Lattice Enthalpy

Question 1

enthalpy change of formation

Answer 1

The standard enthalpy change of formation of a compound is the energy transferred when
1 mole of the compound is formed from its elements under standard conditions (298K and 100kpa), all reactants and products being in their standard states

Question 2

Enthalpy of atomisation

Answer 2

The 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

Question 3

Enthalpy of sublimation

Answer 3

The enthalpy change for a solid metal turning
to gaseous atoms can also be called the
Enthalpy of sublimation and will numerically
be the same as the enthalpy of atomisation

Question 4

First Ionisation enthalpy

Answer 4

The first ionisation enthalpy is the enthalpy change
required to remove 1 mole of electrons from 1 mole
of gaseous atoms to form 1 mole of gaseous ions
with a +1 charge

Question 5

Second Ionisation enthalpy

Answer 5

The second ionisation enthalpy is the enthalpy
change to remove 1 mole of electrons from one mole
of gaseous 1+ ions to produces one mole of gaseous
2+ ions

Question 6

first electon affinity

Answer 6

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

Question 7

is first electron affinity exothermic or endothermic

Answer 7

The first electron affinity is exothermic for atoms that
normally form negative ions because the ion is more
stable than the atom and there is an attraction
between the nucleus and the electron

Question 8

second electron affinity

Answer 8

The second electron affinity is the enthalpy change
when one mole of gaseous 1- ions gains one
electron per ion to produce gaseous 2- ions.

Question 9

is second electron affinity exothermic or endothermic?

Answer 9

The second electron affinity for oxygen is
endothermic because it take energy to overcome
the repulsive force between the negative ion and
the electron

Question 10

Lattice Enthalpy

Answer 10

The Lattice Enthalpy is the standard enthalpy change
when 1 mole of an ionic crystal lattice is formed from
its constituent ions in gaseous form. The lattice enthalpy can be used as a measure of ionic bond strength

Question 11

Enthalpy of Hydration

Answer 11

Enthalpy change when one mole of gaseous ions
become aqueous ions. This always gives out energy (exothermic, -ve)
because bonds are made between the ions and the
water molecules.

Question 12

Enthalpy of solution

Answer 12

The enthalpy of solution is the standard enthalpy
change when one mole of an ionic solid dissolves in
a large enough amount of water to ensure that the
dissolved ions are well separated and do not
interact with one another.

Question 13

Trends in Lattice Enthalpies

Answer 13

1. The sizes of the ions:
   The larger the ions, the less negative the enthalpies of lattice
   formation (i.e. a weaker lattice). As the ions are larger the
   charges become further apart and so have a weaker attractive
   force between them.

Question 14

Trends in Lattice Enthalpies 2

Answer 14

2. The charges on the ion:
   The bigger the charge of the ion, the greater the attraction
   between the ions so the stronger the lattice enthalpy (more
   negative values).

Question 15

free energy change and enthropy change: A problem with ∆H

Answer 15

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

Question 16

Entropy, S

Answer 16

Substances with more ways of arranging their atoms and
energy (more disordered) have a higher 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
∆S entropy: ˚ = Σ S˚products - ΣS˚reactants

Question 17

Predicting Change in entropy ‘∆S’ Qualitatively

Answer 17

An increase in disorder and entropy will lead to a positive entropy change ∆S˚ = +ve
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.

Question 18

examples: predicting if ∆S is negative or positive through balanced equations

Answer 18

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

Question 19

Gibbs Free Energy Change, ∆G

Answer 19

The balance between entropy and enthalpy determines the
feasibility of a reaction.
This is given by the relationship :
∆G = ∆H - T∆S
Gibbs free energy is a term that combines the effect of enthalpy and entropy into one number
A reaction that has increasing entropy (+ve ∆S) and is exothermic (-ve ∆H) will make ∆G be negative and will always be feasible
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.

Question 20

∆G = ∆H - T∆S

Answer 20

for T: units are in K,
units of ∆G: KJ mol-1,
units for S: J K-1 mol-1
need to convert S by dividing by 1000

Question 21

Effect of Temperature on feasibility

Answer 21

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-
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)

Question 22

energy change using a graph

Answer 22

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

Question 23

enthalpies of solution

Answer 23

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)
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

Question 24

What does ΔsolH tell us?

Answer 24

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.
ΔsolH = -ΔleH + (sum of)Δ hyd H

Question 25

For salts where ΔH solution is exothermic

Answer 25

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

Question 26

For salts where ΔH solution is endothermic

Answer 26

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