Transcript Ionic Equations - Geneva Area City Schools

Energy
Energy
• Energy – the ability to do work or produce
heat
• Energy exists in two different forms –
kinetic energy & potential energy
Potential Energy
• Potential energy – energy due to
composition or position of an object
• Potential energy is stored energy that
results from the attractions or repulsions of
other objects
Kinetic Energy
• Kinetic energy – the energy of motion
• Kinetic energy depends on as objects
mass & its velocity
• Atoms has mass & they are in motion;
therefore they will have kinetic energy
Energy
• A roller coaster at the top of a hill has a
great amount of potential energy.
• As the rollercoaster begins to speed down
the hill, the potential energy is turned into
kinetic energy
Energy
• The SI unit for energy is the joule (J)
• 1 J = 1 Kgm2 / s2
• Another unit of energy that you may be
more familiar with is the calorie
• calorie – amount of energy required to
raise 1 g of water 1°C
• 1 cal = 4.18 J
Energy
• The calories that you eat are actually
kilocalories or Calories (with a big C)
• 1000 calories = 1 Kilocalorie = 1 Calorie
Energy Conversions
• Convert 15,500 joules into Calories
• 15500 J x 1 cal x 1 Cal
=
•
4.18 J 1000 cal
• 3.71 Cal
Formulas – Kinetic Energy
•
•
•
•
•
Kinetic energy
KE = ½ mv2
KE = kinetic energy (joules)
m = mass (must be in Kg)
V = velocity (must be in m/s)
Formulas – Potential Energy
•
•
•
•
•
•
Potential Energy
PE = mgh
PE = Potential Energy (J)
m = mass (Kg)
g = gravitational constant = 9.8 m/s2
h = height (m)
Formulas - Work
• Work (w) – the energy used to move an
object against a force
• Force (f) – a push or pull on an object
• W = mgd = fd = PE
• Work and potential energy can be looked
at in the same light
Work
• It is important to understand that if there is
no movement, there is no work done
• If I push and push on the demonstration
table with all of my might. I may get hot
and sweaty and feel like I have done a
TON of work, but in reality I have done NO
work because the table has not moved
Examples
• A bowler lifts a 5.4 kg bowling ball 1.6m
and then drops it to the ground.
• How much work was required to raise the
ball?
• W = mgd
• W = (5.4 kg)(9.8 m/s2)(1.6m)
• 85 Kgm2/s2 = 85 J
Examples
• How much potential energy does that ball
have at this height?
• 85 J
Examples
• If the mass is dropped and we assume
that all of the potential energy is turned
into kinetic energy, at what velocity will the
bowling ball hit the ground?
• KE = PE = 85J
• m = 5.4 Kg
• V=?
Examples
•
•
•
•
KE = ½ mv2
85 J = ½ (5.4 Kg) v2
v2 = 31.5
v = 5.6 m/s
More examples
• What is the kinetic energy of 1 atom of Ar
moving at 650 m/s?
• KE = ½ mv2
• 1atom Ar x
1mol Ar
x 39.95 g Ar x 1 Kg Ar
=
•
6.02 x 1023 atoms
1 mol Ar
1 x 103 g Ar
• 6.64 x 10-26 kg Ar
More Examples
• KE = ½ mv2
• KE = ½ (6.64 x 10-26 Kg)(650m/s) 2
• KE = 1.4 x 10 -20 J
1st Law of Thermodynamics
• 1st Law of Thermodynamics – energy is
conserved
• The law of conservation of energy
states that in any chemical reaction or
physical process, energy can be converted
from one form to another, but it is neither
created nor destroyed.
1st Law of Thermodynamics
• Since energy can neither be gained nor
lost, the change in E can be calculated
using:
• E = Ef – Ei
• In a chemical reaction i indicates reactants
and f indicated products
E
•
E has 3 parts:
1. A # indicating the magnitude
2. A sign (+/-) indicating the direction
3. A unit
Thermochemistry
• Thermochemistry is the study of heat
changes that accompany chemical
reactions and phase changes.
• In thermochemistry, the system is the
specific part of the universe that contains
the reaction or process you wish to study.
Thermochemistry
• Everything in the universe other than the
system is considered the surroundings.
• Therefore, the universe is defined as the
system plus the surroundings.
universe = system + surroundings
Relating E to heat & work
• The system can exchange energy with its
surroundings in 2 ways: as heat or work
• E = q + w
• E = change in energy
• q = heat
• w = work
q&w
• Don’t forget q & w must have signs
• In order to get the sign you must look at
the system as a box and the surroundings
as everything else
System
Surroundings
q&w
• Anything going INTO the box will be +
• Anything going OUT of the box will be –
+
-
q&w
• If heat is transferred from the surroundings
to the system and work is done on the
system what are the signs for q & w?
q=+
w=+
q&w
• If heat is lost to the surroundings and
work is done on the system what are
the signs for q & w?
q=w=+
Summary for q & w
•
•
•
•
q + = heat into system
q - = heat into surroundings
w + = work done on the system
w - = work done on the surroundings
Examples
• A system loses 1150 J of heat to the
surroundings and does 480 J of work on
the surroundings. Calculate E.
• E = q + w
• E = (-1150J) + (-480J)
• E = -1630 J
Examples
• A system absorbs 140 J of heat from the
surroundings and does 85 J of work on the
surroundings. Calculate E.
• E = q + w
• E = (+ 140J) + (-85J)
• E = + 55 J
Endothermic & Exothermic
• Endothermic
– system absorbs heat
– Heat flows into the system
– Temperature goes down
• Exothermic
– Heat flows out of the system and into the
surroundings
– Temperature goes up
• Only look at heat (q) to determine if the
system is endo or exo
for example,
1. Ice melting at room temperature
2. Dissolving sugar in hot coffee
3. Na(s) + H2O(l) Na+(aq) + OH-(aq) +
H2(g)
Entropy
1st Law of Thermodynamics
• Energy is neither created nor destroyed
– The energy of the universe is constant
– Energy just changes from one form to another
• This law lets us know the energy of the
system, but does not give us any
information about the direction of the
energy flow
Spontaneity
Spontaneous process – a process that occurs
without intervention
Spontaneous processes can be fast or slow
Spontaneity tells us the direction of the energy
flow
It tells us NOTHING about the speed of
the reaction
For example…
• A ball spontaneously rolls down a hill
– It does not spontaneously roll up
• If iron is exposed to air, it spontaneously
rusts
– The rust does not spontaneously turn back
into air & iron
Entropy
• Entropy (S) – the measure of molecular
randomness or disorder
– Think of entropy as the amount of chaos
• The driving force for a spontaneous
process is an increase in entropy
Entropy
• The natural progression of things is from order
to disorder
• Or from lower entropy to higher entropy
– Think of a deck of cards…when you drop one it goes
from order to disorder
– Think of your room… it goes from order to disorder
• Think if entropy as a probability…not a certainty
– Nature spontaneously proceeds toward that states
that have the highest probability of existing
2nd Law of Thermodynamics
• 2nd Law of Thermodynamics – In any spontaneous
process there is always an increase in entropy of the
universe
• Energy is conserved…entropy is NOT conserved!
• The entropy of the universe is always increasing
– S univ = + = spontaneous
– S univ = - = not spontaneous (would be spontaneous in the
opposite direction)
– S univ = 0 = no tendency to occur (system is at equilibrium)
• S univ = S sys + S surr
Temperature & Spontaneity
Ssurr = _ H
T
T must be in Kelvin
 H is usually given in KJ/mol
 S will be in KJ/K, but is usually changed to
J/K
Example
Sb4O6 (s) + 6C (s)  4Sb (s) + 6CO (g) H = 778 kJ
Calculate Ssurr for this reaction at 25C and 1 atm
S = - (H/T)
S = - (778KJ/298K)
S = -2.61 KJ/K
S = -2610 J/K
3rd Law of Thermodynamics
• The entropy of a perfect crystal at 0K is
zero
The Surroundings
The
Universe
The System
Lecture - Entropy
Spontaneous Processes
the system
spontaneous change - no action
from outside the system is
necessary
Non-spontaneous reactions...
1. 2 Fe(l) + Al2O3(s) 2 Al(s) + Fe2O3(s)
2. Boiling water at 1 atm, 50oC
3. H2O(l) H2(g) + ½ O2(g) at 25oC
What about the reverse
reactions?
1. Water freezing at room
temperature?
(ice melts)
2. Sugar precipitating out of hot coffee?
(sugar dissolves)
3. Na+(aq) + OH-(aq) + H2(g)  Na(s) +
H2O(l) ?
(reverse reaction occurs)
Entropy
•
Predict the sign of the entropy change for
the following…
1. Sugar is added to water to form a
solution
+
2. Iodine vapor condenses on a cold
surface to produce a liquid
-
so why does this reaction
work?
NH4Cl(s) + H2O(l) NH4+(aq) + Cl(aq)
Ho = + 15.1 kJ/mol
or this one?
H2O(l)  H2O(g)
H = + 44 kJ/mol
C3H8(l) C3H8(g)
H = +16.7
kJ/mol
g
l
cold!
Heat vs. Energy
spontaneous change = heat
released?
NOT ALWAYS!
spontaneous change = energy
released?
ALWAYS!
Entropy, S
- a measure of the disorder of the
system
Entropy tends to increase
breaking a glass
using the library
not cleaning up your house
1st law of thermodynamics:
The energy of the universe is
constant
(The best you can do is break
even)
2nd law of thermodynamics:
The entropy of the universe is
increasing
(You can’t break even)
Low entropy is less probable
according to Boltzmann,
S = k ln W
k = Boltzmann constant
= 1.38 x 10-23 J K-1
W = number of ways that state
can be achieved
royal flush: 4 hands possible
S = 1.38 x 10-23 J K-1 ln(4)
= 1.9 x 10-23 J K-1
pair: 1,098,240 hands possible
S = 1.38 x 10-23 J K-1 ln(1098240)
1.9 x 10-22 J K-1
Entropy is a State Function
S = Sfinal - Sinitial
path taken is irrelevant
rate of change is irrelevant
S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
H2O(s)  H2O(l)

ordered,
less ordered,
low S
high S
S > 0
S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
H2O(l) H2O(g)
high entropy
low entropy
S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
Benzene
low entropy
high entropy
Toluene
Very
unlikely!
More
likely!
recall this one?
NH4Cl(s) + H2O(l)  NH4+(aq) + Cl(aq)
H>0 (unfavorable)
S>0 (favorable!!)
S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
Ba(OH)28H2O(s) + 2 NH4NO3(s)  Ba(NO3)2(aq) + 2 NH3(aq) + 10 H2O(l)
H = +80.3 kJ (unfavorable)
3 moles  13 moles
S > 0 (favorable)
S > 0 for:
- melting
- vaporizing
- making a solution
- a reaction that produces
an increased number of moles
- heating a substance
G
Entropy
Svaporization
Sfusion
L
S
Temperature
Calculating S for a reaction
Srxn = S(ni Soi)
absolute entropy
stoichiometric
coefficient of species i of species i
for example,
C8H18(g) + 12.5 O2(g) 8 CO2(g) + 9 H2O(g)
13.5 moles 17 moles
(expect S > 0)
Srxn = S(ni Soi)
Entropy
• Predict which has the highest entropy
1. CO2 (s) or CO2 (g)
CO2 (g)
2. 1 mol of N2 at 1 atm or 1 mol of N2 at
0.001 atm
1 mol of N2 at 0.001 atm
for example,
= 8 Soi(CO2(g)) + 9 Soi(H2O(g)) Soi(C8H18(g)) - 12.5 Soi(O2(g))
= 8(213.6) + 9(188.6) - 463.2 12.5(204.8)
= +383.0 J K-1 mol-1
The Food Calorie
1 kcal = 1000 cal
1 Cal = 1 kcal
1 Cal = 4180 J
2,000 Cal/day is a meager diet.
what if you drank gasoline?
How much would you need?
1.
2.
Assume gasoline is octane (C8H18(l))
Find Hcombustion
MW = 114.2 g/mol
r = 0.70 g/mL
How much would you need?
1.
2.
Assume gasoline is octane (C8H18(l))
Find Hcombustion
C8H18(l) + 12.5 O2(g) 8 CO2(g) + 9 H2O(g)
Hcombustion = 8(-393)+9(-285.6) - (-208.2)
= -5506.2 kJ/mol
How much would you need?
4.18 J
2000 Cal 1000 cal
x
x
day
Cal
cal
x
1 mol gas
114.2 g gas
1.0 mL gas
x
x
x
5506 x 103 J 1 mol gas
0.7 g gas
= 248 mL gas / day
100
90
80
70
60
50
40
30
20
10
0
U.S. Energy Usage
Wood
Coal
Petroleum/Natural
Gas
1850 1900 1940 1980 1990
Hydro and
Nuclear
Coal
buried plant material
coal
empirical formula = CH2O
MW 500,000
heat,
pressure,
time
Coal gets better with age
Lignite
Subbituminous
Bituminous
Anthracite
71
77
80
92
increasing Hcombustion
%C
Burning coal
Coal + Air CO2(g) + H2O(g) +
heat
+ CO(g) + SO2(g) + ash + NOx(g)
Other Energy Sources
1. Coal conversion
- convert to gaseous fuels
- large molecules small
molecules
- must break C-C bonds
- C-H and C-O bonds are made
... or chemical
2 CaBr2 + 4 H2O(l)  2 Ca(OH)2 + 4
HBr
H>0
2 Hg + 4 HBr 2 HgBr2 + 2 H2(g)
H<0
2 HgBr2 + 2 Ca(OH)2 2 CaBr2 + 2
HgO
+ 2 H2O(l)
H>0
2 HgO 2 Hg + O2(g)
H>0
Ti(s) + x/2 H2(g)
favored at higher
temperature
TiHx(s)
favored at lower
temperature
Free Energy
• Free energy (G) – a thermodynamic
function equal to the enthalpy minus the
product of the entropy and the Kelvin
temperature
• G = H – TS
• A process is only spontaneous in the
direction where G is negative
Example
• At what temperatures is the following process
spontaneous at 1 atm?
– Br2(l)  Br2(g)
– H = 31.0 KJ/mol  31000 J/mol
– S = 93.0 J/ K mol
•
•
•
•
G = H – TS
0 = 31000 – T(93.0)
T = 333K
Above 333K the reaction is spontaneous
Example
Sb2S3 (s) + 3Fe (s)  2Sb (s) + 3FeS (s) H = -125 kJ
Calculate Ssurr for this reaction at 25C and 1 atm
S = - (H/T)
S = - (-125KJ/298K)
S = 0.419 KJ/K
S = 419 J/K
Dependence of H & S on
Spontaneity
G = H-TS
H
-
S
+
+
+
-
-
+
-
Result
Spontaneous at all temperatures
Spontaneous at high temperatures
Spontaneous at low temperatures
Not spontaneous at any temperature
Hydrocarbons and Heat
• Most hydrocarbons are used as fuels.
• Knowing how much energy a fuel provides,
can tell us if it is useful for a certain application.
• For example, the amount of energy a food
releases when burned, can tell us about it’s
caloric content (fats release lots of energy).
• Heat energy released during combustion can
be measured with a calorimeter.
• A “bomb calorimeter” is shown. It includes wate
in a heavily insulated container, a stirrer, valve,
bomb chamber, ignition wires,
& a thermometer
Exothermic and Endothermic changes
• An alternative to the bomb calorimeter is a
“coffee cup” calorimeter, where two nested
polystyrene cups take the place of the container
• In either case, the change in heat of the water
tells us about the reaction of the chemicals.
• An increase in water temperature indicates that
the chemicals released energy when they
reacted. This is called an “exothermic” reaction.
• In an “endothermic” reaction, water temperature
decreases as the chemicals absorb energy.
• We will see that heat is measured in Joules (J)
or kiloJoules (kJ). Before we do any heat
calculations, you should know several terms …
Specific heat capacity
The heat needed to  the temperature of 1 g of
a substance by 1 C.
Symbol: c, units: J/(gC).+
Heat capacity
The heat needed to  the temperature of an
object by 1 C. Symbol: C (=c x m), units:
J/C
Heat of reaction
The heat released during a chemical
reaction. Symbol: none, units: J.
Specific heat (of reaction)
The heat released during a chemical reaction
per gram of reactant.
Symbol: h, units: J/g.
Molar heat of reaction
The heat released during a chemical reaction
per mole of reactant. Symbol: H, units: J/mol.
Heat Calculations
• To determine the amount of heat a substance
produces or absorbs we often use q = cmT
• q: heat in J, c: specific heat capacity in J/(gC),
m: mass in g, T: temperature change in C,
• This equation makes sense if you consider units
J
J=
x g x C
gC
For a list of c values see page 568 (table 3)
Sample problem: (must know water = 4.18 J/gC)
When 12 g of a food was burned in a calorimeter,
the 100 mL of water in the calorimeter changed
from 20C to 33C. Calculate the heat released.
q=cmT = 4.18 J/(gC) x 100 g x 13C = 5.4 kJ
More practice
1. 5 g of copper was heated from 20C to 80C.
How much energy was used to heat the Cu?
q=cmT = 0.38 J/(gC) x 5 g x 60 C = 114 J
2. If a 3.1 g ring is heated using 10.0 J, it’s temp.
rises by 17.9C. Calculate the specific heat
capacity of the ring. Is the ring pure gold?
q=cmT
10.0 J
c = q/mT=
= 0.18 J/(g°C)
3.1 g x 17.9C
The ring is not pure. Gold is 0.13 J/(gC) -pg. 568
5. q=cmT = 4.18 J/(gC) x 2570 g x 92 C
= 988319 J = 988 kJ = 0.988 MJ
6. q=cmT
1 750 000 J
T = q/cm =
= 33.5°C
4.18 J/(gC) x 12500 g
Since the temperature started at 5.0°C, the
final temperature is 38.5°C.
7. q=cmT = 0.86 J/(gC) x 2500 g x 335C
= 720 kJ or 0.72 MJ
8. Total heat = water heat + pot heat
= 4.18 J/(gC) x 1200 g x 53.0C = 265 848 J
= 0.510 J/(gC) x 450 g x 53.0C = 12163.5 J
= 278 kJ
How much heat is required to change 50.0 g of ice at
0 degrees to steam at 100 degrees
You are looking for q
water at
0C
Ice at
0C
Hfmi
+
water at
100 C
mwcwT
+
Hf for water = 334 j/g
= 80 cal
(334j)(50.0 g)
g
Hvmw
Hv = 2257 j
= 540 cal
+ (50.0g)(4.18 j)(100 C)
gC
16700 j +
steam at
100 C
20900 j
+
+ (2257 j) ( 50.0 g)
g
112850 j
Heat Capacity Calculations
• Recall that heat capacity (J/C) is different from
specific heat capacity (J/gC).
• Heat capacity is sometimes a more useful value
• For example, because a calorimeter includes
wires, the stirrer, thermometer, etc. some heat
will be transferred to these other materials.
• Rather than having to calculate q for each
material a J/C value is used.
Sample problem:
q=cmT
A calorimeter has a heat capacity of 2.05 kJ/C.
How much heat is released if the temperature
change in the calorimeter is 11.6C?
q = cm T q= 2.05 kJ/C x 11.6 C = 23.8 kJ
Thermochemical Equations
Thermochemical equations are chemical
equations with an added heat term.
• KBrO3(s) + 42 kJ  KBrO3(aq)
This is endothermic (heat is absorbed/used)
• 2 Mg(s) + O2(g)  2 MgO(s) + 1200 kJ
This is exothermic (heat is produced/released)
Read 12.3 (pages 582 – 585). Do 2-6 (pg. 585).
Sample problem:
3.00 g of octane was burned in a calorimeter
with excess oxygen, the 1000 mL of water in
the calorimeter rose from 23.0C to 57.6C.
Write the thermochemical equation for octane,
representing the molar heat of combustion.
2. Specific refers to mass in grams.
3. Specific heat of reaction: J/g or kJ/g, etc.
Molar heat of reaction: J/mol or kJ/mol, etc.
4. J/mol = J/g x g/mol (molar mass is used)
5. Exothermic.
E.g. CH4 + 2O2  CO2 + 2H2O + x kJ
Other examples include propane, octane, Mg
6.
49.90 kJ 26.04 g
= 1299 kJ/mol
x
g
mol
C2H2 + 2.5O2  2CO2 + H2O + 1299 kJ
or 2C2H2 + 5O2  4CO2 + 2H2O + 2598 kJ
Sample: First, calculate the heat released by the
combustion reaction via q=cmT …
= 4.18 J/(gC) x 1000 g x 34.6 C = 144 628 J
Octane (C8H18) has a molar mass of 114.26 g/mol
We can determine the molar heat of reaction
1) via the specific heat of reaction or 2) directly
h = J = 144 628 J H = 48.2 kJ x 114.26 g
g
3.00 g
g
mol
= 48209 J/g = 48.2 kJ/g
= 5508 kJ/mol
or
J
144 628 J
= 5508 kJ/mol
H =
=
mol
0.0263 mol
C8H18 + 12.5O2  8CO2 + 9H2O + 5508 kJ
For more lessons, visit
www.chalkbored.com
Entropy & Chemical
Reactions
2nd Law of Thermodynamics
• A process will be spontaneous is the
entropy of the universe increases
• Now we will look at entropy regarding to
chemical reactions
What is the sign for S?
• N2(g) + 3H2(g)  2NH3(g)
• First look at the states…if you go from a solid or
a liquid to a gas, you will have a + entropy
• In this case, all of the states are the same, so we
look at the number of moles
• N2(g) + 3H2(g)  2NH3(g)
• 1
+ 3
vs. 2
• The entropy decreases because you go from 4
moles to 2
• S is negative
What is the sign for S?
• 4NH3 + 5O2  4NO + 6H2O
• 9 moles vs. 10 moles
• S increases (+)
Calculating S
• Calculating S is just like calculating H
• Simply use the Appendix…just look at the
column for S instead of H
• S° of any element or diatomic molecule is
NOT zero.
• You must look these up!
Example
•
•
•
•
Calculate S for the following reaction:
2NiS(s) + 3O2(g)  2SO2(g) + 2NiO (s)
2(-53) + 3(-205) + 2(248) + 2(38)
-149 J/K
Example
•
•
•
•
Calculate S for the following reaction:
Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g)
(-51)
+ 3(-131) + 2(28) + 3(189)
179 J/K
Gibbs Free Energy & Chemical
Reactions
•
1.
2.
3.
You can calculate G in 3 ways…
Like Hess’s Law
Like H°
With the equation G = H - T S
Example
• Calculate H, S, & G at 25°C using the
following data…
• 2SO2 + O2  2SO3
Substance
H (KJ/mol) S (J/K mol)
SO2
-297
248
SO3
-396
257
O2
0
205
Example
2SO2 + O2  2SO3
H = 2(+297)+(0) + 2(-396) = -198 KJ
S = 2(-248)+ (-205) + 2(257) = -187J/K
G = H – T S
=(-198) – (298 x -0.187)
=-142 KJ
Calculate G
• Using the following data at 25°C
•
•
•
•
Cdiamond + O2(g)  CO2 (g)
G = -397KJ
Cgraphite + O2(g)  CO2 (g)
G = -394KJ
Calculate G for the reaction:
C diamond  C graphite
• G = -3KJ
Calculating G
• Methanol is a high octane fuel used in high
performance racing engines. Calculate G for the
following reaction
• 2 CH3OH(g) + 3O2(g)  2CO2(g) + 4H2O(g)
• Given the following free energies of formation:
Substance
CH3OH (g)
O2 (g)
CO2 (g)
H2O (g)
G°(KJ/mol)
-163
0
-394
-229
Calculating G
• 2 CH3OH(g) + 3O2(g)  2CO2(g) + 4H2O(g)
• 2(163) + 3(0) + 2(-394) + 4(-299)
• -1378 KJ
Hess’s law
• Hess’s Law states that the heat of a whole
reaction is equivalent to the sum of it’s steps.
• For example: C + O2  CO2
The book tells us that this can occur as 2 steps
C + ½O2  CO
H = – 110.5 kJ
CO + ½O2  CO2
H = – 283.0 kJ
C + CO + O2  CO + CO2 H = – 393.5 kJ
I.e. C + O2  CO2
H = – 393.5 kJ
• Hess’s law allows us to add equations.
• We add all reactants, products, & H values.
• We can also show how these steps add
together via an “enthalpy diagram” …
Steps in drawing enthalpy diagrams
1. Balance the equation(s).
2. Sketch a rough draft based on H values.
3. Draw the overall chemical reaction as an
enthalpy diagram (with the reactants on one
line, and the products on the other line).
4. Draw a reaction representing the intermediate
step by placing the relevant reactants on a line.
5. Check arrows: Start: two leading away
Finish: two pointing to finish
Intermediate: one to, one away
6. Look at equations to help complete balancing
(all levels must have the same # of all atoms).
7. Add axes and H values.
C + ½O2  CO
CO + ½O2  CO2
H = – 110.5 kJ
H = – 283.0 kJ
 CO2
H = – 393.5 kJ
C
+ O2
C + O2
Reactants
Products
Enthalpy
Intermediate
H = – 110.5 kJ
CO + ½O2
H =
– 393.5 kJ
H =
– 283.0 kJ
CO2
Note: states such as (s) and (g) have been ignored to reduce
clutter on these slides. You should include these in your work.
Practice Exercise 6 with Diagram
Using example 5.6 as a model, try PE 6.
Draw the related enthalpy diagram.
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H= –1411.1 kJ
2CO2(g) + 3H2O(l)  C2H5OH(l) + 3O2(g) H= +1367.1 kJ
Intermediate
H =
– 1411.1 kJ
Reactants
Products
Enthalpy
C2H4(g) + H2O(l)  C2H5OH(l)
H= – 44.0 kJ
C2H4(g) + H2O(l) + 3O2(g)
C2H5OH(l)+ 3O2(g) H=
H =
+1367.1 kJ
2CO2(g) + 3H2O(l)
– 44.0 kJ
GeO(s)
 Ge(s) + ½ O2(g) H= + 255 kJ
Ge(s) + O2(g)  GeO2(s)
H= – 534.7 kJ
GeO(s) + ½ O2(g) GeO2(s)
H= – 279.7 kJ
Reactants
Products
Enthalpy
Intermediate
H =
– 534.7 kJ
Ge(s) + O2(g)
H = +255 kJ
GeO(s) + ½ O2(g)
GeO2(s)
H=
– 280 kJ
NO(g)
 ½ N2(g) + ½ O2(g) H= – 90.37 kJ
½ N2(g) + O2(g)  NO2(g)
H= + 33.8 kJ
H= – 56.57 kJ
NO(g) + ½ O2(g)  NO2(g)
NO + ½ O2(g)
Intermediate
H =
– 90.37 kJ
Products
Enthalpy
Reactants
NO2(g)
H = +33.8 kJ
½ N2(g) + O2(g)
H=
– 56.6 kJ
Hess’s law: Example
We may need to manipulate equations further:
2Fe + 1.5O2  Fe2O3 H=?, given
Fe2O3 + 3CO  2Fe + 3CO2 H= – 26.74 kJ
CO + ½O2  CO2
H= –282.96 kJ
1: Align equations based on reactants/products.
2: Multiply based on final reaction.
3: Add equations.
2Fe + 3CO2  Fe2O3 + 3CO H= + 26.74 kJ
CO + ½O2  CO2
H= –282.96 kJ
3CO + 1.5O2  3CO2
H= –848.88 kJ
2Fe
+ 1.5O2  Fe2O3
H= –822.14 kJ
Don’t forget to add states.
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