Chem_370_Handout_04_25_08

Chem_370_Handout_04_25_08 - Chem 370 Handout Spring 2008...

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Unformatted text preview: Chem 370 Handout Spring 2008 Keq(T) for Reaction D+H2 HD+H Note a=b=c=d=1. Then specializing general equation for K eq ( T ) gives for D + H2 HD + H K eq ( T ) = Now q 0 x,m 0 q0 q0 ( T ) HD,m T, V H,m q 0 H 2 ,m ( ) exp E - ( T, V ) q ( T ) 0 0 D,m tr x 0 rot x vib x 0 RT . ( T ) = Q ( T, V ) Q ( T ) Q ( T ) g 1 for all x el 0,x E0 = - N A 0,HD + D0,H - D0,D - D0,H2 D D0,H = D0,D = ground state energy of H. 1 1 H 1 3 % % D0,HD = De - hce,HD = De - hc 2 e,H2 D0,HD = De - hc e,H . % = 2 HD 2 2 2 4 Also 1 % D0,He = De - hce,H2 . 2 3 1 1 % % E0 = - N A hce - 1 + N A hce,H2 ( 0.1340 ) = 2 2 4 Therefore or For x= H or D 2 m kT q 0 ( T, V ) = Q tr T, V 0 g el = Q tr T, V 0 = 2x V 0 x,m x 0,x x h ( ) ( ) 32 q0 H,m q0 D,m m 1 = H = = 0.3536 . m 2 D For x = HD or H2 32 32 VIB VIB Since Q HD = Q H2 B1 2 q HD,m q H2m 32 32 m 3 3 m HD T rot HD HD 2 4 3 HD = = 2 = 2 = 2 = 2.449 . H T 2rot H m H m 2 2 2 1 3 2 2 2 H2 32 32 Lets pull everything together K eq ( T ) = or 1 % N A 2 hce,H 2 K eq ( T ) = 2 ( 2.449 ) ( 0.3596 ) exp - . [ 0.1340] RT Given NA 1 hc % % = e = 1.44e = VIB and RT kT k VIB K eq ( T ) = 1.7604exp -0.1340 H2 2T = q 0 ( T, V ) q 0 HD,m H,m q 0 H 2 ,m ( T, V ) q 0 D,m E 0 exp - RT ( ) Table 13.2 (- = 1.7604exp .1340 ) ( 6338K ) 2T ( K ) or 849.2K K eq ( T ) = 1.7604exp - . 2T ( K ) Lets take T = 300K 849.2 K eq ( T = 300K ) = 1.7604exp - = 1.7604exp ( -1.4155 ) 600 or K eq ( T = 300K ) = 0.4274 . ...
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