Given 1mol of Na+ and 1 mol of Cl, calculate the energy released when the Na and

Question

Given 1mol of Na+ and 1 mol of Cl, calculate the energy released when the Na and Cl ions condense as

For noninteracting ion pairs;i.e, consider only pairwise interactions

Ebond= (z1*z2*e2)/(4*3.14*e0*r0)

z1= 1

z2= -1

e0=8.85*10-12 which is the permittivity of a vacuum

r0= 283 pm or 283*10-12m which is the distance between ions

Ebond= (1*-1*e2)/4*3.14*8.85*10-12*283*10-12)=-490KJ/mol

(a) Noninteracting ion squares; i.e., every four ions, 2Na and 2C1 interact with each other, but not with others. Answer: -633kJ/mol

(b)1/8 unit cell of NaCl; i.e., 8 atoms interact. Answer: -713kJ/mol

(c) Compare with -755kJ/mol for solid NaCl lattice. Hint: Make sure you include all pairwise attractions and repulsions. You can assume the Born exponent n = oo.

Explanation / Answer

GIven
eo = 8.85*10^-12
ro = 283*10^-12 m

a. for non interacting squares
   net energy released = [ke^2][4/ro - 2/ro*sqroot(2) ] = [e^2][4 - sqroot(2) ]/4*pi*eo*ro
   E = (1.6*10^-19)^2[4 - sqroot(2)]/4*pi*8.85*10^-12*283*10^-12 = 2.10049*10^-18 J
   for 1 mole NaCl molecules, this energy is
   6.022*10^23*E /2 = -633.46 kJ/mol

b. For 8 atoms interacting the energy becomes
   E = [ke^2][12 - 4/sqroot(3) ]/ro = [ke^2][12 - 2*sqroot(3)]/ro
   E = 8.98*10^9*(1.6*10^-19)^2[12 - 2*sqroot(3)]/283*10^-12 = 6.933*10^-18 J
   for one mole, it becomes E total = 6.022*10^23*E/8 = -713 kJ/mol

c. comparing this with solid NaCl lattice , enthalapy of formation of NaCl is -788 kJ/mol, so the answer is close to the experimental value

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