Thermodynamics Flashcards
What is the Born-Haber cycle? When is it used?
‘Special type of Hess’s law cycle’
- Used to calculate the lattice enthalpy of ionic compounds
Define atomisation
Formation of 1 mole of gaseous atoms
Define Ionisation
Formation of gaseous ions
Define standard enthalpy change of atomisation
The enthalpy change when 1 mole of gaseous atoms is formed under standard conditions
OR
The enthalpy change for this reaction:
Ca(s) ——> Ca(g)
Is atomisation ENDO or EXO. Why?
ENDO
because, bonds are being broken
Define standard enthalpy change of the 1st I.E
(1st I.E)
the enthalpy change of this reaction:
Ca(g) ——> Ca+ (g) + e-
Is the 1st I.E ENDO or EXO
Why?
ENDO
because, energy is needed to overcome the electrostatic attraction between the nucleus and the electron being removed
Define standard enthalpy change of the 2nd I.E
The enthalpy change for this reaction:
Ca+(g) ——> Ca2+(g) + e-
Is the 2nd I.E ENDO or EXO
Why?
MORE ENDO
because, more energy is needed to overcome the electrostatic attraction between the nucleus and the electron being removed
Define the standard enthalpy of the first electron affinity
The enthalpy change for this reaction:
F(g) + e- ——> F- (g)
Is the first electron affinity ENDO or EXO
Why?
EXO
because, there is an electrostatic force of attraction between the nucleus of the gaseous atom and the electron being added
Define the standard enthalpy of the second electron affinity
The enthalpy change for this reaction:
O-(g) + e- ——> O2- (g)
Is the second electron affinity ENDO or EXO
Why?
ENDO
because energy is needed to overcome the repulsion between the negatively charged gaseous ion and the negatively charged electron being added
Define Enthalpy of Lattice Formation
The enthalpy change when 1 mole of an ionic solid is formed from the gaseous ions under standard conditions
OR
the enthalpy change for this reaction:
K+(g) + F-(g) ——> KF(s)
Is the lattice of formation ENDO or EXO
Why?
EXO
because, there is a strong electrostatic force of attraction between the oppositely charged ions