Hess’s Law

The Born Haber cycle is used to determine the heat of formation of a given compound.

Example 1

i. Draw the Born Haber cycle for the formation of a compound:

Solution:

∴ ΔH°$_f$ = Es + EI + Eat + Eaff + EL

Es: Sublimation energy

EI: Ionization energy

Eat: Atomization energy

Eaff: Electron affinity energy

EL: Lattice energy

Example 2

Draw the Born Haber cycle for the formation of the following compounds:

Aluminium Chloride (AlCl$_3$):

∴ ΔH°$_f$ = Es + EI + Eat + Eaff + EL

Aluminium Oxide (Al$_2$O$_3$):

∴ ΔH°$_f$ = Es + EI + Eat + Eaff + EL, where:

• EI = 1st EI + 2nd EI + 3rd EI
• Eaff = 1st Eaff + 2nd Eaff

Calcium Iodide (CaI$_2$):

∴ ΔH°$_f$ = Es + EI + Eat + Eaff + EL

Potassium Bromide (KBr):

∴ ΔH°$_f$ = Es + EI + Eat + Eaff + EL

Example 3

(a) Define the Born Haber cycle as applied in energetics:

Answer: The Born Haber cycle is used to determine the heat of formation of ionic compounds by involving intermediate changes.

(b) Construct the Born Haber cycle for the formation of KCl:

∴ ΔH°$_f$ = Es + EI + Eat + Eaff + EL

Example 4

(a) Define the following:

1. Atomization Energy: The energy absorbed when a molecule or element is converted to gaseous atoms.
2. Ionization Energy: The energy required to remove an electron from the outermost shell of a gaseous atom.
3. Electron Affinity: The energy released when one mole of gaseous atoms combines with one mole of electrons.

(b) Use the given data to calculate the electron affinity of chlorine:

Solution

∴ E$_{aff}$ = ΔH°$_f$ – (Es + 1st EI + 2nd EI + Eat + EL)

Where

From,

∴ ΔH°$_f$ = Es + 1st EI + 2nd EI + Eat + Eaff + E

Eaff = ΔH°$_f$ – (Es + 1st EI + 2nd EI + Eat + EL)

Example 5

(a) Draw a well-labeled Born Haber cycle for the formation of Al$_2$O$_3$:

∴ ΔH°$_f$ = Es + EI + Eat + Eaff + EL

Where by, Es is the sublimation energy. EI is the Ionization energy. Eat is the Atomization energy. Eaff is the Electronic affinity energy. EL is the lattice energy.

Reactants are atomized (changed to atoms) during a reaction, and product energy is calculated from bond energy.

Example 1

Calculate the heat of formation of CH$_4$ given the enthalpy of atomization of carbon and hydrogen:

$$\begin{aligned} \text{C} - \text{C} &= 348 \ \text{kJ mol}^{-1} \\ \text{C} - \text{H} &= 416 \ \text{kJ mol}^{-1} \\ \text{H} - \text{H} &= 436 \ \text{kJ mol}^{-1} \end{aligned}$$

Solution

Example 2

Use the information provided in Example 1 to calculate the heat of formation of ethane (C$_2$H$_6$):

Solution

F.B.E

Calorimetry: A method to determine the heat change of a reaction using a calorimeter.

Determination of enthalpies of neutralization by calorimetry

Neutralization Reaction: A reaction between an acid and a base to produce salt and water.

Example

In this method the density of salt solution formed is assumed to be equal to the density of water.

Quantity of that in this method can be calculate by using heat capacity ( C ) and by using specific heat capacity ( c ).

Let the quantity of heat be Q.

By using heat capacity

C = heat capacity.

By using specific heat capacity.

Where by $\Delta T$ is change in temperature.

c = specific heat capacity.

If quantity of heat calculated is the same and $\Delta T$ is also the same, then the relationship between c and C can be,

Enthalpy of neutralization is calculate with respect to the number of moles of water producer.

Reaction.

But number of moles of water depends on the moles of limiting reagent

Example

250 cm$^3$ of NaOH of 0.4M were added to 250 cm$^3$ of the HCl of 0.4M in a calorimeter. The temperature of the two solutions and the calorimeter was 17.05 °C. The mass of calorimeter was 50g and its specific heat capacity was after the reaction the temperature rose to 19.5 °C

Assuming the specific heat capacity of all the solution is. Calculate the standard enthalpy of neutralization.

Solution

Data

Reaction

Limiting reagent

Base

Acid The same 0.1 mol Both are limiting reagent. Moles of water produced

.