entropy

what are standard conditions?

what are standard conditions?

a temperature of 298 K and a pressure of 100 kPa

lattice dissociation enthalpy (ΔH_latt^ꝋ)

the standard enthalpy change when 1 mol of a solid ionic lattice is separated into its (constituent) gaseous ions

enthalpy of lattice formation (ΔH_latt^ꝋ)

standard enthalpy change when one mole of ionic lattice is formed from its (constituent) gaseous ions

atomisation enthalpy (ΔH_at^ꝋ)

• enthalpy change when 1 mole of gaseous atoms is formed from its element in its standard state, under standard conditions

Enthalpy of sublimation

The enthalpy change for a solid metal turning to gaseous atoms

• numerically has the same value as enthalpy of atomisation

electron affinity (ΔH_ea^ꝋ)

• the enthalpy change when 1 mol of gaseous atoms gain 1 mol of electrons to form 1 mol of gaseous ions with a -1 charge

• first electron affinity always exothermic

• however second electron affinity can be exothermic

second electron affinity

the enthalpy change when one mol of gaseous -1 ions gain 1 electron per ion to form gaseous -2 ions

why is the second electron affinity for oxygen endothermic?

because it takes energy to overcome the repulsive force between the negative ion and the electron

First Ionisation enthalpy (ΔH_ie^ꝋ)

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 ionisation energy

the enthalpy change required to remove 1 mole of electrons from one mole of gaseous +1 ions to produce 1 mole of gaseous 2+ ions

enthalpy of formation (ΔH_f^ꝋ)

the enthalpy change when 1 mole of a compound is formed from its elements under standard conditions, with all reactants and products being in their standard states

enthalpy of solution

the standard enthalpy change when one mole of ionic solid is dissolved in a large enough amount of water to ensure that the dissolved ions are well separated and do not interact w each other

how to calculate enthalpy of solution

• enthalpy of solution = lattice dissociation enthalpy + enthalpy of hydration

OR

• enthalpy of solution = -(lattice formation enthalpy) + enthalpy of hydration

enthalpy of hydration

the enthalpy change when one mole of gaseous ions is dissolved in water to form one mole of aqueous ions under standard conditions

• always exothermic bc bonds are made between the ions and the water molecules

Bond dissociation energy/bond energy/bond enthalpy (E)

the standard molar enthalpy change when 1 mole of a covalent bond is broken into 2 gaseous atoms (or free radicals)

perfect ionic model

perfect ionic mode assumes that:

• all the ions are perfectly spherical

• the ions display no covalent character

  • (exam question extra) — if an ionic lattice contains covalent character, it means that the forces holding the lattice together are stronger

entropy definition

a measure of disorder in a system

what is entropy measured in?

• symbol:S

JK-1mol-1

• for most calculations, will have to divide by 1000 to convert to kJ

increase in entropy

system becomes energetically more stable

systems with higher entropy

will be energetically more favourable

second law of thermodynamics

entropy of an isolated system always increases

• its overwhelmingly more likely for molecules to be disordered than ordered

equation for entropy

ΔS_system^ꝋ = ΣΔS_products^ꝋ - ΣΔS_reactants^ꝋ

if a reaction is feasible/spontaneous

whether a reaction will occur or not (on its own, without input from us)

• there may be an activation energy barrier

the point at which a reaction become feasible

when ΔG=0

Gibbs-free energy

ΔG^ꝋ = ΔH_reaction^ꝋ - TΔS_system^ꝋ

• units of ΔG^ꝋ = kJ mol^-1

• units of ΔH_reaction^ꝋ = kJ mol^-1

• units of T = K

• units of ΔS_system^ꝋ = J K^-1 mol^-1

  • therefore must be converted to kJ K^-1 mol^-1 by dividing by 1000

what are the limitations of using G as an indicator of whether a reaction will occur?

Gibbs-free energy only indicates if a reaction is feasible

• doesn’t take into account the rate of reaction

for Gibbs-free energy equations that involve graphs

most of the time you have to equate ΔG = ΔH - TΔS. to y=mx+c

ΔS: gradient

ΔH: y-intercept

When ΔG^ꝋ is negative

the reaction is feasible

When ΔG^ꝋis positive

the reaction is not feasible