CHEM 201 - 10.10.07

CHEM 201 - 10.10.07 - CHEM 201 Example from text#107 o...

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CHEM 201 – October 10, 2007 Example from text #107 o Givens Net ε change for CaC 2 = -60 KJ/mol Heat of sublimation = 178 kJ/mol E i1 Ca = 590 kJ/mol E i2 Ca = 1145 kJ/mol Heat of sublimation for C = 717 kJ/mol Bond Dissociation C 2 = 614 kJ/mol E ea1 C 2 (g) = -315 kJ/mol E ea2 C 2 (g) = 410 kJ/mol o Solving the problem Ca(s) Ca(g) 178 2C(s) Ca(g) 2(717) Ca(g) Ca + (g) + e - 590 Ca + (g) Ca 2+ + e - 1145 2C(g) C 2 (g) -614 C 2 (g) + e - C 2 - (g) -315 e -
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