Transcript Q11 Chem Tut 5 - 12S7F-note

Presented by 12S7F Yu Tian :D
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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
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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.
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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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(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
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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
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