a) (i) Describe, using the kinetic theory of matter, what happens when potassium chloride dissolves in water.
(ii) Give a reason why the process in (a) (i) is endothermic.
(b) (i) An underground iron pipe is less likely to corrode if it is bonded at intervals with magnesium rods. Give reasons for this observation.
(ii) State the stages involved in the rusting of iron.
(iii) State the condition for the rusting of iron in water.
(a)(i) The moving, polar water molecules collide with and surround the K+ and Cl- ions at the surface of the crystal, overcoming the electrostatic forces of the lattice. The ions are pulled away (hydrated) and, because of their kinetic motion, diffuse and become evenly spread throughout the water.
(a)(ii) The process is endothermic because the energy absorbed to break up the ionic lattice (lattice energy) is greater than the energy released when the ions are hydrated (hydration energy); the net heat is taken in from the surroundings.
(b)(i) Magnesium is more reactive (more electropositive) than iron, so it acts as a sacrificial anode: it corrodes in preference to the iron, protecting the pipe (cathodic protection).
(b)(ii) Stages of rusting: iron is oxidised at anodic areas, \( \text{Fe} \to \text{Fe}^{2+} + 2e^- \); oxygen and water are reduced at cathodic areas, \( \text{O}_2 + 2\text{H}_2\text{O} + 4e^- \to 4\text{OH}^- \); the Fe2+ is further oxidised by oxygen to hydrated iron(III) oxide (rust).
(b)(iii) Both water and oxygen (air) must be present.
(c)(i) A spontaneous reaction is one that, once started, proceeds on its own without a continuous supply of external energy.
(c)(ii) A negative enthalpy change (exothermic) and an increase in entropy/disorder (giving a negative free-energy change, \(\Delta G < 0\)).
(c)(iii) Sodium is more electropositive (more reactive) than calcium and loses its outer electron more readily; also 1 g of sodium contains more atoms than 1 g of calcium (Na = 23, Ca = 40), so it reacts faster.
(c)(iv) \[ 2\text{Na} + 2\text{H}_2\text{O} \to 2\text{NaOH} + \text{H}_2 \] \[ \text{Ca} + 2\text{H}_2\text{O} \to \text{Ca(OH)}_2 + \text{H}_2 \]
(d) Mass of PbCO3
\( M(\text{PbCO}_3) = 207 + 12 + 48 = 267 \). In 267 g there is 207 g Pb, so:
\[ \text{mass} = \frac{267}{207} \times 35.0 = 45.1\ \text{g} \]
(e) (i) Fluorine: van der Waals (dispersion) forces. (ii) Hydrogen fluoride: hydrogen bonding.