Chemistry Labs

Problem 2

(a) In the NaCl crystal, NaX+\ce{Na+} and ClX−\ce{Cl-} each form a face-centred cubic lattice; the ionic radii are r(NaX+)=0.102r(\ce{Na+}) = 0.102 nm and r(ClX−)=0.181r(\ce{Cl-}) = 0.181 nm. Give the numbers of NaX+\ce{Na+} and ClX−\ce{Cl-} per unit cell, their coordination numbers, and calculate the crystal density. (b) From the Born\–Haber data \— ΔfH(NaCl(s))=−411\Delta_fH(\ce{NaCl(s)}) = -411, ΔsubH(Na)=+109\Delta_{sub}H(\ce{Na}) = +109, IE(Na)=+496IE(\ce{Na}) = +496, D(ClX2)=+242D(\ce{Cl2}) = +242, EA(Cl)=−349EA(\ce{Cl}) = -349 kJ mol−1^{-1} \— write the equations for the formation step and the direct dissociation NaCl(s)→NaX+(g)+ClX−(g)\ce{NaCl(s) -> Na+(g) + Cl-(g)}, and calculate the lattice formation enthalpy. (c) The Solvay process achieves 2 NaCl+CaCOX3→NaX2COX3+CaClX2\ce{2NaCl + CaCO3 -> Na2CO3 + CaCl2} through the cycle: CaCOX3→ΔA+B\ce{CaCO3 ->[\Delta] A + B}; NaCl+NHX3+B+HX2O→C+D\ce{NaCl + NH3 + B + H2O -> C + D}; 2 C→ΔNaX2COX3+HX2O+B\ce{2C ->[\Delta] Na2CO3 + H2O + B}; A+HX2O→E\ce{A + H2O -> E}; E+2 D→CaClX2+2 HX2O+2 NHX3\ce{E + 2D -> CaCl2 + 2H2O + 2NH3}. Identify compounds A\–E.
Step 2 of 3: Born–Haber lattice enthalpy
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ΔlatH=ΔfH−ΔsubH−IE−12D(ClX2)−EA=−411−109−496−121+349=−788 kJ mol−1\Delta_{lat}H = \Delta_fH - \Delta_{sub}H - IE - \tfrac{1}{2}D(\ce{Cl2}) - EA = -411 - 109 - 496 - 121 + 349 = -788\ \text{kJ mol}^{-1}
Analysis

The cycle Na(s)+12ClX2(g)→NaX+(g)+ClX−(g)→NaCl(s)\ce{Na(s)} + \tfrac{1}{2}\ce{Cl2(g)} \rightarrow \ce{Na+(g)} + \ce{Cl-(g)} \rightarrow \ce{NaCl(s)} must equal ΔfH\Delta_fH, so ΔlatH=−788\Delta_{lat}H = -788 kJ mol−1^{-1} (equivalently the reverse dissociation step F is +788+788 kJ mol−1^{-1}).