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)
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Problem statement >.< Al2O3 (s)
HF (g)
AlF3 (s)
H2O (g)
ΔHfo / kJ mol-1
-1676
-271
-1504
-242
ΔSfo / J K-1 mol-1
-313
+7.0
-266
-44.4
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, ΔHo , of this reaction • The standard entropy change, ΔSo , of this reaction
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Problem wracking x.x Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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Problem wracking x.x Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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Problem wracking x.x Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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Problem wracking x.x Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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Problem wracking x.x Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
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. 9
Problem wracking x.x Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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Problem wracking x.x Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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Problem wracking x.x Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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Problem wracking x.x Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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(b) Use the values calculated in (a) to calculate ΔGo .
ΔGo =
ΔHo – T x ΔSo
= (-432) x 103 – (298) x (-394.2) J mol-1
= - 315 J mol-1
(to 3 s.f.)
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Problem statement (c) How will the value of ΔGo for this reaction change with temperature?
What consequences will this have for the conditions used to make AlF3 industrially?
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Problem wracking (c) How will the value of ΔGo for this reaction change with temperature? GRADIANT OF THE GRAPH
Formula: ΔGo = ΔHo – ΔSo x T ΔGo 0
Y- INTERCEPT
ΔSo is negative T
ΔHo 16
Problem wracking (c) How will the value of ΔGo for this reaction change with temperature? • Ans: As value of ΔSo is negative, according to formula Formula: ΔGo = ΔHo – ΔSo x T, the higher the reaction temperature, the more positive the value of ΔGo becomes.
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Problem wracking (c) What consequences will this have for the conditions used to make AlF3 industrially?
Sign of ΔGo
Positive
Reaction is not feasible.
0
The system is at equilibrium
Negative
Reaction is feasible
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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 19
