Question

Using the following information, calculate the heat of sublimation of S(s) to S(g).

?H°f MgS= -343 kJ/mol,

?H°sublimation Mg(s) = 148 kJ/mol,

The lattice energy MgS = 3406 kJ/mol

?Mg(g) ? Mg2+(g) + 2 e ??st + 2nd IE = 2186 kJ/mol,

S(g) +2e ? S2-(g) ????H = 450 kJ/mol,

A. 575 kJ/mol

B. 1179 kJ/mol

C. 279 kJ/mol

D. 965 kJ/mol

E. 4651 kJ/mol

Answer 1

deltaH _f^degree (Mgs) = -343 KJ/mol deltaH _fule (Mg) (s) = 148 KJ/mol lattice energy, mu mgs = 3406 KJ/mol Mg(g) rightarrow mg^2 + (g) + 2e^- IE = 2186 KJ/mol s(g) + 2e rightarrow s62^- (g) delta H_gs 450 KJ/mol 3406 = 148 + 2186 - 450 + deltaH s(s) + 343 delta H s(s) = 1179 KJ/mol therefore Heat of sublimation of s = 1179 KJ/mol