ps2solutions

# ps2solutions - Chemistry 4511 Problem Set 2 1 For the...

This preview shows pages 1–2. Sign up to view the full content.

Chemistry 4511 Problem Set 2 1. For the reaction 2CO 2 Æ 2CO + O 2 a. If there are initially 0.534 moles of CO 2 and 0.42 moles of CO are generated at some time, demonstrate that ξ is the same with respect to each component. b. Once the reaction goes to completion, there is a H value measured at T = 1000 K. Calculate the expected value for H two ways and compare the values (use heats of formation and bond enthalpies and adjust the H value away from standard T using C P ,m(T)) a. ξ = n i / ν i i ν i n i C O 2 -2 -0.42 CO +2 +0.42 O 2 +1 +0.21 ξ CO2 = -0.42/-2 = 0.21 ξ CO = 0.42/2 = 0.21 ξ O2 = 0.21/1 = 0.21 b. H o = Σ∆ f H o prod Σ∆ f H o react = 2(-110.53)+0-2(-393.51) = 565.96 kJ/mol H o = Σ∆ H broken Σ∆ H formed = 2(531) – 497 = 565 kJ/mol C P,m = d + eT + fT -2 d = 2(28.41) + 29.96 – 2(44.22) = -1.44 J K -1 mol -1 e(x 10 3 ) = 2(4.1) + 4.18 – 2(8.79) = -5.2 Æ -5.2 x 10 -3 J K -2 mol -1 f(x 10 4 ) = 2(-4.6) + (-16.7) – 2(-86.2) = 146.5 1.465 x 10 6 J K mol -1 C P,m = -1.44 – 5.2 x 10 -3 T + 1.465 x 10 6 T -2 + = = ∆∆ 1 2 2 1 2 2 1 2 , 1 1 ) ( 2 ) ( ) ( 2 1 T T f T T e T T d dT C T H T T m P + = ∆∆ 15 . 298 1 1000 1 10 465 . 1 ) 15 . 298 1000 ( 2 10 2 . 5 ) 15 . 298 1000 ( 44 . 1 ) ( 6 2 2 3 x x T H 1 69 ) ( = ∆∆ Jmol T H so H 1000 = H o + ∆∆ H H 1000 = 565.96 + 0.069 = 566.029 kJ/mol for formation constants H 1000 = 565 + 0.069 = 565.069 kJ/mol for bond enthalpies We can see that the T dependence is not significant compared to the process, and this is typically the case. Also, we see that the two ways of deriving

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

ps2solutions - Chemistry 4511 Problem Set 2 1 For the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online