Thermodynamics Flashcards
Hess Law
enthalpy change for a chemical reaction is the same regardless the route taken from the reactants to products
standard enthalpy of formation
-enthalpy change when 1 mole of a compound is formed
-from its elements under standard conditions
-all reactants and products in their standard states
standard enthalpy of combusion
-enthalpy change when 1 mole of compound is completely burned in oxygen
-under standard conditions
-all reactants and products in their standard states
standard enthalpy of atomisation
-enthalpy change when 1 mole of gaseous atoms
-formed from element in its standard state
mean bond enthalpy
- enthalpy change when 1 mole of gaseous molecule
-each break covalent bond to form 2 free radicals
-averaged over a range of compounds
relationship between bond enthalpy + atomisation
atomisation x2 = bond enthalpy
first ionisation enthalpy
-enthalpy change when 1 mole of electrons
-removed from 1 mole of gaseous atoms
-to produce 1 mole of gaseous ions
-each which a single positive charge
second ionisation enthalpy
-enthalpy change when 1 mole of electrons
-removed from 1 mole of gaseous 1+ ions
-to produce 1 mole of gaseous ions
-each which a 2+ positive charge
first electron affinity
-standard enthalpy change when 1 mole of gaseous atoms
-converted into 1 mole of gaseous ions
-each with a single negative charge
-under standard conditions
second electron affinity
-standard enthalpy change when 1 mole of electrons is added to
-1 mole of gaseous ions
-each with a single negative charge
-to form 1 mole of gaseous ions with 2- charge
lattice formation enthalpy
- standard enthalpy change when 1 mole of solid ionic compound is formed
-from its gaseous ions
lattice formation enthalpy
-standard enthalpy change when 1 mole of solid ionic compound is formed
-from its gaseous ions
lattice dissociation enthalpy
-standard enthalpy when 1 mole of solid ionic compound
-dissociates into its gaseous ions
standard enthalpy of hydration
-standard enthalpy change when 1 mole of gaseous ions
-converted into 1 mole of aqueous ions
standard enthalpy of solution
-standard enthalpy change when 1 mole of solute dissolves
[-in enough solvent to form a solution
-ions are far apart to not interact with each other]
mean bond enthalpy equation
mean bond enthalpy = sum of [reactants]- sum of [products]
two factors which determine how exothermic a lattice enthalpy is
- charge on ion [greater the charge, greater its attraction to an oppositely charged ion]
- size of ion [smaller the ion, greater the attraction to an oppositely charged ion]
top tip:
the more exothermic, the stronger the ionic bonds
Covalent character eg. [Zn]2+ and [Se]2-
-Zn ion is quite small, strong + charge
-Se ion is quite large, strong - charge
-two form ionic attraction due to electrostatic attraction
-Zn ion is strongly polarising
-Se electron cloud is distorted
-some of the e- density is distorted
-this is covalent character
ionic bonding
strong electrostatic attraction between oppositely charged ions
perfect ionic model and features
-purely ionic bonding w no covalent character
-ions act as point charges
-ions are perfect spheres which cannot be distorted
what is the effect of covalent character on lattice enthalpy
- more exothermic
-stronger bonding
solid to liquid
inc or dec of entropy
increase in entropy
what is happening in terms of bonds when dissolving ionic compounds
- breaking ionic bonds
- form bonds between water molecules + ions
mixing 2 liquids
inc or dec of entropy
increase in entropy
3 factors which determine if a reaction is feasible
-temperature
-enthalpy
-entropy
entropy key knowledge
-units
-definition
-delta?
-it is delta S
-delta S is the change in entropy
-units are JK-1mol-1
-measure of disorder
evaporating a liquid
inc or dec of entropy
increase in entropy
more moles produced
inc or dec of entropy
increase in entropy
entropy equation
change in entropy = sum [products]-sum [reactants]
standard conditions for entropy
100 KPa
298 K
1 mol dm-3
feasible definition [2]
-reaction can happen at a given temp
- feasible when delta G is less then or = to 0
Gibbs free energy equation
*G = H-[TxS}
G= gibbs free energy KJ mol-1
H= enthalpy change KJ mol-1
T= temperature Kelvin
S= entropy change kJK-1mol-1
