Question

Calculate the lattice energy for nacl(s) using a born-haber cycle

Answer 1

Answer:

Explanation:

For NaCl, the lattice dissociation enthalpy is +787 kJ mol-1. You should talk about "lattice formation enthalpy" if you want to talk about the amount of energy released when a lattice is formed from its scattered gaseous ions. For NaCl, the lattice formation enthalpy is -787 kJ mol-1.

Answer 2

Put this all together, with some data, and we get, for 1 mol of NaCl(s) : NaCl(s)→Na(s)+12Cl2(g) , −ΔHf,NaCl(s)=+411 kJ. Na(s)→Na(g) , ΔHsub,Na=107 kJ. Na(g)→Na+(g)+e− , IE1,Na(g)=502 kJ. 12Cl2(g)→Cl(g) , 12ΔHbond,Cl2(g)=12×242 kJ. Cl(g)+e−→Cl−(g) , EA1,Cl(g)=−355 kJ. Na+(g)+Cl−(g)→NaCl(s) ,