Enthalpy, entropy and free energy

• lattice enthalpy more exothermic • more energy required to overcome electrostatic attraction • higher melting point • higher thermal stability

• lattice enthalpy becomes more exothermic • more energy required to overcome electrostatic attraction • higher melting point 4.higher thermal stability

distortion of electron cloud in molecule/ion by nearby positive charge

enthalpy change of formation= atomisation metal + ionisation energy metal + atomisation gas + electron affinity + lattice enthalpy

m(s) -> m(g) 1/2x2(g) -> x(g) enthalpy change when 1 mol of gaseous atoms is formed from the element in its standard state

m(g) -> m+(g) + e- enthalpy change to remove 1 electron from each atom in 1 mol of gaseous atoms to form 1 mol of gaseous 1+ ions

X(g) + e- -> X-(g) The enthalpy change accompanying the gain of 1 mol of electrons by 1 mol of gaseous atoms

M(s) + X2(g) -> MX2(s) Enthalpy change when when 1 mol of a compound is formed from its constituent elements in their standard state

aM^b+(g) + bX^a-(g) -> MaXb(s) enthalpy change when 1 mol of ionic lattice is formed from its constituent gaseous ions 298k and 10KPa

enthalpy change when 1 mol of gaseous ions is completely hydrated by water X^a+(g) -> X^n+(aq)

• smaller ionic radius • larger ionic charge • increased electrostatic attraction to water molecules

compound(s) -> X+(aq) + M-(aq) enthalpy change when 1mol of ionic solid dissolves in water

enthalpy of hydration = lattice enthalpy + enthalpy of solution

measure of dispersal energy in a system ΔS KJmol^-1s^-1

• solid becomes liquid • liquid becomes gas • temperature rises • solid dissolves in liquid to form a solution • products have more molecules in same state

• low • small degree of freedom • stuck vibrating particles

• greater degree of freedom of movement than solid • disperse more energy

• greatest degree of freedom of movement • disperse the most energy

entropy of products - entropy of reactants KJmol^-1K^-1

ΔS system + ΔS surroundings

ΔH -TΔS units KJmol^-1

Δ<0

ΔG=0 0=ΔH - TΔS T = ΔH/ΔS

273K