Chapter 1 Chemical Energetics Flashcards
Enthalpy change (ΔH) refers to
the amount of heat energy transferred during a chemical reaction, at a constant pressure
Enthalpy change of atomisation
The standard enthalpy change of atomisation (ΔHatꝋ) is the enthalpy change when 1 mole of gaseous atoms is formed from its element under standard conditions
Standard conditions in this syllabus are ?
- a temperature of 298 K and a pressure of 101 kPa
The ΔHatꝋ is always endothermic/exothermic?
endothermic as energy is always required to break any bonds between the atoms in the element, to break the element into its gaseous atoms
- Since this is always an endothermic process, the enthalpy change will always have a positive value
The lattice energy (ΔHlattꝋ) is the
enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions (under standard conditions)
The ΔHlattꝋ is always exothermic/endothermic
as when ions are combined to form an ionic solid lattice there is an extremely large release of energy
- Since this is always an exothermic process, the enthalpy change will always have a negative value
- Because of the huge release in energy when the gaseous ions combine, the value will be a very large negative value
The large negative value of ΔHlattꝋ suggests what?
that the ionic compound is much more stable than its gaseous ions
- This is due to the strong electrostatic forces of attraction
- Since there are no electrostatic forces of attraction between the ions in the gas phase, the gaseous ions are less stable than the ions in the ionic lattice
- The more exothermic the value is, the stronger the ionic bonds within the lattice are
The ΔHlattꝋ of an ionic compound
- cannot be determined directly by one single experiment
- Multiple experimental values and an energy cycle are used to find the ΔHlattꝋ of ionic compounds
The lattice energy (ΔHlattꝋ) of an ionic compound can be written as an equation
- For example, magnesium chloride is an ionic compound formed from magnesium (Mg2+) and chloride (Cl-) ions
- Since the lattice energy is the enthalpy change when 1 mole of magnesium chloride is formed from gaseous magnesium and chloride ions, the equation for this process is:
Mg2+(g) + 2Cl-(g) → MgCl2(s)
The first electron affinity (EA1) is the
- enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms, to form 1 mole of gaseous ions each with a single negative charge under standard conditions
X(g) + e- → X-(g)
EA1 is usually exothermic/endothermic
exothermic, as energy is released
- Since this is generally an exothermic process, then the value for EA1 will usually be a negative number
The second and third electron affinities are endothermic/exothermic
as energy is absorbed
- This is because the incoming electron is added to an already negative ion
- Energy is required to overcome the repulsive forces between the incoming electron and negative ion
- Since these are endothermic processes, the values will be positive
Second & third electron affinity table
Factors affecting electron affinity
- Nuclear charge
- Distance
- Shielding
Factors affecting electron affinity
- nuclear charge
the greater the nuclear charge, the stronger the attractive forces between an incoming electron and the nucleus
Factors affecting electron affinity
- distance
the greater the distance between the nucleus and the outermost shell/orbital where the electron is added, the weaker the force of attraction
Factors affecting electron affinity
- Shielding:
the greater the number of shells, the greater the shielding effect and the weaker the force of attraction
The value of the electron affinity depends on
how strongly the incoming electron is attracted to the nucleus
- The greater the attractive forces between the electron and nucleus, the more energy is released and therefore the more exothermic (more negative) the EA1 value will be
Electron affinities of non-metals become endothermic/exothermic?
- more exothermic across a period, with a maximum at Group 17
- There is generally a downwards trend in the size of the electron affinities of the elements in Group 16 and 17
- The electron affinities generally become less exothermic for each successive element going down both Groups, apart from the first member of each Group (oxygen and fluorine respectively)
Electron affinity table Group 16 & 17
Fluorine is an exception and has a lower/higher EA1 than chlorine
has a lower EA1 than chlorine
- Fluorine has a very small atomic radius
- This means that the electron density of fluorine is high
- There is more repulsion between the incoming electron and the electrons that are already present in fluorine
- These repulsive forces reduce the attractive forces between the incoming electron and nucleus
- As a result, the EA1 of fluorine is less exothermic than expected
Going down Group 16 and 17 affinity
- The outermost electrons are held less tightly to the nucleus as they are further away
- The number of electron shells increases causing an increased shielding of the outermost electrons
- It gets more difficult to add an electron to the outer shell
- Less energy is released upon adding an electron to the outer shell
- So generally, the EA1 becomes less exothermic
To calculate the lattice energy (ΔHlattꝋ) of an ionic compound
a Born-Haber cycle is used, which is a special type of energy cycle
To find the ΔHlattꝋ of an ionic compound, the following values must be known:
- Enthalpy change of formation (ΔHfꝋ)
- The various enthalpy changes involved when going from elements in their standard states to their gaseous ions (the sum of all of these can be referred to as ΔH1ꝋ)
The various enthalpy changes involved when going from elements in their standard states to their gaseous ions include:
- Enthalpy change of atomisation of each element
- First ionisation energy of the metal
- Successive ionisation energies of the metal if applicable
- First electron affinity of the non-metal
- Successive electron affinities of the non-metal if applicable
The order that these are written in the Born-Haber cycle is important
- First ionisation energy cannot take place before atomisation, because first ionisation energy is the process of turning gaseous atoms into gaseous ions
- Electron affinity cannot take place before first ionisation energy, since the electrons must be lost from the metal first during ionisation, to be present in order for the non-metal to gain them
Hess’s law states that
‘the enthalpy change in a chemical reaction is the same regardless of the route taken, as long as the final and initial conditions and states of reactants and products are the same for each route’
- Once a Born-Haber cycle has been constructed, it is possible to calculate the lattice energy (ΔHlattꝋ) by applying Hess’s law and rearranging:
ΔHfꝋ = ΔHatꝋ + ΔHatꝋ + IE + EA + ΔHlattꝋ
which becomes: ΔHfꝋ = ΔH1ꝋ + ΔHlattꝋ
sometimes a value may need to be doubled or halved, depending on the ionic solid involved
- For example, with MgCl2 the value for the first electron affinity of chlorine would need to be doubled in the calculation, because there are two moles of chlorine atoms
- Therefore, you are adding 2 moles of electrons to 2 moles of chlorine atoms, to form 2 moles of Cl- ions
The two key factors which affect lattice energy, ΔHlattꝋ
are the charge and radius of the ions that make up the crystalline lattice
Ionic radius
- The lattice energy becomes less exothermic as the ionic radius of the ions increases
- This is because the charge on the ions is more spread out over the ion when the ions are larger
- The ions are also further apart from each other in the lattice
- The attraction between ions is between the centres of the ions involved, so the bigger the ions the bigger the distance between the centre of the ions
- therefore, the electrostatic forces of attraction between the oppositely charged ions in the lattice are weaker
Ionic charge
- The lattice energy gets more exothermic as the ionic charge of the ions increases
- The greater the ionic charge, the higher the charge density
- This results in stronger electrostatic attraction between the oppositely charged ions in the lattice
- As a result, the lattice energy is more exothermic
The lattice energies get less exothermic as the ionic radius of the ions increases
Enthalpy change of solution
- The standard enthalpy change of solution (ΔHsolꝋ) is the enthalpy change when 1 mole of an ionic substance dissolves in sufficient water to form a very dilute solution
- The symbol (aq) is used to show that the solid is dissolved in sufficient water
- ΔHsolꝋ can be exothermic (negative) or endothermic (positive)
Enthalpy change of hydration
The standard enthalpy change of hydration (ΔHhydꝋ) is the enthalpy change when 1 mole of a specified gaseous ion dissolves in sufficient water to form a very dilute solution
When an ionic solid dissolves in water
- positive and negative ions are formed
- Water is a polar molecule with a δ- oxygen (O) atom and δ+ hydrogen (H) atoms which will form ion-dipole attractions with the ions present in the solution
- The oxygen atom in water will be attracted to the positive ions and the hydrogen atoms will be attracted to the negative ions
- Since the ΔHhydꝋ of KCl is -685 kJ mol-1, 685 kJ mol-1 is released in forming these ion-dipole attractions when KCl dissolves in water
- This compensates for the remaining +685 kJ mol-1 which was needed to break down the KCl lattice
Energy cycle involving enthalpy change of solution, lattice energy, and enthalpy change of hydration
- The energy cycle shows that there are two routes to go from the ionic lattice to the hydrated ions in an aqueous solution:
- Route 1: going from ionic solid → ions in aqueous solution (this is the direct route)
ΔHsolꝋ= Enthalpy of solution
- Route 2: going from ionic lattice → gaseous ions → ions in aqueous solution (this is the indirect route)
-ΔHlattꝋ + ΔHhydꝋ = reverse lattice enthalpy + hydration enthalpies
Lattice enthalpy usually means Lattice
- formation* enthalpy, in other words bond forming. If we are breaking the lattice then this is reversing the enthalpy change so a negative sign is added in front of the term (alternatively it is called lattice dissociation enthalpy)
- According to Hess’s law, the enthalpy change for both routes is the same, such that:
ΔHsolꝋ = -ΔHlattꝋ + ΔHhydꝋ
ΔHhydꝋ = ΔHsolꝋ + ΔHlattꝋ
- Each ion will have its own enthalpy change of hydration, ΔHhydꝋ, which will need to be taken into account during calculations
- The total ΔHhydꝋ is found by adding the ΔHhydꝋ values of both anions and cations together
- The energy cycle involving the enthalpy change of solution (ΔHsolꝋ ), lattice energy (ΔHlattꝋ), and enthalpy change of hydration (ΔHhydꝋ) can be used to calculate the different enthalpy values
- According to Hess’s law, the enthalpy change of the direct and of the indirect route will be the same, such that:
ΔHhydꝋ = ΔHlattꝋ + ΔHsolꝋ
- This equation can be rearranged depending on which enthalpy value needs to be calculated
- For example, ΔHlattꝋ can be calculated using:
ΔHlattꝋ = ΔHhydꝋ - ΔHsolꝋ
- Remember: the total ΔHhydꝋ is found by adding the ΔHhydꝋ values of both anions and cations together
- Remember: take into account the number of each ion when completing calculations
The standard enthalpy change of hydration (ΔHhydꝋ) is affected by the amount that
- the ions are attracted to the water molecules
- The factors which affect this attraction are the ionic charge and radius
ΔHhydꝋ becomes more exothermic with (ionic radius)
decreasing ionic radii
- Smaller ions have a greater charge density resulting in stronger ion-dipole attractions between the water molecules and the ions in the solution
- Therefore, more energy is released when they become hydrated and ΔHhydꝋ becomes more exothermic
ΔHhydꝋ is more exothermic for ions with (Ionic Charge)
larger ionic charges
- Ions with large ionic charges have a greater charge density resulting in stronger ion-dipole attractions between the water molecules and the ions in the solution
- Therefore, more energy is released when they become hydrated and ΔHhydꝋ becomes more exothermic
The enthalpy of hydration is more exothermic for smaller ions and for ions with a greater ionic charge
The polar water molecules will form ion-dipole bonds with the ions in solution (a) causing the ions to become hydrated (b)
energy cycle
energy level simple
energy level complicated
