1 / 20

Tutorial 5: Question 11

Tutorial 5: Question 11. Presented by 12S7F Yu Tian :D. Problem statement >.<. Aluminium fluoride is made industrially by reacting aluminium oxide with hydrogen fluoride at a high temperature . Al 2 O 3 (s) + 6HF (g)  2A l F 3 (s) + 3H 2 O (g). Problem statement >.<. UNIT:

fisseha
Download Presentation

Tutorial 5: Question 11

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Tutorial 5: Question 11 Presented by 12S7F Yu Tian :D

  2. Problem statement >.< • Aluminium fluoride is made industrially by reacting aluminium oxide with hydrogen fluoride at a high temperature. Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g)

  3. Problem statement >.< UNIT: J K-1 mol-1 NOTkJK-1 mol-1

  4. Problem statement >.< • Use the data given to calculate: • The standard enthalpy change, ΔHo , of this reaction • The standard entropy change, ΔSo , of this reaction Hess’s Law Energy cycle

  5. Problem wracking x.x ΔHo ? Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g) ΔHfo of Al2O3 (s) 2Al (s) + 3H2 (g) + 3F2 (g) + O2(g)

  6. Problem wracking x.x ΔHo ? Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g) ΔHfo of HF(s) x 6 2Al (s) + 3H2 (g) + 3F2 (g) + O2(g)

  7. Problem wracking x.x ΔHo ? Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g) ΔHfo of AlF3 (s) x 2 2Al (s) + 3H2 (g) + 3F2 (g) + O2(g)

  8. Problem wracking x.x ΔHo ? Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g) ΔHfo of H2O (g) x 3 2Al (s) + 3H2 (g) + 3F2 (g) + O2(g)

  9. Problem wracking x.x ΔHo ? Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g) ΔHfo of HF(s) ΔHfo of AlF3 (s) ΔHfo of H2O (g) x 2 x 6 ΔHfo of Al2O3 (s) x 3 2Al (s) + 3H2 (g) + 3F2 (g) + O2(g) By Hess’s Law, the enthalpy change of a particular reaction is determined only by the initial and final states of the system regardless of the pathway taken.

  10. Problem wracking x.x ? Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g) ΔHfo of HF(s) ΔHfo of AlF3 (s) ΔHfo of H2O (g) x 2 x 6 ΔHfo of Al2O3 (s) x 3 2Al (s) + 3H2 (g) + 3F2 (g) + O2(g) + ΔHo= ΔHfo of AlF3 (s) x 2 ΔHfo of H2O (g) x 3 - - ΔHfo of Al2O3 (s) ΔHfo of HF(s) x 6

  11. Problem wracking x.x ? Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g) ΔHfo of HF(s) ΔHfo of AlF3 (s) ΔHfo of H2O (g) x 2 x 6 ΔHfo of Al2O3 (s) x 3 2Al (s) + 3H2 (g) + 3F2 (g) + O2(g) + ΔHo= ( -1504 ) x 2 ( -242 ) x 3 - - ( - 1676 ) ( -242 )x 6 = -432 kJ mol-1

  12. Problem wracking x.x ΔSo ? Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g) ΔSfo of HF(s) ΔSfo of AlF3 (s) ΔSfo of H2O (g) x 2 x 6 ΔSfo of Al2O3 (s) x 3 2Al (s) + 3H2 (g) + 3F2 (g) + O2(g) + ΔSo= ΔSfo of AlF3 (s) x 2 ΔSfo of H2O (g) x 3 - - ΔSfo of Al2O3 (s) ΔSfo of HF(s) x 6

  13. Problem wracking x.x ? Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g) ΔSfo of HF(s) ΔSfo of AlF3 (s) ΔSfo of H2O (g) x 2 x 6 ΔSfo of Al2O3 (s) x 3 2Al (s) + 3H2 (g) + 3F2 (g) + O2(g) + ΔSo= ( - 266 ) x 2 ( - 44.4 ) x 3 - - ( - 313 ) ( + 7.0 )x 6 = - 394 J mol-1 (to 3 s.f. )

  14. (b) Use the values calculated in (a) to calculate ΔGo. Standard Gibbs Free energy change  ΔGo is measured at 298 K and 1 atm ΔGo= ΔHo – T x ΔSo = (-432) x103 – (298) x (-394.2) J mol-1 = - 315 J mol-1(to 3 s.f.)

  15. Problem statement (c) How will the value of ΔGofor this reaction change with temperature? What consequences will this have for the conditions used to make AlF3 industrially? What is the relationship between T and ΔGo?

  16. Problem wracking (c) How will the value of ΔGofor this reaction change with temperature? GRADIANT OF THE GRAPH Formula: ΔGo= ΔHo – ΔSo x T Y- INTERCEPT ΔSo is negative ΔGo T 0 POSITIVE GRADIENT ΔHo

  17. Problem wracking • Ans: As value of ΔSois negative, according to formula Formula: ΔGo= ΔHo – ΔSo x T, the higher the reaction temperature, the more positive the value of ΔGobecomes. (c) How will the value of ΔGofor this reaction change with temperature?

  18. Problem wracking (c) What consequences will this have for the conditions used to make AlF3 industrially?

  19. Problem wracking (c) What consequences will this have for the conditions used to make AlF3 industrially? • From ΔGo= - 315 J mol-1 • Under standard condition, the reaction is not spontaneous/feasible For the reaction to be feasible, ΔGo> 0 T > 1100 K or T> 823 ℃ - High temperature is needed in industrial production of AlF3

  20. Thank you!

More Related