L25-post - 1 Lecture 25 Make It or Break It: Bond Energies...

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Unformatted text preview: 1 Lecture 25 Make It or Break It: Bond Energies Key Idea: Energy is required to break bonds, and is released when bonds are formed. ∆ H o 2 Al (s) + Fe 2 O 3 (s) → Al 2 O 3 (s) + 2 Fe = ( +824 ) + ( − 1676 ) kJ/mol = − 852 kJ/mol ! (l) The Thermite Reaction Al (s) powder + Fe 2 O 3 (s) Mg (s) fuse Enthalpy Enthalpy Hill Hill Elements in Elements in Standard States Standard States H Standard Enthalpies of Formation ( ∆ H f ): o Compounds Compounds ∆ H f o > 0 ∆ H f o = 0 Compounds Compounds ∆ H f o < 0-242 H 2 O(g)-286 H 2 O (l)-394 CO 2 (g) +717 C(g)-1676 Al 2 O 3 (s)-824 Fe 2 O 3 (s)-1274 C 6 H 12 O 6 (s)-74 CH 4 (g) +473 N(g) +249 O(g) +218 H(g) H 2 (g), O 2 (g) C(s), Al(s) Fe(s), N 2 (g) Standard Enthalpies of Formation ( ∆ H f ): o aA + bB aA + bB → cC + dD cC + dD Calculating ∆ H rxn from Tables of ∆ H f : o ∆ H rxn rxn = = c ∆ H ∆ H f º(C) + + d ∆ H ∆ H f º(D) − a ∆ H ∆ H f º(A) − b ∆ H ∆ H f º(B) o o For CO 2 (g) + 2 H 2 O (l) CH 4 (g) + 2 O 2 (g) → ∆ H rxn rxn = = ∆ H ∆ H f º(CO CO 2 ) + + 2 ∆ H ∆ H f º(H 2 O) − ∆ H ∆ H f º(CH CH 4 ) − 2 ∆ H ∆ H f º(O 2 ) o = − 394 + 2( − 286) – ( − 74.4) − 0 = − 891 kJ/mol CH 4 (g) + 2 O 2 (g) → CO 2 (g) + 2 H 2 O (l) Increasing Enthalpy C (s) + 2 H 2 (g) + 2 O 2 Why using ∆ H f works:...
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This note was uploaded on 04/03/2008 for the course CHEM 1A taught by Professor Nitsche during the Fall '08 term at University of California, Berkeley.

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L25-post - 1 Lecture 25 Make It or Break It: Bond Energies...

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