Q.v. What is the enthalpy of atomization? Give an example.

A combined question spanning thermodynamic definitions, $IF_7$'s hybridisation/geometry, Henry's law with a molality calculation (or, in the OR branch, carbon-reduction metallurgy, sulphur oxoacid structures, and rate-law reasoning for a second-order reaction).

(i)(a) Enthalpy of atomization ($\Delta_aH$): the enthalpy change accompanying the complete conversion of 1 mole of a substance into gaseous atoms, breaking all the bonds present; it is always positive (endothermic), since energy must be supplied to break bonds. E.g. $CH_4(g) \rightarrow C(g) + 4H(g)$.

(i)(b) Enthalpy of vaporization ($\Delta_{vap}H$): the enthalpy change when 1 mole of a liquid is converted completely into its vapour at its boiling point, at constant temperature and pressure; also always positive, since energy is needed to overcome intermolecular forces of attraction in the liquid.

(ii) $IF_7$: iodine heptafluoride has iodine as the central atom surrounded by 7 fluorine atoms with no lone pairs. Geometry: pentagonal bipyramidal — 5 F atoms arranged in an equatorial pentagon around the central I, with the remaining 2 F atoms occupying the axial positions above and below the plane. Hybridisation: $sp^3d^3$ (using the $5s$, three $5p$, and three $5d$ orbitals of iodine to accommodate 7 bond pairs).

(iii)(a) Henry's law: at a constant temperature, the solubility (mole fraction) of a gas in a liquid is directly proportional to the partial pressure of that gas in equilibrium with the solution: $p = K_H\,x$, where $K_H$ is the Henry's law constant (specific to the gas-solvent pair and temperature) and $x$ is the mole fraction of the dissolved gas.

(iii)(b) Molality of urea solution: moles of urea $= \dfrac{22.22\ g}{60\ g\,mol^{-1}} = 0.3703\ mol$. Molality $= \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}} = \dfrac{0.3703}{0.300} \approx 1.234\ mol\,kg^{-1}$.

OR — (i) Action of carbon on metal oxides:

1. In the blast furnace, carbon (via the CO it generates: $2C+O_2\rightarrow 2CO$) reduces $Fe_2O_3$ to metallic iron: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$.
2. In the vertical retort furnace, carbon directly reduces zinc oxide (around 1673 K) to metallic zinc, which distils off as vapour and is condensed: $ZnO + C \rightarrow Zn + CO$. …