Problem 2

(a) In the NaCl crystal,

\ce{Na+}

and

\ce{Cl-}

each form a face-centred cubic lattice; the ionic radii are

r(\ce{Na+}) = 0.102

nm and

r(\ce{Cl-}) = 0.181

nm. Give the numbers of

\ce{Na+}

and

\ce{Cl-}

per unit cell, their coordination numbers, and calculate the crystal density. (b) From the Born\–Haber data \—

\Delta_fH(\ce{NaCl(s)}) = -411

\Delta_{sub}H(\ce{Na}) = +109

IE(\ce{Na}) = +496

D(\ce{Cl2}) = +242

EA(\ce{Cl}) = -349

kJ mol

^{-1}

\— write the equations for the formation step and the direct dissociation

\ce{NaCl(s) -> Na+(g) + Cl-(g)}

, and calculate the lattice formation enthalpy. (c) The Solvay process achieves

\ce{2NaCl + CaCO3 -> Na2CO3 + CaCl2}

through the cycle:

\ce{CaCO3 ->[\Delta] A + B}

;

\ce{NaCl + NH3 + B + H2O -> C + D}

;

\ce{2C ->[\Delta] Na2CO3 + H2O + B}

;

\ce{A + H2O -> E}

;

\ce{E + 2D -> CaCl2 + 2H2O + 2NH3}

. Identify compounds A\–E.

Step 3 of 3: Solvay intermediates

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Intuition

Ammonia is the recyclable catalyst-like reagent: it drives $\ce{NaHCO3}$ precipitation and is released again by $\ce{Ca(OH)2}$ .

A = \ce{CaO},\quad B = \ce{CO2},\quad C = \ce{NaHCO3},\quad D = \ce{NH4Cl},\quad E = \ce{Ca(OH)2}

Analysis

Calcination gives $\ce{CaO}$ and $\ce{CO2}$ ; bubbling $\ce{CO2}$ and $\ce{NH3}$ into brine precipitates $\ce{NaHCO3}$ leaving $\ce{NH4Cl}$ ; heating $\ce{NaHCO3}$ regenerates $\ce{CO2}$ and yields $\ce{Na2CO3}$ ; slaking CaO gives $\ce{Ca(OH)2}$ , which decomposes $\ce{NH4Cl}$ and recycles $\ce{NH3}$ .