May 2019 Paper 2 Question 1

a) Explain the difference between:

i) the standard bond enthalpy and the standard enthalpy of atomisation of chlorine gas

Standard bond enthalpy: $Cl_2\rightarrow 2Cl$

The energy required to break 1 mole of bonds at standard room temperature and pressure.

Standard enthalpy of atomisation: $\frac{1}{2}Cl_2\rightarrow Cl$

The energy absorbed when 1 mole of gaseous atoms are formed from its original state at standard temperature and pressure.

ii) the standard enthalpy of formation and the standard lattice enthalpy of sodium chloride

Standard enthalpy of formation: $Na{(s)}+\frac{1}{2}Cl_2_{(g)}\rightarrow NaCl{(s)}$

The energy released or absorbed when 1 mole of a compound is formed from its components at standard temperature and pressure.

Standard lattice enthalpy: $Na^+_{(g)}+Cl^-_{(g)}\rightarrow NaCl_{(s)}$

The energy released when 1 mole of a compound is formed by reacting the ions at gaseous state at standard temperature and pressure.

b) Construct a Born Haber Cycle for NaCl

c) The enthalpy change of solution od sodium chloride is +3.87 kJ mol^-1. The hydration of sodium ions is -424 kJ mol^-1 whilst that of chloride is -364 kJ mol^-1. From this information, calculate the lattice enthalpy of sodium chloride.

$\Delta H_{solution}=\Delta H_{lattice\; dissociation}+\Delta H_{hydration}$

$3.87=\Delta H_{lattice\: dissociation}+(-424)+(-364)$

$\Delta H_{lattice\: dissociation}=784.13kJ$

The lattice enthalpy is equivalent to the lattice dissociation though it will be in the opposite direction.

Therefore the lattice enthalpy is -784.13kJ

d) Although the enthalpy of solution of sodium chloride is endothermic, the compound is highly soluble. Explain this observation.

$\Delta G=\Delta H-T\Delta S$ When a solid forms a solution the entropy increases. This increase happens for 2 reasons

from 1 mole of solid 2 moles of aqueous ions are formed.
an aqueous ion is more disordered than a solid.

Since $\Delta H$ is small, $T\Delta S$ will be bigger than $\Delta H$ which would give a negative value for $\Delta G$ , a value that would make the process spontaneous.