THermodynamics

Question 1

Enthalpy change of formation

Formula for enthalpy can’t ef formation for NaCl

Answer 1

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]

Question 2

Enthalpy of atomisation
Formula for
Na
O2

Answer 2

The enthalpy of atomisation 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]

Question 3

Bond dissociation enthalpy (bond energy)

Answer 3

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)
Cl2 (g)  2Cl (g) dissH = +242 kJ mol-1 Or
CH4 (g)  CH3 (g) + H(g) dissH = +435 kJ mol-1

Question 4

First ionisation enthalpy

Answer 4

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
Mg (g)  Mg+
(g) + e- [ IE 1H]

Question 5

Second ionisation enthalpy

Formula for Mg +

Answer 5

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.
Mg+(g) -> Mg 2+ (g) + e- [IE 2H]

Question 6

First electron affinity

Answer 6

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

Question 7

Second electron affinity

Answer 7

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 take energy to overcome the repulsive force
between the negative ion and the electron

Question 8

Enthalpy of lattice formation

Answer 8

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

Question 9

Enthalpy of lattice dissociation

Answer 9

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

Question 10

Enthalpy of hydration

Answer 10

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.

Question 11

Enthalpy of solution

Answer 11

The enthalpy of solution is the standard enthalpy change when one mole of an 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 

Question 12

Define ATOMIC MASS

Answer 12

The average mass of ELEMENTS atoms, compared to (/)

1/12th the mass of a carbon-12 atom

Question 13

DEFINE ACTIVATION ENERGY

Answer 13

he energy needed to ‘activate’ the reactant particles in order for them to collide effectively and cause a chemical reaction