Transcript Thermochemistry ppt

THERMOCHEMISTRY
HEAT AND CHEMICAL
CHANGE
THE FLOW OF ENERGY
ENERGY TRANSFORMATIONS
Thermochemistry is concerned with the heat changes that occur
during chemical reactions.
Energy is the capacity for doing work or supplying heat.
Energy stored within the structural units of chemical substances
is called chemical potential energy.
Heat, represented by q, is energy that transfers from one object
to another because of a temperature difference between them.
Heat always flows from the warmer object to a cooler object.
EXOTHERMIC AND ENDOTHERMIC PROCESSES
A system is the part of the universe on which you
focus your attention.
The surroundings include everything else in the universe.
The system and the surroundings make up the universe.
The law of conservation of energy states that in any chemical or
physical process energy is neither created nor destroyed.
A process which absorbs heat from the surroundings is called an
endothermic reaction.
A process that releases energy to its surroundings is called an exothermic
reaction.
HEAT CAPACITY AND SPECIFIC HEAT
A calorie (c)is defined as the quantity of heat needed
to raise the temperature of 1 gram of water 1oC.
A kilocalorie (C) is one thousand calories.
The SI unit for energy is the joule (J). One calorie is equal to
4.184 J.
The amount of heat to increase the temperature of an object 1oC is
the heat capacity of that object.
The specific heat of an object is the amount of energy necessary to
raise one gram of the substance 1oC (J/goC).
At 4.184 J/goC, water has a very high specific heat. Most metals have a
relatively low specific heat.
The vast amount of water that covers Earth (about 3/4ths it surface) results in
slow heating of the Earth’s surface and,corresponding, slow cooling. This
moderates the Earth’s temperature allowing life as we know it to exist.
CALCULATIONS:
The standard formula for the calculation for
changes in energy is:
q = m (T)Cp
Where: q = heat of reaction (heat exchanged in joules)
m = mass of material
T = change in temperature (initial
temperature (ti) - final temperature (tf).
Cp = specific heat of the material
Sample Problem 1.
A 37 mL (37 g) sample of water at an initial temperature
of 13oC is heated to 34o C. What is the amount of heat
gained by the water?
q = m(T)Cp
m = 37 g
ΔT = 34oC – 13oC = 21oC Cp water = 4.184 J/goC
q = 37 g (21oC)4.184 J/goC
q = 3251 JgoC/goC
=
3251 J
Practice Problem 1:
What is the heat gained by 27 grams of a metal with a
specific heat of 0.177 J/goC, if its initial temperature is
44oC and its final temperature is 71oC?
Solving:
q = m(T)Cp
q = 27g(27oC)0.177J/goC
q = 129 J
Sample Problem 2:
What is the mass of zinc which, when heated goes
from an initial temperature of 88OC to 131oC and
gains 1678 J of heat? The Cp of zinc is 0.388 J/goC.
q = m(T)Cp
q = 1678 J
T = 131 – 88 = 43oC
m=g?
Cp = 0.388 J/goC
1678 J = m(43oC)0.388 J/goC
1678 J = m(16.68 oCJ/goC
1678 Jg
16.68 J
=
m
=
=
100.6 g
1678 J = m 16.68 J/g
Practice Problem 2:
What is the mass of a metal with a specific heat of
1.44 J/goC which when heated begins at 25oC and
has a final temperature of 32oC and gains 1145 J?
Solving:
q = m(T)Cp
1145 J = m(7oC)1.44 J/goC
1145 g = m10.08
m = 113.6 g
Sample Problem 3:
What is the specific heat of 50 g of a metal which, when
heated to 97oC and added to 30 g of water with an initial
temperature of 14oC, results in the final temperature of the
water being 27oC.
Since we are dealing with both water and a metal, we need an equation of each
one.
q = m(T)Cpwater
q = m(T)Cpmetal
Entering what we know in each equation and solving were possible we have
the following:
The water equation:
q = 30(13)4.184
q = 1631 J
Note: Because the hot metal will exchange heat with the cooler water until
they both reach the same temperature (equilibrium), the final temperature of
the metal and the water will be the same.
The metal equation: q = 50(70)Cp
The equation appears to be unsolvable because we have two unknowns but
remember the law of conservation of energy states that whatever energy was
gained by the water had to be lost by the metal, therefore the two q’s are equal.
Sample Problem 3 continued:
1631 J = 50 g (70oC)Cp
1631 J = 3500 gOC Cp
1631 J/3500 goC = Cp
0.466 J/goC = Cpmetal
Practice Problem 3:
What is the specific heat of a metal if 42 g of the metal
heated to 99oC is added to 25 g of water with an initial temperature
of 10.0oC? The water final temperature is 18.0oC.
q = m(ΔT)Cpwater
q = m(ΔT)Cpmetal
q = 25(8)4.184
836.8 J = 42(81)Cp
q = 836.8 J
836.8 J = 3402 Cp
836.8 J = 3402 Cp
3402
3402
Cp = 0.246 J/goC
Sample Problem 4:
Copper heated to 86oC is added to 78 g of water at an
initial temperature of 14oC. The water rises to a final temperature
of 25oC. The specific heat of copper is 0.385 J/goC. What is the mass
of copper used in this experiment?
q = m(T)Cpwater
q = m(T)Cpmetal
q = 78 g(11oC)4.184 J/goC
3590 J = m(61oC)0.385 J/goC
q = 3590 J
3590 g/23.5 = m
m = 152.8 g
Practice Problem 4:
Aluminum is heated to 100.0oC and then added
to 47 g of H2O with a Ti of 8.0oC. The water increases
to a Tf of 63.0oC. The Cp of aluminum is 0.9025 J/goC.
What is the mass of aluminum used in this experiment?
q = m(ΔT)Cpwater
q = m(ΔT)Cpaluminum
q = 47(55)4.184
10 816 J = m(37)0.9025
q = 10 816 J
10 816 J = 33.4 m
10 816 J = 33.4 m
33.4
33.4
323.8 g = m
Sample Problem 5:
What is the final temperature (Tf) of a system
in which 60 g of iron heated to 98oC is added to
25 g of water with an initial temperature of 12oC?
The Cp of iron is 0.449 J/goC.
Again we begin by writing a water and iron equation and
inserting the know values.
q = m(T)Cpwater
q = m(T)Cpiron
q = 25(Tf – 12)4.184
q = 60(98 –Tf)0.449
Each equation has two unknowns but we know that the
q of each equation must be equal. We can therefore put
the two equations together as follows:
Sample Problem 5 continued:
25(Tf – 12)4.184 = 60(98 – Tf)0.449
Now simplify using distributive properties as follows:
(25Tf – 300)4.184 = (5880 – 60Tf)0.449
104.6Tf – 1255.2 = 2640 – 26.9Tf
Group like items on each side of the equation:
104.6Tf + 26.9Tf = 2640 + 1255.2
131.5Tf = 3895.2
131.5Tf
131.5
=
3895.2
131.5
Tf = 29.6oC
Practice Problem 5:
What is the Tf of a system in which 34 g of
nickel heated to 99oC is added to 40 g of water
with a temperature of 10oC? The Cp of nickel is 0.444 J/goC.
q = m(T)Cpwater
q = m(T)Cpnickel
q = 40(Tf – 10)4.184
q = 34(99 – Tf)0.444
40(Tf – 10)4.184 = 34(99 – Tf)0.444
(40Tf – 400)4.184 = (3366 – 34Tf)0.444
167.4Tf – 1673.6 = 1494.5 – 15.1Tf
167.4Tf + 15.1Tf = 1494.5 + 1673.6
182.5Tf = 3168.1
182,5Tf
182,5
= 3168,1
182.5
Tf = 17.4oC
Sample Problems 6: J ↔ c and c ↔ C
a. 9.67 c = J
9.67 c
4.184 J
1c
=
40.5 J
=
857.6 c
=
9430 c
b. 3588 J = c
3588 J
1 c___
4.184 J
c. 9.43 C = c
9.43 C
1000 c
1C
Practice Problems 6:
a. 19.6 c = J
19.6 c 4.184 J = 82.0 J
1c
b. 3.44 J = c
3.44 J
1 c__
= 0.82 c
4.184 J
c. 11 864 c = C
11 864 c
d. 24.1 C = c
24.1 C
e. 378.2 J = C
378.2 J
1 C__
1 000 c
1000 c
1C
=
=
11.864 C
24 100 c
1c
1 C__
= 0.0904 C
4.184 J 1000 c
CALORIMETRY
Calorimetry is the accurate and precise
measurement of heat exchange for chemical
and physical processes.
The insulated device used to measure the absorption
or release of heat in a chemical or physical processes
is called a calorimeter.
Calorimeters may be as simple as a styrofoam cup and
lid or a complex device such as a bomb calorimeter.
Calorimetry continued:
Enthalpy (H) is the heat content of
a system at constant pressure.
Because most chemical reactions and physical changes
carried out in the laboratory are open to the atmosphere,
these changes occur at constant pressure.
Therefore:
q = H = m(T)Cp
THERMOCHEMICAL EQUATIONS
If you mix calcium oxide with water, an exothermic
reaction takes place and the water becomes warm.
When 1 mol of calcium oxide reacts with 1 mol of water,
1 mol of calcium hydroxide forms and 65.2 kJ of heat is
released.
You can show this in the chemical equation by including
heat change as a product of the reaction.
CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ
An equation that includes the heat change is called a
thermochemical equation.
Thermochemical Equations continued:
A heat of reaction is the heat change for the
equation exactly as it is written.
It is usually reported as H, the heat change at
constant pressure (the change in enthalpy).
In a thermochemical equation, the physical state of
the reactants and products must also be given.
Thermochemical Reactions continued:
In an exothermic reaction, heat is given off.This means,
according to the law of conservation of energy, that the
chemical potential energy of the resulting product(s) is
less than the energy of the reactant(s). Because of this
loss of energy, the heat of reaction (H) is negative.
CaO(s) + H2O(l) → Ca(OH)2(s) + 65.2 kJ
H = -65.2 kJ
Thermochemical Equations continued:
The decomposition of baking soda, sodium
hydrogen carbonate, by heat is an endothermic reaction.
The products have more chemical potential energy than
the reactants therefore the H is positive.
2NaHCO3(s) + 129 kJ → Na2CO3(s) + H2O(g) + CO2(g)
H = 129 kJ
Mass – Energy Stoichiometry
The mass – energy relationships within a
chemical equation can be determined in a manner
similar to mass – mass stoichiometry.
The steps to mass – energy stoichiometry are:
1. Balance the equation
2. Convert grams of given substance to moles.
3. Give the ratio of energy to moles of given substance.
4. Solve
Mass – Energy Stoichiometry continued:
Using the reaction below, what amount of energy
will be produced by the combination of 155 g of
carbon monoxide (CO) with a excess of iron (III) oxide?
Fe2O3(s) + 3CO(g) → 2 Fe(s) + 3 CO2(g) + 26.3 kJ
Step 1. Balance the equation
Step 2. Convert grams of given to mols
Step 3. Give ratio of mols given substance to energy
Step 4. Solve
155 g CO
1 mol CO 26.3 kJ
28.0 g CO 3 mol CO
=
48.5 kJ
Mass – Energy Stoichiometry continued:
Practice Problem:
Gasohol contains ethanol (C2H5OH)(l), which when burned
reacts with oxygen to produce CO2(g) and H2O(g). How
much heat is released when 12.5 g of ethanol burns?
C2H5OH(l) + 3 O2(g) → 2 CO2(g) +3 H2O(g) + 1235 kJ
12.5 g C2H5OH
1 mol C2H5OH
1235 kJ
=
46.1 g C2H5OH 1 mol C2H5OH
334.9 kJ
HEATS OF FUSION AND SOLIDIFICATION
All solids absorb heat as they melt to become
liquids.
The heat absorbed by one mole of a substance in
melting from a solid to a liquid at constant temperature
is the molar heat of fusion (Hfus).
The heat lost when one mole of liquid solidifies at a
constant temperature is the molar heat of
solidification (Hsolid).
Hfus = - Hsolid
Heats of Fusion and Solidification of Water
The melting of one mole of ice to one mole of water
at 0oC requires the absorption of 6.01 kJ of heat.
Hfus = 6.01 kJ
The solidification of one mole of water at 0oC to one mole of
ice at 0oC releases 6.01 kJ.
Hsolid = - 6.01 kJ
Heat of Fusion – Sample Problem:
How many grams of ice at 0oC and 101.3 kPa
could be melted by the addition of 2.25 kJ of heat?
2.25 kJ
1 mol ice
6.01 kJ
18.0 g ice
1 mol ice
=
6.74 g ice
Heat of Fusion – Practice Problem:
How many grams of ice at 0oC and 101.3 kPa
could be melted by the addition of 0.400 kJ of heat?
0.400 kJ
1 mol ice 18.0 g ice
6.01 kJ 1 mol ice
=
1.2 g ice
HEAT OF VAPORIZATION AND CONDENSATION
The amount of heat necessary to vaporize one mole of a
given liquid is called its molar heat of vaporization (Hvap).
The molar heat of vaporization of water is 40.7 kJ/mol.
This means, to vaporize one mole of water, 40.7 kJ of energy
must be provided.
The amount of heat released when one mole of vapor
condenses is called the molar heat of condensation,
Hcond.
The molar heat of condensation for water vapor is – 40.7 kJ.
Heat of Vaporization Practice Problem:
How much heat (in kJ) is absorbed when 24.8 g H2O(l)
at 100oC is converted to steam at 100oC?
24.8 g H2O
1 mol H2O 40.7 kJ = 56.1 kJ
18.0 g H2O 1 mol H2O
Heat of Vaporization Practice Problem:
How much heat (in kJ) is absorbed when 63.7 g H2O(l)
at 100oC is converted to steam at 100oC?
63.7 g H2O
1 mol H2O 40.7 kJ =
18.0 g H2O 1 mol H2O
144.0 kJ
HEAT OF SOLUTION
The heat change caused by dissolution of one mole
of substance in the molar heat of solution (Hsoln).
Sodium hydroxide is a good example of an exothermic
molar heat of solution. When 1 mol of sodium hydroxide
(NaOH)(s) is dissolved in water, the solution can become so
hot that it steams. The heat is released as the sodium ions
and the hydroxide ions separate and interact with the water.
The temperature of the solution increases, releasing 445.1
kJ of heat as the molar heat of solution (Hsoln = - 445.1 kJ/mol)
A practical application of exothermic heat of solution is the
popular hot packs to relieve pain.
Heat of Solution continued:
The dissolution of ammonium nitrate
(NH4NO3)(s) is an example of an endothermic process.
When ammonium nitrate dissolves in water, the solution
becomes very cool.
The heat absorbed as the ammonium and nitrate ions
of one mole of ammonium nitrate separate is the molar
heat of solution (Hsoln = 25.7 kJ/mol)
Heat of Solution Sample Problem:
How much heat (in kJ) is released when 100.0 g of
NaOH(s) is dissolved in water?
100.0 g NaOH 1 mol NaOH
-445.1 kJ = - 1112.8 kJ
40.0 g NaOH 1 mol NaOH
Heat of Solution Practice Problem:
How many grams of NH4NO3(s) must be
dissolved in water so that 88.0 kJ of heat is released
from the water? (The Hsoln of NH4NO3 = 25.7 kJ/mol)
88.0 kJ 1 mol NH4NO3 80.0 g NH4NO3 = 273.9 g NH4NO3
25.7 kJ
1 mol NH4NO3
CALCULATING HEAT CHANGES
HESS’S LAW
Hess’s Law makes it possible to measure a heat of reaction
indirectly.
Hess’s law of heat of summation states that if you add
two or more thermochemical equations to give a final
equation, then you can also add the heats of reactions to
give the final heat of reaction.
Hess’s Law continued:
Carbon in the diamond form is less stable than
the graphite form therefore you would expect the following
reaction to take place:
C(diamond) → C(graphite)
This enthalpy change is too slow to be measured directly.
You can use Hess’s law to find the enthalpy changes for
the conversion of diamond to graphite by using the
following combustion reactions:
Hess’s Law continued:
a. C(s, graphite) + O2(g) → CO2(g) H = -393.5 kJ
b. C(s, diamond) + O2(g) → CO2(g) H = -395.4 kJ
Write equation a in reverse to give:
c. CO2(g) → C(s, graphite) + O2(g) H = 393.5 kJ
Note: When you reverse the reactions the sign for H
changes.
Adding the equations b and c you get the equation for the
conversion of diamond to graphite.
C(s,diamond) + O2(g) → CO2(g)
CO2(g) → C(s, graphite) + O2(g)
H = - 395.4 kJ
H = 393.5 kJ
C(s, diamond) → C(s, graphite)
H = - 1.9 kJ
Hess’s Law continued:
If you wished to determine the enthalpy change
for the formation of carbon monoxide from its elements,
the reaction would be:
C(s, graphite) + ½ O2(g) → CO(g)
H = ?
During this reaction carbon dioxide (CO2) is also produced so
that the measured heat of reaction relates to both.
Using Hess’s law and two reactions which can be carried out
in the laboratory we have:
a. C(s, graphite) + O2(g) → CO2(g) H = - 393.5 kJ
b. CO(g) + ½O2(g) → CO2(g)
H = - 283.0 kJ
Writing the reverse reaction of equation b and changing the
sign of H gives:
c. CO2(g) → CO(g) + ½O2(g)
H = 283.0 kJ
Hess’s Law continued:
Adding equations a and c yields:
C(s, graphite) + O2(g) → CO2(g)
H = - 393.5 kJ
CO2(g) → CO(g) + ½O2
H = 283.0 kJ
C(s, graphite) + ½O2 → CO(g)
H = - 110 .5 kJ
The formation of CO(g) is exothermic; 110.5 kJ of heat
is given off when 1 mol of CO(g) is formed from its elements.
STANDARD HEATS OF FORMATION
Sometimes it is hard to measure the heat
change for a reaction. In such cases you can
calculate the heat of reaction from standard heats of
formation.
The standard heat of formation (Hf0) of a compound
is the change in enthalpy that accompanies the formation
of one mole of a compound from its elements with all
substances in their standard state at 25oC.
Standard Heats of Formation continued:
The Hf0 of a free element in its standard state is
arbitrarily set at zero. For example, the Hf0 = 0 for the
diatomic molecules H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(l),
and I2(s).
Many values of Hf0 have been measured. The following
table lists Hf0 for some common substances. Standard heats
of formation of compounds are handy for calculating heats
of reaction at standard conditions.
The standard heat of reaction (H0) is the difference between
the standard heats of formation of all reactants and products.
This relationship can be expressed by the following equation.
H0 = Hf0(products) - Hf0(reactants)
Standard Heats of Formation (Hf0) at 25oC and 101.3 kPa
Substance
Al2O3(s)
Br2(g)
Br2(l)
C(s, diamond)
C(s, graphite)
CH4(g)
CO(g)
CO2(g)
CaCO3(s)
CaO(s)
Cl2(g)
F2(g)
Hf0
(kJ/mol)
- 1676.0
30.91
0.0
1.9
0.0
- 74.86
- 110 .5
- 393.5
-1207.0
- 635.1
0.0
0.0
Substance
Fe(s)
Fe2O3(s)
H2(g)
H2O(g)
H2O(l)
H2O2(l)
HCl(g)
H2S(g)
I2(g)
I2(s)
N2(g)
NH3(g)
Hf0
(kJ/mol)
0.0
- 822.1
0.0
- 241.8
- 285.8
- 187.8
- 92.31
- 20.1
62.4
0.0
0.0
- 46.19
Substance
Hf0
(kJ/mol)
NO(g)
90.37
NO2(g)
33.85
Na2CO3(s)
- 1131.1
NaCl(s)
- 411.2
O2(g)
0.0
O3(g)
142.0
P(s, white)
0.0
P)s, red)
- 18.4
S(s, rhombic)
0.0
S(s, monoclinic) 0.30
SO2(g)
- 296.8
SO3(g)
- 395.7
Standard Heats of Reaction continued:
What is the standard heat of reaction (ΔH0)
for the reaction of gaseous carbon monoxide with
oxygen to form gaseous carbon dioxide?
1. Write and balance the equation.
2 CO + O2 → 2 CO2
2. Find and sum the ΔHf0 of all reactants, taking into
account the number of moles of each.
2 mol CO(g)
1 mol O2(g)
- 110.5 kJ
1 mol CO(g)
0 kJ
1 mol O2(g)
= - 221.0 kJ
=
0 kJ
- 221.0 kJ
3. Find the ΔHf0 of the product
2 mol CO2
- 393.5 kJ
= - 787.0 kJ
1 mol CO2
4. Find the difference between ΔHf0(products)
and ΔHf0(reactants).
ΔH0 = (- 787.0 kJ) – ( - 221.0 kJ)
ΔH0 = - 566.0 kJ