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Calculate the lattice energy of NaCl. Heat of sublimation is 108kj/mol and Heat

ID: 612057 • Letter: C

Question

Calculate the lattice energy of NaCl. Heat of sublimation is 108kj/mol and Heat of NaCl is -411kj/mol. Energy to dissociate .5 mole of Cl2to an atom of Cl is121.4 kj/mol.

Explanation / Answer

These are the changes for each individual element that must occur before formation of a salt: Na(s) --> Na(g) ---> Na+(g) 109kJ + 496kJ = 605kJ for this process 1/2 Cl2(g) --> Cl(g) --> Cl-(g) 1/2(243)kJ + (-349kJ) = -227.5kJ for this process Finally, both ionic gases combine together to form NaCl(s) and this is known as the lattice energy, which is what we're solving for. The overall equation will look like this: -411kJ = 605kJ - 227.5 + E(lattice) where the term on the left side corresponds to the enthalpy of formation and it is the overall process (summation of all the steps): Na(s) + 1/2Cl2(g) --> NaCl(s) H = -411kJ So, the lattice energy will be: E(lattice) = -411kJ - 605kJ + 227.5kJ = -788.5kJ

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