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L25-post - The Thermite Reaction Lecture 25 Make It or...

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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 0 Standard Enthalpies of Formation ( H f ): o 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) 0 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 = = 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 = H f º( CO CO 2 ) + 2 H f º( H 2 O ) H f º( CH 4 ) 2 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: 0 o CO 2 (g) + 2 H 2 O (l) H o (products) = H f º(CO 2 ) + 2 H f º(H 2 O) = 394 + 2( 286) = 966 kJ/mol + 2 O 2 CH 4 (g) H o (reactants) = H f º(CH
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