G
Some chemical reactions begin as soon as the reactants are brought together and continue until at least one reactant (the limiting reactant) has been totally consumed. A reaction of this type would be the reaction of sodium metal and chlorine gas to form sodium chloride:
2 Na(s) + Cl
2
(g) è 2 NaCl(s) + heat
In the reaction above energy in the form of heat is released. Reactions of this type are described as being PRODUCT FAVORED.
Other reactions will not occur unless some form of outside intervention is added. An example of this type of process is the decomposition of dintirogen tetroxide by adding heat energy. Reactions of this type are described as REACTANT FAVORED.
Heat + N
2
O
4
(g) è N
2
(g) + 2 O
2
(g)
What do we mean by outside intervention? Usually it is some flow of energy. Energy is a central idea when describing a reaction as product or reactant favored For this reason it is important to know something about energy and its interactions with matter. The most common interaction is the transfer of energy, heat energy to be specific, when a chemical process occurs. Transfer of heat energy is the major theme of THERMOCHEMISTRY, the chemistry of heat and work.
Energy: Its Forms and Units.
ENERGY is defined as the capacity to do work. For example if you climb a mountain, you do some work against the force of gravity as you carry yourself and your equipment up the mountain. You can do this because you have the energy, or capacity to do so, the energy being supplied by the food that you have eaten. Food energy is chemical energy – energy stored in chemical compounds and released when the compounds undergo the chemical process of metabolism.
Energy can be assigned to one of two classes or types, and energy in each class can take on several different forms. An object has KINETIC ENERGY because it is moving.
Examples are:
•
Thermal Energy of atoms, ions or molecules in motion at the sub-microscopic level. All matter has thermal energy because, according to the kinetic-molecular theory, all particles of matter in in constant motion.

•
Mechanical Energy of a macroscopic object like a thrown baseball or car in motion.
•
Electrical Energy of electrons in motion through a copper wire.
•
Sound which is the compression or expansion of the spaces between molecules.
Potential Energy results form an objects position. Examples are:
•
Chemical potential energy results form the attractions between electrons and atomic nuclei in molecules.
•
Gravitational Potential Energy such as water at the top of a waterfall.
Potential energy is stored energy- it can be converted to kinetic energy. For example as water droplets fall over a waterfall, the potential energy of the water is converted to kinetic energy as the droplets are accelerated to high and higher velocities by the force of gravity.
Similarly this kinetic energy falling water can be converted to electrical energy by a turbine.
The LAW OF CONSERVATION OF ENERGY tells us that energy cannot be created or destroyed only transformed from one class or form and transferred from one place to another. For example, when natural gas is combusted in the presence of oxygen gas, the chemical potential energy that was stored within the bonds of methane and oxygen is released as heat energy, the random kinetic motion of particles on the submicroscopic scale. We consider the molecules of natural gas and oxygen and the products of the reaction to comprise a system of interest or simply a system. Energy in the form of thermal motion or heat is transferred from the system to the surroundings (the rest of the universe). It is important to remember that the energy released in combustion is not destroyed, but simply transformed to a less accessible form.
Energy Units
One of the earliest units of energy to be devised was the unit of calorie, or cal. The calorie is defined as the amount of energy required to raise the temperature of 1.0 gram of water by 1.0 ˚C. Since the calorie represents a small amount of heat energy and we usually work with larger quantities of matter the unit of kilocalorie or kcal is used instead.
Most of us have heard of the term calorie before in the context of food and diet. This
Calorie is different from the unit describe above and is spelled with a capital C. One dietary Calorie is equal to 1 kcal.
Currently most chemists use the Joule, J, the SI unit of energy. One calorie is defined as being equal to 4.184 Joules.
• cal = 4.184 J

Heat Transfer
We say that energy has left the system as heat if the energy flows out as a result of a temperature difference between the system and the surroundings. Heat energy only flows in one direction, from the hotter to the cooler object. We will symbolize heat energy transferred by the letter Q.
Q
HOT OBJECT COLD OBJECT
Heat transfer occurs by way of collisions between randomly moving particles of matter.
The particles with higher thermal energy are moving more quickly, when they collide with slower moving particles, some of their energy is transferred to the slower particle as heat energy increasing its speed
Let us consider a simple experiment. We have two identical insulated containers, one container holds100.0 g of liquid water, the other contains 100.0 g of ethylene glycol (a clear colorless liquid used as antifreeze). The liquids in both containers are at the same initial temperature of 22.2 ˚C. We use a simple electrical wire heating apparatus to add the same amount of heat energy to both containers, 4,184 Joules. After the heat energy has been added the final temperature of the liquid water is 32.2˚C whereas the final temperature of the ethylene glycol is 42.2˚C. Why are the final temperatures different if the same amount of energy was added to each?
The temperatures were different in the example above because liquid water and ethylene glycol have different HEAT CAPACITIES. The heat capacity of an object is its ability to capture heat energy without a subsequent rise in temperature. Heat capacity is the ratio of energy transferred per temperature change. Thus in the example above:
Temperature change =
∆ t = t
(final)
– t
(initial)
Heat Capacity
(water)
= 4,184 J/10.0˚C = 418 J/˚C
Heat Capacity
(ethylene glycol)
= 4,184 J/20.0˚C = 209 J/˚C
Since heat capacity depends on the mass of an object(an extrinsic property) an intrinsic property has been derived called the Specific Heat, which is symbolized by the letter C:
Specific Heat = heat transferred/ mass in grams x temperature change ˚C or
C = Q / m
• ∆Τ
Using the equation above:
Specific Heat
(liquid water)
= C
(liquid water)
= 4.18 J/g ˚C
Specific Heat
(ethylene glycol)
= C
(ethylene glycol)
= 2.09J/g ˚C
Other specific heat values are listed in the table on page 186 of your text.

Liquid water has one of the highest know values of specific heat. The high specific heat of water tells us that a considerable amount of heat energy must be transferred into or out of a body of water before a temperature change takes place.
The specific heat of a substance is determined experimentally:
Example 1:
A 15.5 g piece of chromium metal, heated to 100.0˚C, is dropped into 55.5 g of water at 16.5 ˚C. The final temperature of the metal and water is 18.9˚C. What is the specific heat of chromium?
Solution:
Keep in mind that the heat energy transferred from the hot chromium metal to the cooler water is equal to the heat energy absorbed by the water in undergoing a temperature change from 16.5˚C to 18.9˚C.
Heat lost from chromium = heat absorbed by water
Heat lost from chromium = C
(chromium)
Heat absorbed by water = C
(water)
x mass
x mass
(water)
(chromium)
g x
∆ t
(chromium)
g x
∆ t
(water)
C
(chromium)
x mass
(chromium)
g x
∆ t
(chromium)
= C
(water)
x mass
(water)
g x
∆ t
(water)
C
(chromium)
= C
(water)
x mass
(water)
g x
∆ t
(water)
/ mass
(chromium)
g x
∆ t
(chromium)
C
(chromium)
= 4.18 J/g˚C x 55.5g x 2.4˚C/ 15.5g x 83.5˚C
C
(chromium)
= 0.43 J/g˚C
Example 2:
A 400. 0 g piece of iron (C
(iron)
= 0.38 J/g˚C) is heated in a flame and dropped into a beaker containing 1000. 0 g of water at 20.0˚C. The final temperature is 32.8˚C.
What was the initial temperature of the iron bar?
Solution:
Keep in mind that the energy transferred from the iron bar is equal to the energy absorbed by the water.
C
(iron)
x mass
(iron)
g x
∆ t
(iron)
= C
(water)
x mass
(water)
g x
∆ t
(water)
Solve for
∆ t
(iron)
:
∆ t
(iron)
= C
(water)
x mass
(water)
g x
∆ t
(water)
/ C
(iron)
x mass
(iron)
g

∆ t
(iron)
= 4.18 J/g˚C x 1000.0g x
12.8
/0.38 J/g˚C x 400.0g
∆ t
(iron)
= 352˚C
∆ t = t
(final)
– t
(initial)
t
(initial)
= t
(final
+
∆ t = 385˚C
Energy and Changes of State
Energy transfers occur when matter changes from one state to another in a physical change. For example, when a solid melts, its atoms, molecules or ions move about vigorously enough to break free of the forces which constrains its position in the solid.
When a liquid boils, thermal energy overcomes the relatively strong forces holding the atoms, molecules or ions in the liquid phase, and the particles move much farther apart from one another. In both cases, attractive forces between the particles must be overcome, which requires the input of energy. The changes of state described as melting or boiling always take place at a constant temperature. For example, if we were to heat water, initially at 25˚C until it begins to boil at 100˚C, the temperature will remain constant at
100˚C until all of the liquid water has been convert to a gas. An ice cube initially at 0˚C, the melting point of ice, will remain at this temperature until it has completely melted.
Any liquid water in contact with the ice will also be at 0˚C.
The quantity of heat required to melt a solid at its melting point is called the HEAT OF
FUSION and is represented by the symbol
∆
H fusion
. For example. if we take an 18.0g ice cube initially at 0˚C and add heat energy, the temperature will remain constant at 0˚C.
However the ice will undergo a state change forming liquid water as the added energy overcomes the strong forces holding the water molecules together in the solid phase.
When 6.01 kJ of energy has been added, the ice will have been completely melted into liquid water at 0˚C. Now any added heat energy will begin to raise the temperature of the water. If we recognize that the molar mass of H
2
O is 18.0 gram per mole, then we can specify the heat of fusion of water as:
∆
H fusion
= 6.01 kJ/mole
You should recognize that when a solid melts to a liquid the heat of fusion is
ENDOTHERMIC, that is heat energy must be added to the system. When a liquid freezes to a solid, the heat of fusion is EXOTHERMIC, or energy must be lost from the system.

If we continue to heat the 18.0 g of liquid water obtained from the ice cube above, the temperature will begin to rise. At 100˚C, the water will begin to boil, that is, it will begin to form water vapor or steam. The temperature will remain constant at 100˚C until all of the liquid has been converted to a gas. During this time any added heat energy is used to overcome the forces holding the water moleculesin the liquid phase. When 40.7 kJ of energy has been added, the water will have been completely converted into steam at
100˚C. The quantity of energy required to completely convert a liquid at its boiling point into a gas at its boiling point is called the HEAT OF VAPORIZATION and is represented by the symbol
∆
H vapor
. The heat of vaporization for liquid water is:
∆
H = 40.7 kJ/mole
EXAMPLE 3
How much heat energy is required to convert 500.0 g of ice, initially at –12.0˚C to liquid water at 25.0˚C?
Solution:
We first need to determine the amount of energy needed to raise the temperature of 500.0g of ice, initially at –12.0˚C to 0˚C. To do this we need to know the heat capacity of ice C ice
= 2.01 J/g˚C
Heat required = C ice
x mass ice
x
∆ t ice
= 2.01 J/g˚C x 500.0 g x 12˚C
= 12.1 kJ
Now determine how much heat is required to melt 500.0g of ice at 0˚C to
500g of liquid water at 0˚C…
∆
H fusion
= 6.01 kJ/mole
Heat required = 500.0 g ice x 1.0 mole ice/18.0 g ice x 6.01 kJ/mole ice
= 167 kJ
Now determine how much heat is required to raise the temperature of 500g of water by 25.0˚C. C liquid water
= 4.18 J/g˚C
Heat required = C iwater
x mass water
x
∆ t iwater
= 4.18 J/g˚C x 500.0 g x 25.0˚C
= 52.3 kJ
Total heat energy required = 12.1 kJ + 167 kJ + 52.3 kJ = 231 kJ
The energy/phase change process describe in example 3 can be illustrated by a temperature vs energy plot such as that shown below.

25
20
E
R
A
T
U
R
E
T
E
M
P
˚C
15
10
5
0
-5
-10 melting freezing
0
Temp-Energy Plot
25 50 75 100 125
ENERGY kJ
150 175
200 225 230
Example 4
How much energy is required to convert 144 g of liquid benzene at 25.0 ˚C (C6H6,) into benzene vapor at its boiling point?
Solution:
Data we need to know: C benzene
= 1.05 J/g˚C,
∆
H vapor
benzene = 30.8 kJ/mole
Molar Mass benzene
= 72.0 g/mole
Boiling Point benzene
= 80.0˚C
Heat required to raise temp of benzene at 25.0˚C to its boiling point:
Heat = C benzene
x mass benzene
x
∆ t benzene
= 1.05 J/g˚C x 144 g x 55.0˚C = 8.32 kJ
Heat required to vaporize 144 g of liquid benzene:
Heat =
∆
H vapor
benzene x mass benzene
x 1/ molar mass benzene
= 30.8 kJ/mole x 144 g x 1 mole/ 72 g
= 61.6 kJ
Total energy = 69.9 kJ/mole

Calorimetry
The heat evolved in a chemical reaction can be determined by a process called calorimetry.
Reactions are carried out in a calorimeter, an apparatus designed to isolate a system of interest from its surroundings. A very simple but effective calorimeter is the coffee cup calorimeter shown below.
Two styrofoam cups are placed one inside the other and covered with a lid. The exothermic reaction occurs in the aqueous solution within and the cups provide enough insulation so that the heat evolved is confined to the aqueous solution, raising the temperature of the water. Knowing the mass of water and calorimeter, its specific heat
(when a reaction is carried out in a dilute aqueous solution, the specific heat is very close to that of water itself) and any temperature change allows the heat energy released to be easily calculated.
Example 5:
When a student mixes 50.0 mL of a 0.0100 M HCl solution and50.0 mL of a 0.0100
M NaOH solution in a coffee-cup calorimeter the temperature of the resultant solution rises from 21.0˚C to 21.4˚C. Calculate the enthalpy change for the
reaction.
Solution:
The final volume of solution is 100. mL. Since the solution is dilute the density of solution will be very close to that of water or 1.00 g/mL . The specific heat will also be very close to that of water (4.18 J/g˚C). Because the temperature increases by
0.4˚C, the reaction must be exothermic
Enthalpy change of reaction = heat absorbed by aqueous solution
∆
H reaction
= C
•
m
• ∆ t = 4.18 J/g˚C x (100. ML x 1.00 g/mL) x 0.4˚C
= - 167 J

To put the reaction on a molar basis we must find the moles of reactant and products:
Moles
Moles
(HCl)
= M
(HCl)
x V
(HCl)
= 0.0100 mol/L x 0.050 L = 5.00 x 10
(NaOH)
= M
(NaOH)
x V
-4
(NaOH)
= 0.0100 mol/L x 0.050 L = 5.00 x 10
-4
Enthalpy per mole of reactant consumed = -167J/ 5.00 x 10
-4
moles = -334. kJ/mole
Reaction Enthalpy
In virtually every chemical process heat is either given off or absorbed by the system. The difference in energy between the initial stage and final stage of the system is called the
ENTHALPY CHANGE of the system and is represented by the symbol
∆
H. Another term is simply heat of reaction.
∆
H = Heat products
– Heat reactants
The standard method of convenience when discussing reaction enthalpies is to write the balanced equation for the reaction along with the
∆
H. For example:
2H
2
(g) + O
2
(g) è 2H
2
O(g)
∆
H = - 483.6 kJ
The negative sign of the reaction tells us that the reaction is exothermic. Notice that the reaction enthalpy is reported with no explicit mention of the amount of reactants involved.
Usually in such cases it is understood that coefficients in the balanced equation represent the number of moles or reactants and products producing the associated enthalpy change.
If we halved the number of moles in the above reaction we would evolve half the heat energy.
H
2
(g) + 1/2O
2
(g) è H
2
O(g)
∆
H = - 241.8kJ
The enthalpy change accompanying a reaction can also be represented by an enthalpy diagram such as that shown below:
2H
2
(g) + O
2
(g)
∆
H = - 483.6 kJ
2H
2
O(g)
The enthalpy of a system is often viewed as a measure of how much potential energy is stored as heat in the system. Thus in an exothermic reaction, the stored heat energy is lost to the surroundings and the product(s) is(are) lower in heat energy than the reactants.

Consider the decomposition of water vapor into hydrogen gas and oxygen gas:
2H
2
O(g) è 2H
2
(g) + O
2
(g)
∆
H = +483.6 kJ
This is the reverse of the reaction discussed above. Note that this reaction is now endothermic, as indicated by the positive reaction enthalpy. Note also that the magnitude of the heat change is the same only the sign is different. The enthalpy diagram for this reaction si shown below:
2H
2
(g) + O
2
(g)
∆
H = 483.6 kJ
2H
2
O(g)
The fact that the reaction is endothermic (heat must be absorbed from the surroundings) indicates that the products of hydrogen gas and oxygen gas have a higher enthalpy content then the reactant. The heat that was absorbed from the surroundings is stored in the products.
The enthalpy of a reaction depends on the state( i.e. solid, liquid or gas) of the reactants and products. For example if we measured the heat required for the endothermic decomposition of liquid water, we would find that it differed from the enthalpy of reaction of water vapor by 81.4 kJ. The state of water, either solid, liquid or gas, is a function of temperature and/or pressure
2H
2
O(l) è 2H
2
(g) + O
2
(g)
∆
H = +565.0 kJ
The state of water, either solid, liquid or gas, is a function of temperature and/or pressure.
Thus it is necessary to specify both of these when a table of enthalpy data is made.
Usually a pressure of 1.0 atmosphere and a temperature of 25˚C are specified. The
STANDARD STATE of an element or compound is the most stable form of a substance in the physical state it would exist in at 1.0 atm and 25˚C. When the enthalpy of reaction is determined at the standard states of the species involved it is represented by the symbol
∆
H˚ At 1.0 atm and 25˚C the standard state of H
2
O is liquid water and the standard states of hydrogen and oxygen are gasses, thus:
2H
2
O(l) è 2H
2
(g) + O
2
(g)
∆
H˚ = +565.0 kJ
The following guidelines are helpful in using thermochemical equations and enthalpy diagrams.

1. This fact means that the magnitude of
∆
H is directly proportional to the amount
of reactant consumed.
The enthalpy change for a reaction is equal in magnitude but of opposite sign to the reverse reaction.
The enthalpy change for a reaction depends on the state of reaction.
Hess’s Law
Energy conservation is the basis of Hess’s Law which states that, if a reaction is the sum of two or more other reactions, the
∆
H overall is the sum of the
∆
H values of the
constituent reactions.. For example in the preceding section, we illustrated the decomposition of water, in two different states, into hydrogen and oxygen gas and showed that the enthalpy of reaction depended on the states of reactants and products. We will now show how we can use Hess’s Law to determine a reaction enthalpy.
The thermochemical equation for the state change of liquid water into water vapor is shown below. From our knowledge of the Heat of Vaporization (
∆
H vap
= 40.7 kJ/mole) of water we can determine the enthalpy of this state change. In addition, we know the enthalpy of reaction of the decomposition of water vapor. Adding the two reaction together we get:
2 H
2
2H
2H
2
2
0(l) è 2 H
O(g) è 2H
2
O(l) è 2H
2
2
O(g)
(g) + O
2
(g) + O
2
∆
H = 81.4 kJ (2 x 40.7 kJ/mole)
(g)
∆
H = 483.6 kJ (determined calorimetrically)
(g)
∆
H = 565.0 kJ
Since H
2
O(g) is the product of the first reaction and is the reactant in the second reaction, we can, just as in an algebraic equation which has the same variables on both sides of the equation, cancel out this term.
Example 6:
Calculate the enthalpy change (
∆
H) for the formation of 1 mole of strontium carbonate (the material that gives bright red color in fireworks) from its elements
Sr(s) + C(s) + 3/2O
2
(g) è SrCO
3
(s)
Given the following data:
Sr(s) + 1/2O2(g) è SrO(s)
SrO(s) + CO
2
(g) è SrCO
3
(s)
C(s) + O
2
(g)
è
CO
2
(g)
Solution:
∆
∆
∆
H˚ = -592 kJ
H˚ = -234 kJ
H˚ = -394 kJ
The constituent steps as given have the reactants and products on the correct sides of the reaction arrows. Cancel out any compound that occurs on both sides of the equation arrow(in this case both CO
2
and SrO). Make sure the stoichiometric coefficients match the overall reaction given. Thus:

Sr(s) + 1/2O2(g) è
SrO(s) + CO
2
(g)
è
SrCO
3
C(s) + O
2
(g)
è
CO
2
Sr(s) + C(s) + 3/2O
2
SrO(s)
(s)
(g)
(g) è SrCO
3
(s)
∆
H˚ = -592 kJ
∆
∆
∆
H˚ = -234 kJ
H˚ = -394 kJ
H˚ = -1220 kJ
Example 7:
Enthalpy changes have been determined experimentally for the following two
processes.
Pb(s) + Cl
2
(g) è PbCl
2
Pb(s) + 2Cl
2
(s)
(g) è PbCl
4
(s)
∆
∆
H˚ = -359.4 kJ
H˚ = -329.3 kJ
What is the enthalpy change of the reaction below?
PbCl
2
(s) + Cl
2
(g) è PbCl
4
(s)
Solution:
∆
H˚ = ?
Reverse the first reaction in order to get PbCl
2
(s) as a reactant and add this to the second reaction. Cancel out any reagents on both sides of the equation(in this case one mole of Cl
2
on the right cancels out one of the two moles of Cl
2 on the left).
PbCl
2
(s)
è
Pb(s) + Cl
Pb(s) + 2Cl
2
2
(g)
(g) è PbCl
PbCl
2
(s) + Cl
2
4
H˚ = +359.4 kJ
(s)
∆
H˚ = -329.3 kJ
(g)
è
PbCl
4
(s)
∆
H˚ = +30.1 kJ
Example 8:
Calculate the enthalpy of combustion of propane gas (C3H8) using the reactions shown below.
Calculate
∆
H° for: C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)
GIVEN the following standard enthalpies of formation:
3 C(s) + 4 H2(g) C3H8 (g)
∆
H°f = 103.85
C(s) + O2(g)
2 H2(g) + O2(g)
CO2(g)
2 H2O(g) kJ
∆
H°f = -393.5
∆
H°f = -285.8
mol kJ mol kJ mol
Solution:
Reverse the first reaction in order to get C
3
H
8
(g) as a reactant and add this to the second and third reaction. Multiply both sides of the second reaction by three and both sides of the third reaction by two in order to get the correct stoichiometric coefficients for the products. Cancel out any reagents on both sides of the equation ( in this case the H
2
(g) and C(s) cancel out)

C3H8 (g)
3 C(s) + 3 O2(g)
4 H2(g) + 2 O2(g)
3 C(s) + 4 H2(g)
3 CO2(g)
4 H2O(g)
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)
∆
H°f = -103.85
∆
H°f = -1180.5
kJ mol kJ mol kJ ∆
H°f = -571.6
∆
H°f = -1856.0
mol kJ mol
Enthalpies of Formation
Hess’s Law makes it possible for chemists to tabulate enthalpy of reaction data for a few reactions and to then use this information to derive many other reaction enthalpies. The enthalpy change which occurs for a reaction which forms one mole of a compound from its constituent elements in their standard states at standard conditions is called the
STANDARD MOLAR ENTHALPY OF FORMATION or Enthalpy of Formation for short and is signified by
∆
H˚ f
. Some of the reaction enthalpies already discussed involve enthalpies of formation. For example, the formation of one mole of liquid water from one mole of hydrogen gas and one-half mole of oxygen gas releases 285.8 kJ :
H
2
(g) + 1/2O
2
(g) è H
2
O(l)
∆
H˚ f
= -285.8 kJ
Similarly, the formation of one mole of solid lead(IV) chloride from one mole of metallic lead and two moles of chlorine gas releases 329.3 kJ of heat:
Pb(s) + 2Cl
2
(g) è PbCl
4
(s)
∆
H˚ f
= -329.3 kJ
When considering enthalpies of formation we observe the convention that the enthalpies of formation of elements in their standard states are always zero. This is because forming and element in its standard state from the same element in its standard state would involve no energy change.
Standard enthalpies of formation are very useful. For example, the enthalpy of reaction for any reaction can be easily calculated if the enthalpies of formation for all reactants
and products are known.. The following equation is used:
Enthalpy change for reaction =
∆
H˚ =
Σ ∆
H˚ f
(products) –
Σ ∆
H˚ f
(reactants)
In the above equation, the symbol
Σ
(the Greek letter sigma) to “take the sum of.” Thus to find the reaction enthalpy you add up the molar enthalpies of formation of the products and subtract from this the sum of the molar enthalpies of the reactants.

Example 9:
Calculate the enthalpy of reaction for the decomposition of limestone given the following enthalpies of formation.
CaCO
3
(s) è CaO(s) + CO
2
(g)
CaCO
3
(s)
∆
H˚ f
= -1206.9 kJ
CaO(s)
CO
2
(g)
∆
H˚ f
= -635.9 kJ
∆
H˚ f
= -393.5 kJ
Solution:
Remember:
∆
H˚ =
Σ ∆
H˚ f
(products) –
∆
H˚ = 177.5 kJ
Σ ∆
H˚ f
(reactants)
∆
H˚ = (-635.9 + -393.5)kJ - (-1206.9 kJ)
Example 10:
Calculate the enthalpy of reaction for the detonation of nitroglycerin given by the following reaction:
2 C
3
H
5
N
3
O
9
(l) è 3N
2
(g) + 1/2O
2
(g) + 6CO
2
(g) + 5H
2
O(g)
∆
H˚f kJ/mole
CO
2
(g)
H
2
O(g)
C
3
H
5
N
3
O
9
(l)
-393.5
- 241.8
-364.0
Solution:
To solve this problem first note that two of the products (N
2
and O
2
) are elements in their standard states and thus have
∆
H˚ f
= 0. We must also multiply the enthalpies of formation for the other products and reactants by their respective stoichiometric coefficients (remember that enthalpies of formation are for one mole). Thus:
∆
H˚ =
Σ ∆
H˚
= [5(
∆
H˚ f
(products) – f
H
Σ
2
O) = 6(
∆
H˚
∆
H˚ f
CO f
(reactants)
2
) – 2(
∆
H˚ f
C
3
H
5
N
3
O
9
)
= [5(-241.8 kJ) + 6(-393.5 kJ)] – 2(-364kJ)
= -2842 kJ

1. Using the heats of formation listed in appendix 2 at the back or your text book calculate the
∆
H° for the reactions below.
Pb (s) + 2 HBr (g)
∆
H° = _________
H2 (g) + PbBr2 (s)
Sn(s) + O
2
(g)
∆
H° = ________
SnO
2
(s)
2. Calculate the standard reaction enthalpy for the following reaction,
CH4 (g) + 4 (S) (s) from the following data
CS2 (l) + 2 H2S (g)
∆
H° = _________
C (s) + 2 H2 (g)
C (s) + 2 S (s)
S (s) + H2 (g)
CH4 (g)
CS2 (l)
H2S (g)
∆
H° = –74.8
∆
H° = +87.9
∆
H° = –20.6 kJ mol kJ mol kJ mol
3. If 5 grams of CH4 is burned in a calorimeter with a heat capacity of 66.0 kJ
K
, the temperature of the calorimeter assembly rises by 4.21 K. What is the enthalpy of combustion of methane?
∆
H = __________
4. A piece of gold of mass 25.0 g at 88.0°C was placed in a calorimeter that contained
75.0 g of water at 25.00°C and the temperature of the water rose to 25.70°C. What is the specific heat capacity of he gold?
Cpgold = _________