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Homework5 Answer (Due Friday, 02/27/2009) P4.1, P4.2, P4.3, P4.6, P4.8, P4.18, P4.19, P4.21, P4.29, P4.31 P4.1) Calculate Δ H reaction ° and Δ U reaction ° at 298.15 K for the following reactions: a. 4NH 3 ( g ) + 6NO( g ) 5N 2 ( g ) + 6H 2 O( g ) b. 2NO( g ) + O 2 ( g ) 2NO 2 ( g ) c. TiCl 4 ( l ) + 2H 2 O( l ) TiO 2 ( s ) + 4HCl( g ) d. 2NaOH( aq ) + H 2 SO 4 ( aq ) Na 2 SO 4 ( aq ) + 2H 2 O( l ). Assume complete dissociation of NaOH, H 2 SO 4 , and Na 2 SO 4 . e. CH 4 ( g ) + H 2 O( g ) CO( g ) + 3H 2 ( g ) f. CH 3 OH( g ) + CO( g ) CH 3 COOH( l ) Table 4.1 4.2 Appendix B a) () ( ) ( ) ( ) g NO, Δ H 6 g , NH Δ H 4 g O, H Δ H 6 g , N Δ H 5 Δ H f 3 f 2 f 2 f reaction o o o o o + = ( ) ( ) ( ) ( ) 1 - -1 -1 -1 -1 reaction mol kJ 1815.0 - mol kJ 1.3 9 6 mol kJ 5.9 4 4 mol kJ 41.8 2 6 mol kJ 0 5 Δ H = × × × + × = o ( ) ( ) ( ) 1 - -1 -1 -1 reaction reaction mol kJ 1817.5 K 298.15 mol K J 8.314472 1 mol kJ 1815.0 T R n Δ H Δ U = × × = Δ = o o b) ( ) ( ) g NO, Δ H 2 g , O Δ H - g , NO Δ H 2 Δ H f 2 f 2 f reaction o o o o = ( ) ( ) ( ) 1 - -1 -1 -1 reaction mol kJ 116.2 mol kJ 1.3 9 2 mol kJ 0 mol kJ 3.2 3 2 Δ H = × × = o ( ) ( ) ( ) 1 - -1 -1 -1 reaction reaction mol kJ 113.7 K 298.15 mol K J 8.314472 1 mol kJ 116.2 T R n Δ H Δ U = × × = Δ = o o

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c) () ( ) ( ) ( ) l l o o o o o O, H Δ H 2 , TiCl Δ H g HCl, Δ H 4 s , TiO Δ H Δ H 2 f 4 f f 2 f reaction + = ( ) ( ) ( ) -1 -1 -1 -1 reaction -1 Δ H 944.0 kJ mol 4 92.3 kJ mol 804.2 kJ mol 2 285.8 kJ mol 62.6 kJ mol =− − × + + × = o ( ) ( ) ( ) -1 -1 -1 reaction reaction -1 Δ U Δ H n R T 60.8 kJ mol 4 8.314472 J K mol 298.15 K 52.7 kJ mol Δ = × × = oo d) The overall reaction is: () () ( ) l O H 2 aq H 2 aq OH 2 2 ⎯→ + + ( ) ( ) aq , OH Δ H 2 aq , H Δ H 2 O, H Δ H 2 Δ H f f 2 f reaction + = o o o o l ( ) ( ) ( ) 1 - -1 -1 -1 reaction mol kJ 111.6 mol kJ 230.0 2 mol kJ 0 2 mol kJ 85.8 2 2 Δ H = × × × = o ( ) 1 - -1 reaction reaction mol kJ 111.6 0 mol kJ 111.6 T R n Δ H Δ U = = Δ = o o e) CH 4 ( g ) + H 2 O( g ) CO( g ) + 3H 2 ( g ) ( ) ( ) g g O, H Δ H , CH Δ H g , H Δ H 3 g CO, Δ H Δ H 2 f 4 f 2 f f reaction o o o o o + = ( ) ( ) ( ) ( ) 1 - -1 -1 -1 -1 reaction mol kJ 05.9 2 mol kJ .8 1 4 2 mol kJ 4.6 7 mol kJ 0 3 mol kJ 5 . 10 1 Δ H = × + = o ( ) ( ) ( ) 1 - -1 -1 -1 reaction reaction mol kJ 00.9 2 K 298.15 mol K J 8.314472 2 mol kJ 05.9 2 T R n Δ H Δ U = × × = Δ = o o
f) CH 3 OH( g ) + CO( g ) CH 3 COOH( l ) () ( ) ( ) g CO, Δ H g OH, CH Δ H COOH, CH Δ H Δ H f 3 f 3 f reaction o o o o l = ( ) ( ) ( ) 1 - -1 -1 -1 reaction mol kJ 172.8 mol kJ 10.5 1 mol kJ 0 . 201 mol kJ 3 . 484 Δ H = = o ( ) ( ) ( ) 1 - -1 -1 -1 reaction reaction mol kJ 167.8 K 298.15 mol K J 8.314472 2 mol kJ 172.8 T R n Δ H Δ U = × × = Δ = o o P4.2) Calculate Δ H reaction ° and Δ U reaction ° for the oxidation of benzene. Also calculate Δ H reaction ° −Δ U reaction ° Δ H reaction ° The chemical equation for the oxidation of benzene is: () () ( ) l O H 3 g CO 6 g O 7 l H C 2 2 2 2 1 6 6 + + The standard enthalpy for this reaction is: ( ) ( ) ( ) 1 1 1 1 i f,i i mol kJ 3268 mol kJ 49.1 mol kJ 285.8 3 mol kJ 393.5 6 H ν Δ H = × + × = = o o Δ U is calculated as: {} 1 1 1 1 3 mol kJ 3264 K 298.15 K mol J 8.314472 1.5 - mol J 10 3268 T R n Δ H Δ U = × × × = Δ = o o And: ( ) 00122 . 0

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