Chemistry 30
Thermochemistry
Objectives
1.
Differentiate between heat and temperature (17-1).
2.
Define the term specific heat (17-1).
3.
Use heat equations to calculate :
a)
b)
c)
changes in heat content of a substance as a result of heating and cooling, and changes of state
(17-1).
mass of substances given heat and temperature change data (17-1).
specific heat of a substance given heat, temperature change and mass data (17-1).
4.
Discuss applications of the heat of vaporization and the heat of fusion (12-3).
5.
Perform calculations involving temperature change through one or more phase changes (in-class
notes).
6.
Discuss the concept of heat of reaction in terms of reorganization of bonds (17-1).
7.
Discuss Hess's Law (conservation of energy) (17-1).
8.
Use Hess's Law and standard heat of formation to calculate energy changes resulting from chemical
reactions (17-1).
9.
Explain the relationship between enthalpy change and the tendency of a reaction to occur (17-2).
10.
Explain the relationship between entropy change and the tendency of a reaction to occur (17-2).
11.
Discuss the concept of free energy, and explain how the value of this quantitiy is calculated and
interpreted. (17-2)
12.
Use the concepts of enthalpy and entropy to discuss Gibb's free energy and the spontaneity of chemical
reactions (17-2).
13.
Explain the concept of reaction mechanism (17-3).
14.
Discuss collision theory (17-3).
15.
Discuss reaction mechanisms, including an explanation of the activated complex (17-3).
16.
Compare and contrast between threshold energy and activation energy (in-class notes).
17.
Draw and interpret potential energy diagrams (17-3).
18.
Define chemical kinetics, and explain the two conditions necessary for chemical reactions to occur
(17-4).
19.
Identify the five major factors which affect the rate of a chemical reaction, and explain the effects in
terms of collision theory (17-4).
20.
Discuss catalysts and inhibitors. Explain how catalysts and inhibitors work (17-4).
21.
Explain and write laws for chemical reactions (17-4).
1
Vocabulary:
thermochemistry
heat
specific heat
fusion
condensation
vaporization
heat of fusion
endothermic
kilocalorie
exothermic
molar heat of formation
enthalpy
threshold energy
activated complex
activation energy
intermediates
catalyst
chemical kinetics
temperature
alorimeter
melting
heat of combustion
heat of vaporization
heat of reaction
calorie
Hess’ Law
Joule
free energy
entropy
free energy change
inhibitor
collision theory
reaction mechanism
catalysis
rate-determining step
rate law
2
Formulas:
Changes in heat content as a result of temperature change:
Amount of heat = Specific heat x mass x temperature change
Q = C x m x T = CmT
temperature change can be written as:
final temperature - initial temperature = Tf - Ti
Changes in heat content as a result of change of state:
For melting and fusion:
Amount of heat = heat of fusion x mass
Q = Hf x m
(for water Hf = 334 kJ/kg or 6.01 kJ/mol)
For condensation and vaporization:
Amount of heat = heat of vaporization x mass
Q = Hv x m
(for water Hv = 2260 kJ/kg or 40.7 kJ/mol)
3
CHEMISTRY 30
THERMOCHEMISTRY
Energy comes in many forms:
a)
kinetic energy is the energy of motion.
b)
potential energy is the energy stored in stressed objects, like a bent bow or compressed spring,
or water at the top of a waterfall.
c)
chemical energy is the energy stored in chemical bonds, which is released when a chemical
reaction, like burning, occurs.
d)
nuclear energy is the energy stored within the nucleus of an atom, which can be released under
certain conditions when the nucleus is made to break apart (fission), or when two nuclei are
forced together (fusion).
Heat is one form of energy. It is often the end product of other forms of energy. Heat is thermal
energy. Heat is the product of motion. Even if we cannot see the motion of an object, it is there in the
particles that make up the object: atoms and molecules. These are always in motion, speeding along, rotating,
or vibrating. Since heat is the product of motion, all matter has heat in it, since all matter is made up of
particles which move. The only we can say that something has no heat is if all motion has stopped inside it.
This point is called absolute zero.
Absolute zero is the lowest temperature that can be reached:
- it is equal to - 273 degrees Celsius
- it is also equal to 0 Kelvins (K)
The Kelvin scale is the same as the Celsius scale; that is, a difference of 10 K and a difference of 10C is the
same thing. The only difference is that 0C is the freezing point of water, while 0 K is the point at which all
motion stops.
Heat always flows from a warmer to a colder object. It continues to flow until both objects are at the
same temperature. On a cold day heat moves from your hands to the outside air and your hands become cold.
Heat moves through the walls of your house to the outside air; insulation is designed to slow down that flow.
Different masses or volumes of a substance require different amounts of heat to produce a similar rise
in temperature. The greater the mass or volume, the more heat that is required for a given temperature
change. If two different volumes of a substance are heated at the same rate, the smaller volume will
experience a greater temperature rise. IT DOES NOT ABSORB ENERGY ANY FASTER OR SLOWER. IT JUST
TRANSLATES THAT ENERGY INTO A GREATER TEMPERATURE RISE.
Different substances of the same mass require different amounts of heat to produce a similar rise in
temperature. Given equal inputs of heat energy, antifreeze will rise much more rapidly in temperature than
water will. Put another way, if water and antifreeze are heated, water requires more heat energy per unit mass
to produce a similar temperature rise. Put a third way, in equal masses of water and antifreeze at the same
temperature, the water contains more heat. This is called the SPECIFIC HEAT CAPACITY (also known as
specific heat). All substances have their own specific heat capacity, measured in kilo Joules per kilogram
degree Celsius (kJ/kgC). See the accompanying handout for some values.
Heat and temperature are related, but are not the same things. Temperature is a measure of the
average kinetic energy of a substance; it looks at how fast the particles of a substance are moving. The faster
the particles are moving, the higher is the temperature.
4
Heat, on the other hand, involves temperature, but it also involves mass since we know that the greater the
mass, the more energy that is required for a given temperature rise. It also involves the amount of heat a
given substance needs per unit mass for a given rise in temperature -- the specific heat capacity. To determine
the amount of heat a substance can hold, the following formula is used:
Q = C x M x (Tf - Ti)
where
Q = the amount of heat in kilo joules (kJ)
C=
M=
the heat capacity of the substance in kilo joules per kilogram degree
Celsius (kJ/kgC)
the mass of the body in kilograms (kg)
Tf =
the final temperature in degrees Celsius (C)
Ti =
the initial temperature in degrees Celsius (C)
Note that we are concerned with the temperature change, not the temperature at which the substance happens
to be.
This is an equation of four variables. If you have three of the variables you should be able to calculate
the fourth. Remember a few things, though:
1.
Make sure the units are correct. If heat or mass are given to you in units other than kilojoules
or kilograms, respectively, you must use a metric conversion.
2
Heat can go into or come out of a substance. If heat goes in, the temperature goes up and the
change is positive. If heat comes out, the temperature goes down and the change is
negative. So watch your signs.
3.
Use the specific heat table for the C values (see handout).
Examples:
1. Calculate the heat required to change the temperature of 1.5 kg water from 40.0°C to 80.0°C.
Since Q = C x M x (Tf – Ti) and the substance is water, therefore
C=
Q
4.200 kJ/kgC
M=
1.5 kg
Tf =
80.0 C
Ti =
40.0 C
= (4.200 kJ/kgC)(1.5 kg)(80.0 C - 40.0 C)
= 250 kJ (2 significant digits)
5
2. Calculate the heat required to change the temperature of 0.300 kg sucrose from 80.0°C to 19.0°C.
Since Q = C x M x (Tf – Ti) and the substance is sucrose, therefore
C=
Q
1.250 kJ/kgC
M=
0.399 kg
Tf =
19.0 C
Ti =
80.0 C
= (1.250 kJ/kgC)(0.300 kg)(19.0 C - 80.0 C)
= -22.9 kJ (a negative value, since the temperature dropped)
3. Calculate the heat required to change the temperature of 650 g copper by 12.6°C.
Since Q = C x M x ΔT and the substance is copper, therefore
Q
C=
0.390 kJ/kgC
M=
650 g x (1 kg / 1000 g) = 0.65 kg
ΔT =
12.6 C (only 1 temperature is given)
= (0.390 kJ/kgC)(0.65 kg)(12.6 C)
= 3.2 kJ (2 significant digits)
4. Calculate the temperature change that occurs when 2500 kJ of heat is added to 4.45 kg of asbestos.
Since Q = C x M x ΔT
ΔT =
Q
C x M
and the substance is asbestos, therefore
C =
0.820 kJ/kgC
M =
4.45 kg
Q = 2500 kJ
ΔT =
2500 kJ
(0.820 kJ/kgC)(4.45 kg)
= 690 C
6
5. Calculate the temperature change that occurs when -652 J of heat is added to 241 g of mercury.
Since Q = C x M x ΔT
ΔT =
Q
C x M
and the substance is mercury, therefore
C =
0.140 kJ/kgC
M =
241 g x (1kg / 1000 g) = 0.241 kg
Q = -652 J x (1 kJ / 1000 J) = -0.652 kJ
ΔT =
-0.652 kJ
(0.140 kJ/kgC)(0.241 kg)
= -19.3 C (the temperature dropped)
6. Calculate the mass of wood that would undergo a temperature change of 26.5°C when 1360 kJ of heat
is added.
Since Q = C x M x ΔT
M
=
Q
C x ΔT
and the substance is wood, therefore
C
= 1.760 kJ/kgC
ΔT = 26.5 C
Q
ΔT =
= 1360 kJ
1360 kJ
(1.760 kJ/kgC)(26.5 C)
= 29.2 kg
7
Phase Changes
During a phase change the temperature does not change, even though heat continues to be applied. Where
does the heat go? It goes into the breaking of physical bonds between the molecules of the substance, as well
as putting distance between the molecules. When a substance melts it goes from a rigid, hard solid to a liquid.
Bonds must be broken so that the molecules have the freedom of movement that a liquid has. When a liquid
becomes a gas the molecules get much further apart.
In either case during a phase change there is no change in temperature, so how can we calculate the heat
involved? The formula we have involves temperature change, so it is of no value. We need something new. It
involves two related concepts:
1. Heat of Fusion (Hf) - This is the heat required to melt a quantity of a substance. It is usually
expressed in kilojoules per kilogram or kilojoules per mole. If a substance is freezing the Hf value is
expressed as a negative.
2. Heat of Vapourization (Hv) - This is the heat required to vapourize or boil a quantity of a substance.
It is expressed in the same units as heat of fusion. If a substance is condensing the H v value is
expressed as a negative.
The formula of each is very simple:
Heat Required to Melt or Freeze = Heat of Fusion x Mass of Substance
a Substance
Q = HfM
Heat Required to Boil or = Heat of Vapourization x Mass of Substance
Condense a Substance
Q = HvM
Examples:
Where the Hf for water is 334 kJ/kg and the Hv for water is 2260 kJ/kg, perform the following calculations.
1. Calculate the heat required to melt 1.5 kg ice.
Since the phase change is melting the formula is
Q = Hf M
where Hf = 336 kJ/kg and
Thus
M = 1.5 kg
Q = (336 kJ/kg)(1.5 kg)
=
5.0 x 102 kJ
8
2. Calculate the heat required to freeze 4.50 x 10 2 kg water.
Since the phase change is freezing the formula is
Q = Hf M
where Hf = - 336 kJ/kg (heat is taken out) and
M = 4.50 x 10 2 kg
Thus
=
Q = (- 336 kJ/kg)(4.50 x 102 kg)
- 1.50 x 105 kJ
(the answer is negative)
3. Calculate the heat required to boil 245 g water.
Since the phase change is boiling the formula is
Q = Hv M
where Hv = 2260 kJ/kg and
Thus
M = 245 g x (1 kg/1000 g) = 0.245 kg
Q = (2260 kJ/kg)(0.245 kg)
=
554 kJ
4. Calculate the heat required to condense 1.2 t steam.
Since the phase change is condensing the formula is
Q = Hv M
where Hv = - 2260 kJ/kg and
Thus
M = 1.2 t x (1000 kg/ 1 t) = 1200 kg
Q = (- 2260 kJ/kg)(1200 kg)
= 2.7 x 106 kJ
(the symbol t stands for tonne, or 1000 kg)
5. Calculate the heat required to melt 162 mol ice.
Since the phase change is melting the formula is
Q = Hf M
where Hf = 336 kJ/kg and
M = is not given, but is 162 mol
The molar mass of water is 18.02 g/mol (from H2O) and
Mass = Molar mass x Number of moles
= (18.02 g/mol)(162 mol) = 2920 g x (1 kg / 1000 g) = 2.92 kg
Thus
Q = (336 kJ/kg)(2.92 kg)
=
975 kJ
9
Combined Heat Problems:
It's now time to combine what you learned on the last few pages. What if I ask how much heat is required
to take 2.00 kg of ice at -20.0°C and turn it into steam at +110.0°C ?
This problem is surprisingly simple. You just have to remember a few things:




Ice only exists at temperatures below 0°C. At zero the ice begins to melt and you need to use the
equation that involves the heat of fusion.
Water only exists between 0°C and 100°C. Below this temperature it freezes and above this
temperature it boils.
Steam only exists above 100°C. At 100°C the steam begins to condense and you use the equation
that involves the heat of vapourization.
Ice, water and steam all have unique specific heat values. For temperature changes involving
these substances you need to use separate equations.
Thus this problem divides into five parts:
1. Heat the ice from -20.0°C to 0°C. Use the formula Q = C x M x (Tf - Ti). Use the specific heat of ice.
2. Melt the ice to form water. Use the formula Q = Hf x M.
3. Heat the water from 0°C to 100°C. Use the formula Q = C x M x (Tf - Ti). Use the specific heat of
water
4. Boil the water to form steam. Use the formula Q = Hv x M.
5. Heat the steam from 100°C to 110.0°C. Use the formula Q = C x M x (Tf - Ti). Use the specific
heat of steam.
Try it. The correct answer, to the correct number of significant digits is 6150 kJ.
Complete the assignment “Measuring Quantities of Heat. Show all work.
10
Heat Changes in Chemical Reactions
The difference between physical change and chemical change is that chemical change involves bond
breakage, while physical change does not. As mentioned before, chemical bonds involve energy, and energy is
the main reason why chemical reactions occur.
This energy relationship in chemical reactions is summarized by the term enthalpy. Enthalpy is the
energy difference between reactants and products in a chemical reaction; that is, the difference in bond
energies between reactants and products. Nature's aim is to achieve the lowest possible energy state, or the
lowest possible enthalpy. In chemical reactions this means that compounds with the lowest possible energy in
the bonds are favoured.
In a chemical reaction if the products have a lower energy in their bonds (enthalpy) than the reactants,
energy is released to the environment as the reaction progresses. This reaction is exothermic. If the
products have a higher enthalpy than the reactants, energy from the environment flows into the system as the
reaction progresses. The reaction is endothermic.
Enthalpy is represented by the symbol H. A change in enthalpy is represented by the symbol ΔH. In
an exothermic reaction energy is released into the environment; energy is lost from the system, Therefore, ΔH
is negative. In an endothermic reaction energy is absorbed from the environment; energy is gained by the
system. In this case, ΔH is positive.
It is impossible to calculate the absolute enthalpy (H) of any molecule. It is possible only to determine
enthalpy change (ΔH). This can be done in two ways, by calorimetry or through the use of tables. Calorimetry
involves actually mixing the chemicals and measuring the energy released or absorbed (this will be the method
used in the lab for this unit). When tables are used we are taking advantage of the fact that other people have
made the measurements for us already:
11
Heats of Formation of Several Chemical Reactions (kJ/mol)
Elements
Product Name
Heat of reaction
(kJ/mol of product)
H2(g) + 1/2 O2(g)  H2O(g)
water vapor
- 242
H2(g) + 1/2 O2(g)  H2O(l)
water
- 286
sulfur dioxide
- 297
H2(g) + S(s) + 2 O2(g)  H2SO4(l)
sulfuric acid
- 812
S(s) + 3/2 O2(g)  SO3(g)
sulfur trioxide
- 395
1/2 N2(g) + 1/2 O2(g)  NO(g)
nitric oxide
+ 90.4
1/2 N2(g) + O2(g)  NO2(g)
nitrogen dioxide
+ 33.1
1/2 N2(g) + 3/2 H2(g)  NH3(g)
ammonia
- 46.0
C(s) + 1/2 O2(g)  CO(g)
carbon monoxide
- 110
carbon dioxide
- 394
methane
- 74.5
ethane
- 84
3 C(s) + 4 H2(g)  C3H8(g)
propane
- 104
1/2 H2(g) + 1/2 I2(g)  HI(g)
hydrogen iodide
+ 25.9
4 C(s) + 4 H2(g) + O2(g)  C3H7COOH(l)
propionic acid
- 522
2 C(s) + 2 H2(g) + O2(g)  CH3COOH(l)
acetic acid
- 487
2Al(s) + 3/2 O2(g)  Al2O3(s)
aluminum oxide
-1676
Cu(s) + 1/2 O2(g)  CuO(s)
copper (II) oxide
- 155
Mg(s) + 1/2 O2(g)  MgO(s)
magnesium oxide
- 602
Na(s) + 1/2 Cl2(g)  NaCl(s)
sodium chloride
- 411
S(s) + O2(g)
C(s) + O2(g)


C(s) + 2 H2(g)
2 C(s) + 3 H2(g)
SO2(g)
CO2(g)
 CH4(g)
 C2H6(g)
12
This table lists the heats of formation of several compounds from their elements. Some reactions are
exothermic, others are endothermic. This table has uses, however, far beyond simply looking at the ΔH of the
listed reactions. To deal with this we must first discuss Hess's law of constant heat summation, or the law
of additivity of heats of reaction.
For instance, carbon and oxygen can react to form carbon monoxide:
C(s) + 1/2 O2(g)  CO(g)
(1)
Δ H = - 110 kJ
Carbon monoxide can react further with oxygen to form carbon dioxide:
CO(g) + 1/2 O2(g) 
(2)
CO2(g)
ΔH = - 294 kJ
A third possible reaction involves carbon and oxygen reacting to form carbon dioxide directly:
(3)
C(s) + O2(g)

CO2(g)
Δ H = - 394 kJ
You will note that the heat of reaction from (1) plus the heat of reaction from (2) exactly equals the heat of
reaction from (3). This makes sense when you consider what is going on; chemically, reactions (1) and (2)
perform in two steps what reaction (3) does in one step. Note what happens when reactions (1) and (2) are
added up:
C(s) + 1/2 O2(g)  CO(g)
ΔH = - 110 kJ
CO(g) + 1/2 O2(g)  CO2(g)
ΔH = - 294 kJ

ΔH = - 394 kJ
C(s) + O2(g)
CO2(g)
This result is stated by Hess's Law:
The enthalpy change for any reaction depends only on the products and reactants and is
independent of the pathway or the number of steps between the reactant and product.
This law allows us to take several reactions from the heats of formation table, add them up and get ΔH values
for reactions that do not appear directly on the table.
Example 1:
2 Al(s) + 3 CuO(s)
3 Cu(s) + Al2O3(s)
1. Find an equation on the Heat of Formation Table to match each of the compounds in the given
equation.
2 Al(s) + 1½ O2(g)
Cu(s) + ½ O2(g)
Al2O3(s)
CuO(s)
ΔH = - 1676.0 kJ/mol
ΔH = - 155.0 kJ/mol
2. Flip the CuO equation to make the compound a reactant as well:
Cu(s) + ½ O2(g)
CuO(s)
ΔH = - 155.0 kJ/mol
Becomes
CuO(s)
Cu(s) + ½ O2(g)
ΔH = + 155.0 kJ/mol
13
3. Multiply the CuO equation by 3 to have the same number of molecules as in the original equation.
2 Al(s) + 1½ O2(g)
3x
CuO(s)
Al2O3(s)
ΔH = - 1676.0 kJ/mol
Cu(s) + ½ O2(g)
ΔH = 3 x (+ 155.0 kJ/mol)
Gives
2 Al(s) + 1½ O2(g)
3 CuO(s)
Al2O3(s)
ΔH = - 1676.0 kJ/mol
3 Cu(s) + 1½ O2(g)
ΔH = + 465.0 kJ/mol
4. Add the two equations together. You treat the equations just like you would in math; add the material
to the left of the arrow together and the material on the right of the arrow together. The enthalpy
changes are also added:
2 Al(s) + 1½ O2(g)
3 CuO(s)
Al2O3(s)
ΔH = - 1676.0 kJ/mol
3 Cu(s) + 1½ O2(g)
2 Al(s) + 1½ O2(g) + 3 CuO(s)
ΔH = + 465.0 kJ/mol
Al2O3(s) + 3 Cu(s) + 1½ O2(g)
ΔH = - 1211 kJ
You will notice that there are similar terms on each side of the reaction equation. These can be treated
again just like mathematical terms and cancelled out where you can:
2 Al(s) + 3 CuO(s)
Al2O3(s) + 3 Cu(s)
ΔH = - 1211 kJ
You know you are right if the net equation you end up with is the same as the original equation you started
with.
Example 2:
2 C2H6(g) + 7 O2(g)
4 CO2(g) + 6 H2O(l)
1. Find an equation on the Heat of Formation Table to match each of the compounds in the given
equation.
2 C(s) + 3 H2(g)
C(s) + O2(g)
H2(g) + ½ O2(g)
C2H6(g)
CO2(g)
H2O(l)
ΔH = - 84.0 kJ/mol
ΔH = - 394.0 kJ/mol
ΔH = - 286.0 kJ/mol
2. Flip the C2H6 equation to make the compound a reactant:
2 C(s) + 3 H2(g)
C2H6(g)
ΔH = - 84.0 kJ/mol
Becomes
C2H6(g)
2 C(s) + 3 H2(g)
ΔH = + 84.0 kJ/mol
14
3. Multiply each equation to have the same number of molecules as in the original equation.
2x
C2H6(g)
2 C(s) + 3 H2(g)
4x
C(s) + O2(g)
6x
H2(g) + ½ O2(g)
CO2(g)
ΔH = 2 x(+ 84.0 kJ/mol)
ΔH = 4 x (- 394.0 kJ/mol)
H2O(l)
ΔH = 6 x (- 286.0 kJ/mol)
Gives
2 C2H6(g)
4 C(s) + 6 H2(g)
ΔH = + 168.0 kJ/mol
4 CO2(g)
ΔH = - 1576.0 kJ/mol
4 C(s) + 4 O2(g)
6 H2(g) + 3 O2(g)
6 H2O(l)
ΔH = - 1716.0 kJ/mol
4 C(s) + 6 H2(g)
ΔH = + 168.0 kJ/mol
4 CO2(g)
ΔH = - 1576.0 kJ/mol
4. Add the equations together.
2 C2H6(g)
4 C(s) + 4 O2(g)
6 H2(g) + 3 O2(g)
6 H2O(l)
ΔH = - 1716.0 kJ/mol
2 C2H6(g) + 4 C(s) + 4 O2(g) + 6 H2(g) + 3 O2(g)
4 C(s) + 6 H2(g) + 4 CO2(g) + 6 H2O(l)
ΔH = - 3124 kJ
Cancel like terms:
2 C2H6(g) + 7 O2(g)
4 CO2(g) + 6 H2O(l)
ΔH = - 3124 kJ
15
Example 3:
SiO2(s) + C(s)
CO2(g) + Si(s)
Where the heat of formation of SiO2(s) = - 861 kJ/mol
1. Find an equation on the Heat of Formation Table to match each of the compounds in the given
equation.
C(s) + O2(g)
CO2(g)
ΔH = - 394.0 kJ/mol
The equation for SiO2 is not on the table, but you are given the heat of formation of the substance from
its elements:
Si(s) + O2(g)
SiO2(s)
ΔH = - 861.0 kJ/mol
2. Flip the SiO2 equation to make the compound a reactant as well:
Si(s) + O2(g)
SiO2(s)
ΔH = - 861.0 kJ/mol
Becomes
SiO2(s)
Si(s) + O2(g)
ΔH = + 861.0 kJ/mol
3. Since there is only one of each molecule in the original equation, no multiplication is necessary.
4. Add the two equations together. You treat the equations just like you would in math; add the material
to the left of the arrow together and the material on the right of the arrow together. The enthalpy
changes are also added:
C(s) + O2(g)
SiO2(s)
CO2(g)
ΔH = - 394.0 kJ/mol
Si(s) + O2(g)
C(s) + O2(g) + SiO2(s)
ΔH = + 861.0 kJ/mol
CO2(g) + Si(s) + O2(g)
ΔH = + 467 kJ
Cancel like terms:
C(s) + SiO2(s)
CO2(g) + Si(s)
ΔH = + 467 kJ
Complete assignment titled “Enthalpy Change Calculations”
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Enthalpy and Entropy
For any chemical reaction we can ask a basic question; will it go forward or not? Just because we can
write a chemical equation does not mean it is a reaction that actually happens spontaneously in nature. To
answer this question we must discuss enthalpy and entropy.
Enthalpy has been discussed. It is the heat content of a compound, found in the bonds. When a
chemical (or physical) reaction occurs a heat change (ΔH) is involved. Nature tends towards a minimum
energy, so a reaction with a negative ΔH will be more likely to go forward spontaneously from reactant to
product.
Entropy (S) is the amount of disorder in a system. Nature does not like to be organized (much like
bedrooms and teacher's countertops). Thus a reaction which involves substances going from a less disordered
state to one which is more disordered will be favoured. Note we are discussing disorder; an increase in
entropy involves an increase in the disorder of a system. A reaction with a positive entropy change (ΔS) will be
more likely to go forward spontaneously from reactant to product. To know how to recognize a change in
entropy (ΔS), the following rules apply:
a)
b)
c)
a gas has more disorder than a liquid which has more disorder than a solid.
the side with more particles is more disordered.
a dissolved substance is more disordered than a solid.
Thus for any reaction it is possible to determine whether the enthalpy and entropy are positive or
negative, but how does this help to determine whether a reaction will be spontaneous or not? To answer this
we need a third function, called Gibb's Free Energy
(ΔG). This is a measure of the maximum possible work that can be obtained by a reaction. It is related to
entropy and enthalpy change by the function:
ΔG = ΔH - T ΔS
T stands for temperature in Kelvins.
If the free energy as calculated using this formula is negative, the reaction will proceed spontaneously.
If the ΔG is positive, the reaction cannot proceed on its own.
energy:
There are four combinations of enthalpy and entropy possible, each with a distinct effect on Gibb's free
ΔH
ΔS
Effect on the Reaction
positive
positive
ΔG will be negative only if TΔS is very large; this reaction will be
spontaneous only at high temperature.
positive
negative
ΔG cannot be negative under any conditions. This reaction will not
proceed spontaneously.
negative
positive
ΔG will be negative under all conditions. This reaction will proceed
spontaneously.
negative
negative
ΔG will be negative only if TΔS is very small; this reaction will be
spontaneous only at low temperature.
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Examples
For each of the following equations you need to deterimine the following:
a. Whether the enthalpy is positive or negative.
b. Whether the entropy is positive or negative.
c. Using the Gibbs Free Energy equation, under what conditions of temperature the reaction will likely be
spontaneous.
1. H20(l)
H20(s)
ΔH = - 6 kJ
ΔH is negative. The reaction is exothermic. This is favourable for spontaneity.
ΔS is negative. This is a phase change and the solid is more ordered than the liquid, so
entropy is decreasing. This is not favourable for spontaneity.
Since
ΔG = ΔH - TΔS
this reaction will be spontaneous only if TΔS is as small as possible, so as to be smaller
than the enthalpy change. This reaction will only proceed at low temperatures.
2. CaCO3(s) + 181 kJ
CaO(s) + CO2(g)
ΔH is positive. The energy term is on the reactant side, which means it is going into the
reaction, so it is endothermic. This is not favourable for spontaneity.
ΔS is positive. This is a decomposition reaction producing more molecules and one
product is a gas. This is favourable for spontaneity.
Since
ΔG = ΔH - TΔS
this reaction will be spontaneous only if TΔS is as large as possible, so as to be bigger
than the enthalpy change. This reaction will only proceed at high temperatures.
3. CS2(g) + 3 O2(g)
CO2(g) + 2 SO2(g)
ΔH = + 1110 kJ
ΔH is positive. The reaction is endothermic. This is not favourable for spontaneity.
ΔS is negative. All the substances are gases, but the reactants have more particles than
the products. This is not favourable for spontaneity.
Since
ΔG = ΔH - TΔS
this reaction will be not be spontaneous at any temperature because Gibbs Free Energy
cannot be made to be positive.
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Na1+(aq) + Cl1-(aq) + energy
4. NaCl(s)
ΔH is negative. The reaction is exothermic, since it appears on the product side of the
equation. This is favourable for spontaneity.
ΔS is positive. Since the salt is dissolving the entropy is increasing. This is favourable
for spontaneity.
Since
ΔG = ΔH - TΔS
since both factors are favourable for a spontaneous reaction this reaction will be
spontaneous at any temperature
Complete the assignment titled “Enthalpy and Entropy”
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The Reaction Process
- is based on collision theory; atoms and molecules can react only if they collide with one another.
- not all collisions result in reaction;
i)
the collisions must have activation energy. Activation energy is defined as the energy
difference between the reactants and the energy needed to the reaction to occur (threshold
energy). Generally, the higher the activation energy, the slower the chemical reaction.
ii)
the collisions must have the correct orientation. The molecules must contact each other just
right. (see text, p. 533)
-
chemical kinetics is the area of chemistry that is concerned with reaction rates and reaction mechanisms.
-
reaction rate is the change in concentration of reactants per unit time as the reaction proceeds; the larger
the change, the higher the rate.
Any chemical reaction consists of reactants becoming products. The rate of the chemical reaction can be
measured in terms of how fast the reactants are consumed or how fast the products are produced:
reaction rate = mass of reactant lost
time
reaction rate = mass of product produced
time
Reaction rate can be determined not only with mass; if a product is a gas, the volume of gas production can
be measured; if a reactant is a liquid, the loss of liquid can be monitored; if the reaction occurs in solution
the change in concentration of reactant or product can be measured.
-
any factor that increases the number of collisions, the energy of collisions or improves the orientation of
collisions will increase the rate.
-
several factors influence the rate of chemical reactions:
i)
nature of the reactants - the more complex the reaction, the slower the rate. Reactions that only
involve a change in charge will proceed more rapidly than a reaction that involves breaking and reforming
chemical bonds (they require more energy or the orientation becomes more difficult).
ii) surface area - the more particles exposed to reaction, the higher the rate (more collisions). Gases react
more quickly than solids; powders react more quickly than larger lumps.
iii) temperature - the higher the temperature, the greater the kinetic energy. This increases both the
number and energy of collisions, which increases the rate.
iv) concentration - the more particles present, the more collisions and the higher the rate. This includes
increasing the molar concentration of solutions, as well as the pressure of gases.
v) catalyst - a substance that increases the rate of a reaction without itself being permanently consumed or
changed. It works by lowering the activation energy of the reaction, either by becoming part of the
activated complex or providing a surface where the reaction can happen.
vi) inhibitor - a substance that reduces the rate of a reaction by increasing the activation energy. They
often work by binding to one of the reactants.
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Energy Diagrams
-
are a way of visualizing a chemical reaction:
Every substance has an average kinetic energy. We call it temperature. The individual molecules have
a variety of energies, some higher than average, some lower. The graph above illustrates this energy
distribution. The mean energy is the temperature of the substance. The threshold energy is the energy needed
for a chemical reaction to happen. If this diagram was for the energy of oxygen molecules in the air, a very
small number have the energy required to react with paper, causing it to burn. As a result, paper degrades
very slowly at room temperature.
If we were to increase the temperature, however:
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Increasing the temperature moves the whole diagram and all the molecules higher. The threshold energy does
not change, so a greater proportion of atoms now have the energy needed to react; the rate of reaction
increases.
Another way to increase the rate of a chemical reaction is to add a catalyst:
The threshold energy is decreased by the catalyst. The overall effect is the same as increasing temperature;
more molecules now have the energy they need to react, and the reaction rate increases.
When two or molecules collide and participate in a chemical reaction, we often discuss reactants and
products. What actually occurs is slightly more complicated; the reactants do not magically turn into products.
There is in fact a series of simple steps which occur between reactants and products called the reaction
mechanism. These steps allow a complex reaction to occur quickly by taking it in little bites, rather than all at
once. Intermediate molecules are produced between reactants and products called the activated complex.
For example, consider the reaction of hydrogen and oxygen to form water:
2 H2 + 1 O2  2 H2O
This reaction can also be illustrated by way of an energy diagram which plots potential energy on the y axis
and progress of the reaction on the x axis:
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The diagram has 3 parts. The first part is the energy of the reactants. The last part is the energy of the
products. In the middle is the reaction itself, which involves the acquisition of activation energy, and the
formation of the activated complex. Note the scale on the left allows determination of the energy of
reactants, products, the threshold energy, the enthalpy change and the activation energy.
The next graph is similar but illustrates an endothermic reaction; energy is absorbed from the environment by
the system.
Potential Energy Diagram of an Endothermic Reaction
Threshold Energy
400
300
H
200
Activation
Energy
100
Reactants
Activated
Complex
Products
Progress of the Reaction
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These diagrams are also useful to indicate the effect of a catalyst: or inhibitor (see in class).
Reaction Mechanism
-
is the series of steps required for a reaction to go from reactants to products.
-
often there is more than one activated complex between the initial reactant and the final product. The
substances between the initial reactants and the final products are called intermediates.
-
each step requires its own activation energy.
-
the step which requires the largest activation energy is the slowest in the reaction mechanism; it
is called the rate-determining step.
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