Calculate the enthalpy change for the combustion of ethane (C2H6) using bond enthalpy data.

Bonds broken (reactants):

• 1 C-C bond: $1 \times 347 \text{ kJ/mol} = 347 \text{ kJ}$
• 6 C-H bonds: $6 \times 411 \text{ kJ/mol} = 2466 \text{ kJ}$
• $\frac{7}{2}$ O=O bonds: $3.5 \times 498 \text{ kJ/mol} = 1743 \text{ kJ}$ Total energy absorbed to break bonds = $347 + 2466 + 1743 = 4556 \text{ kJ}$

Bonds formed (products):

• 4 C=O bonds (in $2\text{CO}_2$): $4 \times 805 \text{ kJ/mol} = 3220 \text{ kJ}$
• 6 O-H bonds (in $3\text{H}_2\text{O}$): $6 \times 464 \text{ kJ/mol} = 2784 \text{ kJ}$ Total energy released when forming bonds = $3220 + 2784 = 6004 \text{ kJ}$

$$\Delta H = \sum (\text{Bond energies of bonds broken}) - \sum (\text{Bond energies of bonds formed})$$ $$\Delta H = 4556 \text{ kJ} - 6004 \text{ kJ} = -1448 \text{ kJ/mol}$$

The enthalpy change for the combustion of ethane is $\boxed{-1448 \text{ kJ/mol}}$.

d) (ii) Suggest why this value differs from the data book value of $-1560 \text{ kJ/mol}$ for the enthalpy of combustion of ethane.
The calculated value uses average bond enthalpies, which are values averaged across many different compounds. The actual bond energies in a specific molecule like ethane can differ slightly from these average values due to the specific molecular environment. Additionally, standard enthalpy of combustion values often refer to water in its liquid state, whereas bond enthalpy calculations typically assume all species are in the gaseous state.

Part (e): Unit Cells

e) (i) What is a centered unit cell?
A centered unit cell is a type of unit cell in a crystal lattice that contains constituent particles (atoms, ions, or molecules) at positions other than just the corners. These additional particles are located either at the center of the unit cell or at the center of its faces.

e) (ii) List two types of centered unit cells.

• Body-centered cubic (BCC)
• Face-centered cubic (FCC)