Topic 15: Energetics and Themochemistry (HL)

Question 1

Born-Haber cycle

Answer 1

A series of reactions that can be combines to determine the enthalpy of formation of an ionic compound

Question 2

Constructing a Born-Haber cycle

Answer 2

Step 1: Lattice enthalpy
MX (s) → M^(+)(g) + X^(-)(g) ∆Hlat > 0
Process is endothermic

Step 2: Enthalpy of atomization
M(s) → M(g) ΔHat>0
1/2X2(g) → X(g) ΔHat>0

Step 3: Ionization energy
IE1: M(g) → M+(g) +e ∆HIE>0
IE2: M+(g) → M2+(g) + e ∆HIE2>0

Step 4: Electron affinity
X(g) + e → X (g) ∆HEA<0

Question 3

Lattice enthalpy

Answer 3

The standard enthalpy change that occurs on the formation of 1 mol of gaseous ions from the solid lattice

MX (s) → M^(+)(g) + X^(-)(g) ∆Hlat > 0

Directly affected by both the charge on the ions and the atomic radii, ie. an increase in ionic charge equals greater attraction between oppositely charged ions, increasing the energy required to break apart the ionic lattice.
As the ionic radii increases, the electrostatic attraction decreases, ie decrease in lattice enthalpy

Question 4

Enthalpy of atomization

Answer 4

The standard enthalpy change
that occurs on the formation of 1 mol of separate gaseous atoms of an element in its standard state

M(s) → M(g) ΔHat>0

1/2X2(g) → X(g) ΔHat>0

Question 5

Electron affinity

Answer 5

The standard enthalpy change

on the addition of 1 mol of electrons to 1 mol of atoms in the gaseous phase

Question 6

Standard enthalpy change of solution

Answer 6

The change in

enthalpy when 1 mol of a substance is dissolved in a large excess of a pure solvent

Question 7

Enthalpy change of hydration

Answer 7

For an ion is the enthalpy
change when 1 mol of the gaseous ion is added to water to form a dilute solution.

Always has a negative value

M+(g) → M+(aq) ∆Hhyd= - kJ mol 1

X (g) → X (aq)
∆Hhyd= - kJ mol 1

Question 8

Solvation

Answer 8

term used in place of hydration for solvents other than water

Question 9

Spontaneous reaction

Answer 9

When a reaction moves towards either completion or equilibrium under a given set of conditions without external intervention.

Occur at different rates.

Either endothermic or exothermic

Lead to an increase in the total entropy within the system and surroundings.

Question 10

Entropy, S

Answer 10

A measure of the distribution of total available energy between the particles.

The greater the shift from energy being localized to being widespread amongst the particles, the lower the chance of the particles returning to their original state and the higher the entropy of the system.

Increase in entropy as the particles gain more freedom of movement, ie solid to liquid to gas.

Question 11

How to determine entropy?

Answer 11

It’s a state function, so a change in entropy is determined by the difference between its final and initial values:

∆ S 298(reaction) = ∑∆ S 298 (products) - ∑∆ S 298 (reactants)

Question 12

Second law of thermodynamics

Answer 12

Chemical reactions that result in an overall increase in the entropy of the universe are spontaneous.

When the overall entropy of the universe remains unchanged, the system is in equilibrium.

∆Stotal = ∆Ssystem + ∆Ssurroundings

Question 13

∆Stotal > 0

Answer 13

spontaneous

Question 14

∆Stotal = 0

Answer 14

equilibrium

Question 15

∆Stotal < 0

Answer 15

non-spontaneous

Question 16

Gibbs free energy, G

Answer 16

A state function, along with enthalpy H, entropy S, and absolute temperature T.

G = H - TS 
∆G = ∆H - T∆S

For a reaction to be spontaneous the Gibbs free energy must have a negative value (∆G < 0) .

Question 17

Gibbs free energy change of formation, ΔGf

Answer 17

Free energy change when 1 mol of a compound is formed from its elements under standard conditions of 298 K and a pressure of 100 kPa:

ΔGr = ΣΔGf(products) - ΣΔGf (reactants)