Chem 1101
A/Prof Sébastien Perrier
Room: 351
Phone: 9351-3366
Email: [email protected]
Prof Scott Kable
Unless otherwise stated, all images in this file have been reproduced from:
Room: 311
Blackman, Bottle, Schmid, Mocerino and Wille,
Chemistry, 2007 (John Wiley)
ISBN: 9 78047081 0866
Phone: 9351-2756
Email: [email protected]
A/Prof Adam Bridgeman
Room: 222
Phone: 9351-2731
Email: [email protected]
Slide 23-1
Slide 23-2
What’s wrong with ∆Hatom?
Highlights of last lecture
For example, consider the ‘halon’: CFClBr2:
∆H = ∆E + P ∆V
Will the following reaction occur?
• ∆H is called the “enthalpy” or “heat of reaction”
CFClBr2 + light → CFCl + 2 Br?
Atomisation Enthalpies
Enthalpy (kJ mol−1)
Hess’s Law
560
•calculate heats of reaction from atomisation enthalpies
1840
•If you add up chemical equations to form overall equation, you can add up
∆H to get overall ∆H
•This allows you to estimate ∆H of an unknown reaction from ∆H of known
reactions
Limit of
stratosphere
radiation
Br is strongly ozone depleting (50
x more than Cl);
Br + O3 →BrO + O2
BrO + O3 →BrO + 2O2
limit of stratospheric radiation:
210 nm
E=hc/λ => 560 kJ/mol
Will this reaction proceed in the
atmosphere?
Work out ∆Hr0 from bond enthalpies
and compare with energy from light.
Slide 23-3
Slide 23-4
What’s wrong with ∆Hatom?
What’s wrong with ∆Hatom?
Br
∆Hr0 = 2 x 285
= 570 kJ/mol
Br
The truth…
+ light → CFCl + 2 Br
570
Enthalpy (kJ mol-1)
C
570
560
Enthalpy (kJ mol-1)
F
Cl
Limit of
solar
radiation
Bond
enthalpy
Therefore CFClBr2 should NOT decompose in stratosphere, yielding 2 x
Br atoms
Slide 23-5
CFCl + 2 Br
Estimate using BE
560
∆Hr = 436 kJ/mol
179
CFClBr + Br
257
CFClBr2
Limit of
solar
radiation
CFClBr2 certainly DOES photodissociate in the
atmosphere providing 2 Br
Slide 23-6
1
What went wrong?
Standard states
Bond energies depend on the molecule. The
values in the table are just averages.
Looks like we need to list bond enthalpies for all
common compounds, but…
Referencing common compounds to atoms is
difficult to measure experimentally (atoms are
hard to isolate!)
Maybe we need to use atoms in a state other than
isolated gas phase. Why don’t we use the state that
the atom is most commonly found in. We call this
the “standard state”
Slide 23-7
Heat of formation (∆Hf)
or
• Gas:
1 atm
• Temperature:
298 K
• Solution:
1 mol L-1
• Elements:
usual form at std cond’ns
The definition of standard states is a convenience so that all
chemists refer to the same quantities.
We use a superscript “0” to indicate std conditions
Slide 23-8
Example problem…
So instead of breaking bonds to form atoms, lets form compounds
in the standard state. We call this the “enthalpy of formation”
or “heat of formation”
∆Hf0 (kJ mol-1)
e.g.
Clearly, if we are defining an element at “standard state” we
need also to define the standard conditions we are referring to:
C(graphite) + 2 H2 (g) → CH4(g)
-74.8
C(graphite) + ½ O2 (g) + 2 H2 (g) → CH3OH (l )
-238.6
Calculate the heat of combustion of ethylene
(ethene) using the standard heats of formation.
Balanced equation:
C2H4(g) + 3 O2(g) → 2CO2(g) + 2 H2O(l)
The heat of formation of an element in its standard
state is defined as zero
e.g. ∆Hf0 [O2 (g)] = 0; ∆Hf0 [Fe(s)] = 0; ∆Hf0 [Hg(l )] = 0;
Q: What are the standard states of: H, Na, Cl, Br ?
Slide 23-9
Slide 23-10
Example problem…
-393.5
Enthalpies
of
formation
Calculate the heat of combustion of ethylene
(ethene) using the standard heats of formation.
Balanced equation:
52.284
C2H4(g) + 3 O2(g) → 2CO2(g) + 2 H2O(l)
∆H(c) = ∆Hf0(prod.) – ∆Hf0 (react.)
= (2 x -393.5) + (2 x -285.9) – (52.3+ 3x 0)
-285.9
= -1411 kJ/mol
Slide 23-11
Slide 23-12
2
Thermite reaction
-1669.8
Predict (using the heat of formation tables), whether the
following reaction is exothermic or endothermic.
Fe2O3 (s) + 2 Al (s) → Al2O3 (s) + 2 Fe (s)
Enthalpies
of
formation
-822.2
Slide 23-13
Slide 23-14
Thermite reaction
Predict (using the heat of formation tables), whether the
following reaction is exothermic or endothermic.
Fe2O3 (s) + 2 Al (s) → Al2O3 (s) + 2 Fe (s)
Fe2O3 (s) + 2 Al (s) → Al2O3 (s) + 2 Fe (s)
∆Hr0 = ∆Hf0(prod.) – ∆Hf0 (react.)
= -1669.8 – (-822.2)
= -847.6 kJ mol-1
which is highly exothermic
Slide 23-15
Thermite reaction
The thermite reaction is
used to weld railroad tracks
in absence of electricity to
power arc welders.
Slide 23-16
Space shuttle booster engines
Space shuttle solid fuel:
A variant of this reaction is used
as igniter/catalyst in solid fuel for
rockets
Component
Compound
Weight percent
Fuel
Powdered Al
16%
Oxidant
Ammonium perchlorate
70%
Catalyst
Iron oxide
0.07%
Binder
Polymer
12%
Curing agent
Epoxy
2%
Unbalanced equation:
Al (s) + NH4ClO4 (s) → Al2O3 (s) + AlCl3 (s) + NO (g) + H2O (l)
Balance it…
Al (s) + NH4ClO4 (s) → 1/3 Al2O3 (s) + 1/3 AlCl3 (s) + NO (g) + 2H2O (l)
Slide 23-17
Slide 23-18
3
Which is the better fuel?
Efficient fuels?
Thermite reaction:
Look to rocket industry (they care most about
efficient fuels)!
Fe2O3 (s) + 2 Al (s) → Al2O3 (s) + 2 Fe (s)
∆Hr = -848 kJ
0
mol-1
= 4.0 kJ g-1 reactants
4 main types of rocket fuel:
Space shuttle solid fuel:
• solid fuel (e.g. above)
Al (s) + NH4ClO4 (s)
• petroleum (~kerosene = C12 fraction)
→ 1/3 Al2O3 (s) + 1/3 AlCl3 (s) + NO (g) + 2H2O (l)
∆Hr0 = -978 kJ mol-1
= 6.8 kJ g-1 reactants
• hypergolic (autoigniting)
• cryogenic (cold liquids)
Slide 23-19
Efficient fuels?
Slide 23-20
Efficient fuels?
Solid fuels:
Petroleum fuels:
∆Hc (kJ mol-1 )
∆Hc /m (fuel)
Fuel
Oxidant
Al
Fe2O3
-848
-
Al
NH4ClO4
-978
-
∆Hc/m (tot)*
-4.0
-6.7
* Including oxidant
Solid fuels have their oxidant included in the solid phase.
∆Hc (kJ mol-1 )
∆Hc /m (fuel)
∆Hc/m (tot)*
Fuel
Oxidant
Ethane
Liq O2
-1561
-52.0
-11.0
Butane
Liq O2
-2877
-49.6
-10.8
Hexane
Liq O2
-4163
-48.4
-10.7
Octane
Liq O2
-5430
-47.6
-10.6
* Including oxidant, Liq O2
All fractions have about the same efficiency, so choose fuel for physical
properties: liquid/gas, vapour pressure, etc
Advantages: easily handled, long-term stability
Advantages: cheap, efficient, easily handled
Disadvantages: can’t stop once it’s started. Therefore used
for booster rockets
Slide 23-21
Disadvantages: Liq O2 is difficult to handle, low density
means larger fuel tank
Slide 23-22
Efficient fuels?
Hypergolic fuels:
Efficient fuels?
* Including oxidant
Hypergolic fuels:
* Including oxidant
∆Hr (kJ mol-1 )
∆Hr /m (fuel)
∆Hr/m (tot)*
Fuel
Oxidant
-9.7
Hydrazine (N2H4)
N2O4
-632
-
-5.1
-28.4
-10.4
Methylhydrazine
N2O4
-1281
-
-8.0
-33.0
-10.5
1,2dimethylhydrazine
N2O4
-1940
-
-8.0
∆Hc (kJ mol-1 ) ∆Hc /m (fuel)
Fuel
Oxidant
Hydrazine (N2H4)
Liq O2
-622
-19.4
Methylhydrazine
Liq O2
-1305
1,2-dimethylhydrazine
Liq O2
-1979
∆Hc/m (tot)*
Hypergolic fuels autoignite upon contact. Dinitrogen tetroxide is preferred to Liq O2
because it does not require cryogenic cooling. The lower energy release is worth the trade
off. Because both fuel and oxidant are liquids this rocket is therefore easy start / easy
stop. This fuel is used in the space shuttle for the orbital maneuvering system (OMS) .
Advantages: easy start / easy stop, no cryogenics
Hydrazine = N2H4
Slide 23-23
Disadvantages: not as much thrust as other fuels, so not as
good for lifting payload from surface.
Slide 23-24
4
Efficient fuels?
Efficient fuels?
So after the lessons from the rocket industry, what might we consider for
efficient Earth-bound engine fuels?
Cryogenic fuels:
Fuel
Oxidant
H2
O2
∆Hc (kJ mol-1 )
∆Hc /m (fuel)
-242
∆Hc/m (tot)*
-121 (!)
-13.4
* Including oxidant
Biggest “bounce per ounce” of any fuel. Although both fuel and
oxidant are cryogenic liquids, the superior efficiency makes this
fuel worth persisting with. This fuel is used in both the main
engines and external tank of the shuttle.
First note that we don’t have to carry the oxidant (O2 obtained from
the air!)
Petroleum:
Hypergolic:
Current choice.
Detrimental to
atmosphere
Hydrazine toxic and
explosive
Solid fuel:
OH&S: ignite on
contact!
Transport and transfer
problems (how do you
fill the tank?)
Advantages: best efficiency
Disadvantages: cryogenic liquids, difficult to store for long
periods, low density makes tanks larger.
Slide 23-25
Example problem…
How do you stop?
Hydrogen:
Good environmental properties
Great efficiency
Explosive
Low density in gas phase means
big fuel tanks
The answer….
Maybe fuel cells?
See later topic!
Slide 23-26
Summary
6.0 g of urea, (NH2)2CO, was burnt in the presence of O2 to yield H2O (l), N2 (g)
and CO2 (g). The reaction was performed in an open container and 63.4 kJ
of heat was produced.
CONCEPTS
a) Write a balanced equation for the reaction.
(NH 2)2CO(s) + 1.5 O2(g) -> CO2(g) + 2 H 2O(l) + N2(g) and q=-63.4 kJ
b) What is ∆Hc for urea?
(MW(urea) = 60.0 g/mol)
Standard states and standard enthalpies
How heat of formation is defined
How various fuels work
Positives and negatives of different fuels
MW(urea) = 60.0 g/mol, therefore 0.100 mol burned
CALCULATIONS
Heat of combustion, ∆H c = -634 kJ/mol
c) What is the standard heat of formation of urea?
∆Hc = Σ [∆Hf(products)] – Σ[∆Hf(reactants)]
–634 = [–285.9 × 2 – 393.5] – [∆Hf(urea)]
=>
Given ∆Hf work out ∆Hr (or vice versa)
Efficiency of fuels
∆Hf(urea) = –331 kJ/mol
Slide 23-27
Slide 23-28
5