What is the enthalpy change, in kJ/mol, for the process below?
K+(g) + Br -(g) --> KBr(s)
Enter your answer rounded to the nearest integer (e.g. -827 or 1547). Do not include units as part of your answer. (Hint: Use the Born-Haber cycle. Note carefully: The reference state for Br2 is Br2(l) not Br2(g)!)
Data (in kJ/mol):
Enthalpy of sublimation of K(s): 89
Standard enthalpy of formation of Br2(g): 32
Bond dissociation energy of Br2(g): 194
Ionization energies (1st, 2nd, 3rd) of K(g): 419; 3,051; 4,411
Ionization energies (1st, 2nd, 3rd) of Br(g): 1,140; 2,103; 3,473
First electron affinity of Br(g): -326
Standard enthalpy of formation for KBr(s): -394
Jan 10, 2018EXPERT
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