230_F13_L09_enthalpy3_post

38 kj h 565 kj need 3fg bg bf3 g kj bond

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Unformatted text preview: s) " BF3( g ) #H = -1135 kJ 1/2 F2(g) " F( g ) B(s) " B( g ) #H = 79.38 kJ ! #H = 565 kJ Need : 3F(g) + B(g) " BF3( g ) = ? kJ Bond enthalpies are the formation of a bond from its atoms in the gas phase. Divide this equation by 3 and you will have average bond enthalpy for an B - F bond 3 / 2 F2( g ) + B( s) " BF3( g ) #H = -1135 kJ 3F( g ) " 3/2 F2(g) #H = -3(79.38 kJ) B(g) " B( s) #H = -565 kJ 3F(g) + B(g) " BF3( g ) = - 1938.05 kJ B - F bond enthalpy = 1938.05 kJ/3 = 646 kJ Which has the smallest (most negaRve) ΔHof ? A)  ΔHof Br2(l) = 0 B)  ΔHof Br2(g) Br2(l) → Br2(g) ΔH >0 C)  ΔHof Br(g) ½ Br2(l) → Br(g) ΔH >0 Note bond enthalpy: 2Br(g)  ­> Br2(g) Phase changes: only potenRal energy change; ΔU changes, but ΔT does not HeaRng Curve for water from  ­20oC to 110oC q=mCs,steamΔΤ Temperature (oC) 110oC 100oC va...
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This note was uploaded on 02/13/2014 for the course CHEM 230 taught by Professor Sharp during the Winter '08 term at University of Michigan.

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