + Cl-(g) ∆Hʅie1(Na) Route 2 Na(g) + ½ Cl2(g) ∆HʅLE(NaCl) ∆Hʅat(Na) Na(s) + ½ Cl2(g) Route 1 ∆Hʅf(NaCl) NaCl(s)

Born-Haber Cycle For Sodium Chloride
Route 1 = Route 2 (Hess’ Law) ∆Hʅf(NaCl) = ∆Hʅat(Na) + ∆Hʅie1(Na) + ∆Hʅat(Cl) + ∆Hʅea1(Cl) + ∆Hʅlatt(NaCl) Na+(g) + e- + Cl(g) enthalpy +122 kJ ∆Hʅat(Cl) -349 kJ ∆Hʅea1(Cl) Na+(g) + ½ Cl2(g) + e- Na+(g) + Cl-(g) +496 kJ ∆Hʅie1(Na) Route 2 Na(g) + ½ Cl2(g) -788 kJ mol-1 ∆HʅLE(NaCl) +108 kJ ∆Hʅat(Na) ∆HʅLE(NaCl) = ∆Hʅf(NaCl) – (∆Hʅat(Na) + ∆Hʅie1(Na) + ∆Hʅat(Cl) + ∆Hʅea1(Cl)) Na(s) + ½ Cl2(g) Route 1 -411 kJ ∆Hʅf(NaCl) NaCl(s)

Tips Σdown arrow processes = Σup arrow processes
Get the expression for the enthalpy change you are trying to find right BEFORE you start putting numbers in Another way to do the calculation is to use the expression: Σdown arrow processes = Σup arrow processes Check your calculation using: HLEʅ (salt) = Hfʅ (salt) – Σ(all other enthalpy changes)

Tips It doesn’t really matter whether you atomise the metal or the non-metal first, but… …enthalpy of atomisation must come before ionisation energy and electron affinity This is because those enthalpy changes are for gaseous atoms Make sure you don’t lose or gain anything (atoms or electrons) when drawing the cycle Pay attention to how many moles of everything you have – you might need to scale the enthalpy values

enthalpy Worked example CsCl ΔH ʅf CsCl = -433 kJ mol-1
ΔH ʅat Cs = +79 kJ mol-1 ΔHie1 Cs = +376 kJ mol-1 ΔH ʅat Cl = +121 kJ mol-1 ΔH ea1 Cl = -346 kJ mol-1 Worked example CsCl enthalpy -663

ΔH ʅf CuO = -155 kJ mol-1 ΔH ʅat Cu = +339 kJ mol-1 ΔH ʅat O = +249 kJ mol-1 ΔHie1 Cu = +745 kJ mol-1 ΔHie2 Cu+ = kJ mol-1 ΔH ʅlatt CuO = kJ mol-1 ΔH ea1 O = -141 kJ mol-1 Worked example: CuO to work out 2nd electron affinity of oxygen enthalpy +790

∆H1=∆H2+ ∆H3+ ∆H4+∆H5+ ∆Hea1+∆H6 +150 ∆H3
Spot the 7 errors in this Born-Haber cycle and calculate the correct value for the electron affinity of chlorine! ∆Hf MgCl2 = -642 kJ mol-1 ∆HLE MgCl2 = kJ mol-1 ∆Hie1 Mg = +736kJ mol-1 ∆Hie2 Mg = +1450kJ mol-1 ∆Hat Cl = +121 kJ mol-1 ∆Hat Mg = +150 kJ mol-1 Mg2+(g) + 2Cl(g) +1450 ∆H5 Mg+(g) + 2Cl(g) ∆Hea1 Mg2+(g) + 2Cl-(g) + 2e- +121 ∆H4 Mg+(g) + Cl2(g) E ∆H1=∆H2+ ∆H3+ ∆H4+∆H5+ ∆Hea1+∆H6 +150 ∆H3 ∆Hea1 = ∆H1 – (∆H2+ ∆H3+ ∆H4 +∆H5+∆H6) Mg+(s) + Cl2(g) -2493 ∆H6 ∆Hea1 = 642 – ( ) +736 ∆H2 ∆Hea1 = kJ mol-1 Mg(s) + Cl2(g) -642 ∆H1 MgCl2(s)

Factors affecting the lattice enthalpy

Lattice enthalpy values
Can you spot any trends? Remember, the bigger the lattice enthalpy, the stronger the attraction between ions

Ionic Radius

Ionic Charge Lattice enthalpy becomes more exothermic as the ionic charge increases This is because the higher the charge, the greater the force of attraction between ions increasing lattice enthalpy

Key points so far The bigger the lattice enthalpy, the stronger the electrostatic force of attraction between ions and the more stable the compound The higher the charge on the ions, the more they will attract each other and the bigger the lattice enthalpy The larger the ionic radius, the smaller the lattice enthalpy (because there is greater distance and hence less attraction between the centres of the ions)

Dissolving ionic solids

Dissolving ionic solids
Lattice enthalpy/kJmol-1 LiF -1047 MgBr2 -2434 BeO -4298 BeCl2 -3017 BaO -3034 CaCl2 -2634 KO -2245 Typically very high, indicating high bond strength Despite this, all of the above readily dissolve in water at room temperature There must be energetically favourable processes that occur when substances are dissolve

Dissolving ionic substances
Ionic lattices dissolve in polar solvents – each ion is surrounded by water molecules and becomes ‘hydrated’ These processes are energetically favourable This helps to overcome the lattice enthalpy 1)ionic lattice breaks down 2)ions are hydrated The enthalpy changes associated with dissolving ionic substances are the enthalpy of solution and hydration

Dissolving ionic substances
Enthalpy of solution, hydration and the lattice enthalpy can be related using a Born-Haber cycle or a standard Hess’s cycle These cycles enable any of the three enthalpy changes to be calculated from the other two values (using Hess’s law again) Two more definitions

Using a Born-Haber cycle
K+(g) + Cl-(g)  KCl(s) ΔH ʅLE KCl(s) + aq  K+(aq) + Cl-(aq) ΔH ʅsol = +26 kJ mol-1 K+(g) + aq  K+(aq) ΔH ʅhyd = -322 kJ mol-1 Cl-(g) + aq  Cl-(aq) ΔH ʅhyd = -363 kJ mol-1 = kJ mol-1

Problem with this method
The Born-Haber method of calculating lattice enthalpy from ΔHʅsol and ΔHʅhydr can be confusing as ΔHʅsol can be exothermic or endothermic Salt ΔHʅS at 25°C in kJmol-1 LiCl -37 NaCl +3.9 KCl +26.0 LiBr -48.8 NaBr -0.6 KBr +19.9 KOH -57.6

Using an energy (Hess’s) cycle
Ionic solid Aqueous ions Gaseous ions ΔHʅsol ΔHʅLE ΔHʅhyd ΔHʅsol = ΣΔHʅhyd - ΔHʅLE

Using Hess’s cycles Calculate the enthalpy change of solution of SrF2 from the following data: ΔH LEʅ [SrF2(s)]=-2492 kJmol-1, ΔHhyd ʅ [Sr2+(g)]=-1480 kJmol-1, ΔHhyd ʅ [F-(g)]=-506 kJmol-1.

Worked example Show that the enthalpy change for solution for MgCl2 is -86 kJmol-1, given that ΔH LEʅ [MgCl2]=-2526 kJmol-1, ΔHhydʅ [Mg2+(g)]=-1920 kJmol-1, ΔHhyd ʅ [Cl- (g)]=-346 kJmol-1

Factors affecting ΔH ʅ hyd
The size of the ion. The smaller the ion, the larger the enthalpy of hydration. (ΔH ʅ hyd more exothermic) ΔH ʅ hyd less exothermic The charge on the ion. The larger the charge on the ion, the larger the enthalpy of hydration.

The charge density concept
The values for melting points and lattice enthalpies of ionic solids depends on The charges on the ionic species Their ionic radii It is helpful to know the concept of charge density charge density = 𝑖𝑜𝑛𝑖𝑐 𝑐ℎ𝑎𝑟𝑔𝑒 𝑣𝑜𝑙𝑢𝑚𝑒

Charge density Charge density is greater for:
ions with a higher charge (e.g. Mg2+ > Li+) ions with a smaller ionic radius (because they have a smaller volume) Ions with greater charger density have: higher melting points (because more energy is required to overcome the electrostatic forces of attraction) more exothermic lattice enthalpies more exothermic hydration enthalpies due to having strong attraction for, and more stable bonds with, water molecules

Sample exam question

charge density = 𝑖𝑜𝑛𝑖𝑐 𝑐ℎ𝑎𝑟𝑔𝑒 𝑣𝑜𝑙𝑢𝑚𝑒
F- Mg2+ F- has a greater charge density than Cl- because it’s the smaller ion Mg2+ has a greater charge density than Na+ because its ionic radius is smaller and it has a bigger charge Strongest attraction is between oppositely charged Mg2+ and F- ions in MgF, and Na+ and F- in NaF Na+ Cl- Can you estimate a value for MgCl2? -2523 kJmol-1

Comparing enthalpies of solution
Tricky question from 2011 The key to this question is realising that there is insufficient data at the moment to answer it – there are no general rules about the sizes of enthalpy changes of solution You need to consider the balance between enthalpy of solution, hydration enthalpy, and lattice enthalpy

This expression tells us how the balance between lattice enthalpy and hydration enthalpies determines enthalpy of solution

What do we know about Rb+ versus K+?
The charge density of K+ is greater than Rb+, because it is the smaller ion ORA (1 mark) The lattice enthalpy of KF will therefore be more exothermic/more -ve than RbF because K+ has a stronger attraction to F- than Rb+ ORA (1 mark) The hydration enthalpy of K+ will be more exothermic/more -ve than Rb+ because K+ is more strongly attracted to water than Rb+ ORA (1 mark)

The sign and value of ΔHθsol is determined by the sizes of ΔHθhydr and ΔH θLE
If ΔHθhydr is the dominant factor, as it becomes more exothermic/more -ve, ΔHθsol will become more exothermic/more -ve. ΔHθsol for KF would be more exothermic than RbF If ΔH θLE were the dominant factor, as it becomes more exothermic/more -ve, ΔHθsol will become less exothermic/more +ve ΔHθsol for KF would be less exothermic than RbF

Put another way… If ΔHθsol is most affected by ΔHθhydr, a compound with a more – ve ΔHθhydr will have a more –ve ΔHθsol If ΔHθsol is most affected by ΔH θLE , a compound with a more - ve ΔH θLE will have a more +ve ΔHθsol

Prediction 2 fits the data given
KF has the more +ve enthalpy of solution and both its enthalpies of solution and hydration are more –ve/more exothermic than RbF We can therefore say that enthalpy of solution is more affected by lattice enthalpy than hydration enthalpy

ΔH LEθ : Experimental versus Theoretical
Born-Haber gives experimental values – lattice enthalpy is calculated using other enthalpy values that have been measured Lattice enthalpy can also be calculated - theoretical values are obtained by treating the ions as point charges that come together to form a lattice For NaCl, experimental and theoretical values for ΔH LEθ agree to within 1%

ΔH LEθ : Experimental versus Theoretical
However, for AgCl, the experimental values are larger (more negative) than the theoretical values This is because AgCl has significant covalent bonding – the difference in electronegativity between Ag and Cl is not enough for complete transfer of an electron from the Ag to Cl. Differences between experimental and theoretical ΔH LEθ are also due to purity for ionic solids (impurities lead to lower values)

Self-test questions Describe and explain the effect of ionic radius and ionic charge on the size of LE The LE of NaCl is more exothermic than the lattice enthalpy of CsCl. State and explain the relative strengths of ionic bonding of these 2 chlorides. Describe how, and explain why, the enthalpy change of hydration of Na+ differs from that of Rb+ The lattice enthalpies of BeO and MgCl2 are -4,298 and -3,800 kJ mol-1 respectively. Their solubilities in water are g/100g and 55g/100g respectively. Explain the relative solubilities.

OCR A2 F Lattice enthalpy

Similar presentations

Presentation on theme: "OCR A2 F Lattice enthalpy"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

OCR A2 F Lattice enthalpy

Similar presentations

Presentation on theme: "OCR A2 F Lattice enthalpy"— Presentation transcript:

Similar presentations

About project

Feedback