Thermodynamics Flashcards
Enthalpy change of formation
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
Na (s) + ½Cl2 (g) NaCl (s) [fH = - 411.2 kJ mol-1]
Enthalpy of atomization
The enthalpy of atomization of an element is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state
Na (s) Na(g) [atH = +148 kJ mol-1]
½ O2 (g) O (g) [atH = +249 kJ mol-1]
Enthalpy of sublimation
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 atomization
Na (s) Na(g) [subH = +148 kJ mol-1]
Bond dissociation enthalpy (bond energy)
The bond dissociation enthalpy is the standard molar enthalpy change when one mole of a covalent bond is broken into two gaseous atoms (or free radicals)
First ionization enthalpy
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
Second ionization enthalpy
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
First electron affinity
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
O (g) + e- O- (g) [ea 1H] = -141.1 kJ mol-1 ]
The first electron affinity is exothermic for atoms that normally form negative ions
This is because the ion is more stable than the atom, and there is an attraction between the nucleus and the electro
Second electron affinity
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
O – (g) + e- O2- (g) [ea 2H = +798 kJ mol-1]
The second electron affinity for oxygen is endothermic because it takes energy to overcome the repulsive force between the negative ion and the electron.
Enthalpy of lattice formation
The Enthalpy of lattice formation is the standard enthalpy change when 1 mole of an ionic crystal lattice is formed from its constituent ions in gaseous form
Na+ (g) + Cl- (g) NaCl (s) [LattH = -787 kJ mol-1]
Enthalpy of lattice dissociation
The Enthalpy of lattice dissociation is the standard enthalpy change when 1 mole of an ionic crystal lattice form is separated into its constituent ions in gaseous form
NaCl (s) Na+ (g) + Cl- (g) [ LattH = +787 kJ mol-1]
Enthalpy of hydration hydH
Enthalpy change when one mole of gaseous ions become aqueous ions
X + (g) + aq X + (aq) For Li+ hydH = -519 kJ mol-1 or X - (g) + aq X- (aq) For F- hydH= -506 kJ mol-1
This always gives out energy (exothermic, -ve) because bonds are made between the ions and the water molecules
Enthalpy of solution
The enthalpy of solution is the standard enthalpy change when one mole of 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
NaCl (s) + aq Na+ (aq) + Cl-(aq)
Born Haber cycles
The lattice enthalpy cannot be determined directly. We calculate it indirectly by making use of changes for which data are available and link them together in an enthalpy cycle the Born-Haber cycle
DRAW Born-Haber cycle: sodium Chloride
DRAW Born-Haber cycle: magnesium Chloride
Second electron affinity
second electron affinity for oxygen is endothermic because it take energy to overcome the repulsive force between the negative ion and the electron
Trends in Lattice Enthalpies
The strength of enthalpy of lattice formation depends on the following factors:
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
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).
Perfect Ionic Model
Theoretical lattice enthalpies assumes a perfect ionic model where the ions are 100% ionic and spherical and the attractions are purely electrostatic
When the negative ion becomes distorted and more covalent we say it becomes polarised. The metal cation is called polarising if it polarises the negative ion
When 100 % ionic the ions are spherical. The theoretical and the Born Haber lattice enthalpies will be the same
The charge cloud is distorted. The theoretical and the experimental Born Haber lattice enthalpies will differ
There is a tendency towards the covalent character in ionic substances when
- the positive ion is small
- the positive ion has multiple charges
- the negative ion is large
- the negative ion has multiple negative charges.
Differences between theoretical and Born Haber (experimental) lattice enthalpies
The Born Haber lattice enthalpy is the real experimental value
When a compound shows a covalent character, the theoretical and the born Haber lattice enthalpies differ. The more the covalent character the bigger the difference between the values.
When a compound has some covalent character- it tends towards giant covalent so the lattice is stronger than if it was 100% ionic. Therefore the Born-Haber value would be larger than the theoretical value.
Why does Calcium chloride have the formula CaCl2 and not CaCl or CaCl3?
We need to calculate an enthalpy of formation for each case. The one with the most exothermic enthalpy of formation will be the one that forms as it will be the most thermodynamically stable
Spontaneous process
A SPONTANEOUS PROCESS (e.g. diffusion) will proceed on its own without any external influence
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 a number of molecules between products and reactants.
Calculate entropy
∆S entropy. ˚ = Σ S˚products - ΣS˚reactants
Calculate ∆S˚ for the following reaction at 25˚C: 2Fe2O3 (s) + 3C (s) 4Fe (s) + 3CO2 (g)
S [Fe2O3] = 87.4 J K-1 mol-1
S [C] = 5.7 J K-1 mol-1
S [Fe] = 27.3 J K-1 mol-1
S [CO2] = 213.6 J K-1 mol-1
∆S˚ = S˚products - S˚reactants
= (3 x 213.6 + 4 x 27.3) – (2 x 87.4 + 3 x 5.7) = + 558.1 J K-1 mol-1 = + 558 J K-1 mol-1 (3 S.F.)
∆G
∆G = ∆H - T∆S
For any spontaneous change, ∆G will be
negative
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.
High activation energy
Catalyst
Calculating the temperature a reaction will become feasible
T = ∆H / ∆S
Hsolution
2 equation
Hsolution = HL dissociation + sum hydH
H solution = - HL formation + sum hydH
Why hydration enthalpies are exothermic
Hydration enthalpies are exothermic as energy is given out when 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
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.
When a solid dissolves
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
