Presented by 12S7F Yu Tian :D 1

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)  2AlF 3 (s) + 3H 2 O (g) 2

Problem statement >.< Al 2 O 3 (s)HF (g)AlF 3 (s)H 2 O (g) ΔH f o / kJ mol ΔS f o / J K -1 mol UNIT: J K -1 mol -1 NOT kJ K -1 mol -1 UNIT: J K -1 mol -1 NOT kJ K -1 mol -1 3

Problem statement >.< (a)Use the data given to calculate: The standard enthalpy change, ΔH o, of this reaction The standard entropy change, ΔS o, of this reaction 4

Problem wracking x.x Al 2 O 3 (s) + 6HF (g)  2AlF 3 (s) + 3H 2 O (g) 2Al (s) + 3H 2 (g) + 3F 2 (g) + O 2 (g) 5

Problem wracking x.x Al 2 O 3 (s) + 6HF (g)  2AlF 3 (s) + 3H 2 O (g) 2Al (s) + 3H 2 (g) + 3F 2 (g) + O 2 (g) 6

Problem wracking x.x Al 2 O 3 (s) + 6HF (g)  2AlF 3 (s) + 3H 2 O (g) 2Al (s) + 3H 2 (g) + 3F 2 (g) + O 2 (g) 7

Problem wracking x.x Al 2 O 3 (s) + 6HF (g)  2AlF 3 (s) + 3H 2 O (g) 2Al (s) + 3H 2 (g) + 3F 2 (g) + O 2 (g) 8

Problem wracking x.x Al 2 O 3 (s) + 6HF (g)  2AlF 3 (s) + 3H 2 O (g) 2Al (s) + 3H 2 (g) + 3F 2 (g) + O 2 (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. 9

Problem wracking x.x Al 2 O 3 (s) + 6HF (g)  2AlF 3 (s) + 3H 2 O (g) 2Al (s) + 3H 2 (g) + 3F 2 (g) + O 2 (g) 10

Problem wracking x.x Al 2 O 3 (s) + 6HF (g)  2AlF 3 (s) + 3H 2 O (g) 2Al (s) + 3H 2 (g) + 3F 2 (g) + O 2 (g) 11

Problem wracking x.x Al 2 O 3 (s) + 6HF (g)  2AlF 3 (s) + 3H 2 O (g) 2Al (s) + 3H 2 (g) + 3F 2 (g) + O 2 (g) 12

Problem wracking x.x Al 2 O 3 (s) + 6HF (g)  2AlF 3 (s) + 3H 2 O (g) 2Al (s) + 3H 2 (g) + 3F 2 (g) + O 2 (g) 13

(b) Use the values calculated in (a) to calculate ΔG o. ΔG o = ΔH o – T x ΔS o = (-432) x 10 3 – (298) x (-394.2) J mol -1 = J mol -1 (to 3 s.f.) 14

Problem statement (c) How will the value of ΔG o for this reaction change with temperature? What consequences will this have for the conditions used to make AlF 3 industrially? 15

Problem wracking (c) How will the value of ΔG o for this reaction change with temperature? Formula: ΔG o = ΔH o – ΔS o x T GRADIANT OF THE GRAPH ΔGoΔGo T 0 ΔHoΔHo Y- INTERCEPT POSITIVE GRADIENT ΔS o is negative 16

Ans: As value of ΔS o is negative, according to formula Formula: ΔG o = ΔH o – ΔS o x T, the higher the reaction temperature, the more positive the value of ΔG o becomes. (c) How will the value of ΔG o for this reaction change with temperature? Problem wracking 17

(c) What consequences will this have for the conditions used to make AlF 3 industrially? Problem wracking Sign of ΔG o Positive Reaction is not feasible. 0 The system is at equilibrium Negative Reaction is feasible 18

(c) What consequences will this have for the conditions used to make AlF 3 industrially? Problem wracking From ΔG o = J mol -1 - Under standard condition, the reaction is not spontaneous/feasible For the reaction to be feasible, ΔG o > 0 T > 1100 K or T> 823 ℃ - High temperature is needed in industrial production of AlF3 19

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