1. Lab 2 • Heats of Neutralization and Hess’ Law
PURPOSE: To determine the quantity of heat, ∆H, evolved in some acid-base reactions.
CONCEPTS: calorimetry, Hess Law, enthalpy, entropy, free energy
Text References: Kotz and Treichel Chapter 6 Principles Of Reactivity: Energy and Chemical Reactions
and Chapter 19 Principles of Reactivity: Entropy and Free Energy
INTRODUCTION:
Every chemical change is accompanied by a change in energy, usually in the form of heat. The
energy change of a reaction that occurs at constant pressure is termed the heat of reaction or the enthalpy
change. The symbol ∆H is used to denote the enthalpy change. If heat is evolved, the reaction is
exothermic (∆H < 0 or negative); and if heat is absorbed the reaction is endothermic (∆H > 0 or positive).
Although the enthalpy change for a chemical reaction tells us whether heat will be released or ab-
sorbed, it cannot be used to determine if a reaction can spontaneously proceed. There are examples of
spontaneous reactions that have either positive or negative enthalpy changes. In addition to enthalpy,
entropy and free energy functions must be used to determine if a particular reaction is possible.
A chemical reaction may also involve energy changes other than a flow of heat to or from the
surroundings. For example, an electrochemical cell can produce electrical energy equivalent to the product
of the cell voltage multiplied by the electric charge transferred to an external circuit. A chemical reaction
that produces a gas must push back the atmosphere, and this mechanical work is another form of energy.
(If the reaction is carried out in containers open to the atmosphere, the work of pushing back the
atmosphere is included in the enthalpy change, ∆H.)
Enthalpy changes may be classified into more specific categories: (1) The heat of formation is the
amount of heat involved in the formation of 1 mol of the substance directly from its constituent elements;
(2) the heat of combustion is the amount of heat produced when a mole of a combustible substance reacts
with excess oxygen; (3) the heat of solution of a substance is the thermal energy change that accompanies
the dissolving of a substance in a solvent; (4) the heats of vaporization, fusion, and sublimation are related
to the thermal energy changes that accompany changes in state: (5) the heat of neutralization is the enthalpy
change associated with the reaction of an acid and a base.
The heat or enthalpy change for a chemical reaction is called the heat of reaction (∆H rxn) The
enthalpy change—defined as the difference in enthalpy between the products and reactants—is equal to the
amount of heat transferred at constant pressure and does not depend on how the transformation occurs:
state function. This definition of enthalpy makes it possible to determine the heats of reaction for reactions
that cannot be measured directly. According to Hess’s Law, if the same overall reaction is achieved in a
series of steps, rather than in one step, the enthalpy change for the overall reaction is equal to the sum of the
enthalpy changes for each step in the reaction series. There are two basic rules for calculating the enthalpy
change for a reaction using Hess’s Law.
• Equations can be “multiplied” by multiplying each stoichiometric coefficient in the balanced
chemical equation by the same factor. The heat of reaction (∆H) is proportional to the amount
of reactant. Thus, if an equation is multiplied by a factor of two to increase the number of moles
of product produced, then the heat of reaction must also be multiplied by a factor of two.
• Equations can be ‘subtracted” by reversing the reactants and products in the balanced chemical
equation. The heat of reaction (∆H) for the reverse reaction is equal in magni tude but opposite
in sign to that of the forward reaction.
In this experiment you will measure the heat of neutralization (or the enthalpy of neutralization)
when an acid and a base react to form water. This quantity of heat is measured experimentally by allowing
Lab 2. Thermodynamics page 1

2. the reaction to take place in a thermally insulated vessel called a calorimeter. The heat liberated in the
neutralization will cause an increase in the temperature of the solution and of the calorimeter. If the
calorimeter were perfect, no heat would be radiated to the laboratory. Calorimetry is the measurement of
heat changes in physical and chemical processes. A calorimeter is an insulated vessel used in the
measurement of heat changes in chemical and physical processes, while minimizing the heat exchange
between the system being observed and its surrounding. The insulated vessel or calorimeter contains a
temperature measuring device and a stirrer.
Because we are concerned with the heat of the reaction and because some heat is absorbed by the
calorimeter itself, we must know the amount of heat absorbed by the calorimeter. This requires that we
determine the heat capacity of the calorimeter. By “heat capacity of the calorimeter” we mean the amount
of heat (that is, the number of joules) required to raise its temperature 1 Kelvin, which is the same as 1 ˚C.
In this experiment, the temperature of the calorimeter and its contents is measured before and after the
reaction. The change in the enthalpy, ∆H, is equal to the product of the temperature change, ∆T, times the
heat capacity of the calorimeter and its contents.
Specific heat and heat capacity are two ways for expressing the energy associated with the
temperature change in a substance. Specific Heat describes the amount of energy required to change the
temperature of a unit of mass of that substance by one degree. The units of energy most commonly used are
the calorie and the Joule (1 cal = 4.184 J).
Specific Heat = (amt of heat, J)
(mass of substance, g) (Temp. change, 0C) Equation I
Heat capacity describes the amount of energy required to raise the temperature of a particular
object, such as a specified calorimeter that is being used repeatedly, by one degree. The equation defining
the heat capacity is:
Heat capacity = (amt of heat, J)
(Temp. change, 0C) Equation II
The heat capacity of the calorimeter is determined by measuring the temperature change that occurs
when a known amount of hot water is added to a known amount of cold water in the calorimeter. The heat
lost by the warm water is equal to the heat gained by the cold water and the calorimeter. (We assume no
heat lost to the laboratory.) The difference between the heat lost by the hot water and the heat gained by
the cold water is the heat lost to the calorimeter. The lost to the calorimeter is equal to the its temperature
change times its heat capacity.
If one assumes that the specific heat of the dilute reaction solution is the same as the specific heat of
the solvent, measuring heat changes for reactions in dilute solution becomes simpler. Calculations of the
energy change, and heat of reaction, utilize the temperature change of the calorimeter and of the dilute
reaction solution. Neutralization reactions of acids and bases are examples of reactions in solution.
Common acid-base reactions involve the reaction of equal numbers of equivalents of the acid and
the base to form a salt and water. The reaction of perchloric acid (HC1O 4) and potassium hydroxide (KOH)
to form the salt, potassium perchlorate (KC1O4) and water (Eq. 1) is an example. The total ionic equation
(Eq. la) and the net ionic change (Eq. lb) are given below. H+ ions in water are shown as H3O+
HClO4(aq) + KOH(aq)  KClO4(aq)+ H2O(l) Equation 1
H3O (aq) + ClO4-(aq) + K+(aq) + OH-(aq)  K+(aq) + ClO4-(aq) + 2H2O(l)
+
Equation 1a
H3O+(aq) + OH-(aq)  2H2O(l) Equation 1b
Enthalpy is defined as the change in the heat content of a substance under constant pressure (∆H =
qp). Hess’ Law states that the enthalpy change, ∆H˚rxn, for a given chemical or physical process depends
only on the conditions in the initial and final states. The enthalpy change is the same regardless of the
pathway by which the process takes place.
Lab 2. Thermodynamics page 2

3. ∆H0rxn = ∑ n∆H˚f (products) — ∑ n∆H˚f (reactants) Equation III
The coefficients m and n are the number of moles of each reactant and each product. The net ionic equation
(Eq. lb) for strong acid, HC1O4, and strong base, KOH, is the same for any strong acid—strong base
neutralization.
Another example is the reaction of solutions of H2S04 and Ca(OH)2 (Equations 2, 2a and 2b).
H2S04(aq) + Ca(OH)2(aq)  CaSO4(aq) + 2H2O Equation 2
2H3O+(aq) + SO42-(aq) + Ca2+(aq) + 2OH-(aq)  Ca2+(aq) + SO42-(aq) + 4H2O(l) Equation 2a
2H3O+(aq) + 2OH-(aq)  4H2O(l)
H3O+(aq) + OH-(aq)  2H2O(l) Equation 2b
The neutralization process for any strong acid-strong base follows Hess’ Law. Somewhat less heat
is evolved when either the acid or base is weak. This is due to the requirement of energy to break the
stronger bonds in the weak electrolytes. Consider the example of the reaction of the weak acid, HNO 2, with
the strong base, KOH, to form the soluble salt, KNO2, and water (Equation 3).
HNO2(aq) + KOH(aq)  KNO2(aq) + H2O(l) Equation 3
HNO2(aq) + K+(aq) + OH-(aq)  K+(aq) + NO2-(aq) + H2O(l) Equation 3a
HNO2(aq) + OH-(aq)  NO2-(aq) + H2O(l) Equation 3b
Some of the heat evolved in the neutralization is consumed in the ionization of HNO2. In this experiment
you will measure the heat of reaction per mole for several acid-base reactions, both strong and weak, using
the calorimeter. ∆Hr will be called ∆H4 , ∆H5 , ∆H6 and ∆H7 in reference to Equations 4,5,6, and 7.
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) ∆H4 =? Equation 4
CH3COOH(aq) + NaOH(aq)  NaCH3COO(aq) + H2O(l) ∆H5 =? Equation 5
CH3COOH(aq) + NH3(aq)  NH4CH3COO(aq) + H2O(l) ∆H6 =? Equation 6
According to Hess’ Law if several equations can be manipulated to give another equation such as Eq. 4 –5
+ Eq 6 = Eq. 7, then the sum of the ∆H’s of the manipulated equations will be equal to the value of ∆H7.
HCl (aq) + NH3(aq)  NH4C1(aq) ∆H7 =? Equation 7
In this experiment you will calculate ∆H7 from ∆H7 (energies of formation) using the table in your
text and Formula III. You will compare your experimentally determined ∆H7 value for Eq. 7 with the
calculated value.
The heat capacity of your calorimeter must be determined before you use it to measure the heat
transferred by the reactions. The following examples illustrate both procedures.
Example 2.1 A 50.0 mL sample of H20 at 41.2 0C is mixed with 50.0 mL of water which is at 21.6 0C in a
calorimeter. The final temperature of the mixture is 30.8 ˚C. Calculate the heat capacity of the calorimeter.
The specific heat of H2O is 4.18 J/g˚C . The density of water is 1.00 g/mL.
Solution. If there is no transfer of heat to the calorimeter, the solution retains all of the heat, and
therefore the heat lost by the warm H20 [(10.4 ˚C)(50.0 g)(4.18 J/g˚C) or 2170 J] equals the heat
gained by the cool H20. [(9.2 ˚C)(50.0 g)(4.18 J/g˚C) or 1920 J].
Since this is not true, the difference must be due to heat lost to the calorimeter.
2170 J — 1920 J = 250 J of heat unaccounted
The heat capacity of the calorimeter is 250 J/(30.8˚C — 21.6 ˚C) = 27J/˚C. When you have calibrated
your calorimeter, you can use it to collect data on the neutralization reactions. An example of the
use of these data is given in Example 2.2.
Lab 2. Thermodynamics page 3

4. Example 2.2: Suppose you mixed 50.0 mL of a solution of an acid, HA, with 50.0 mL of a base, 50 in
the calorimeter in Example 2.1. Both solutions were initially at 21.7 ˚C. The temperature rises to
27.8 ˚C. Calculate ∆Hr.
1st Calculate the amount of heat absorbed by the solution. Assume the specific heat of the solution is
the same as H20, 4.18 J/g˚C and that its density is the same as H20, 1.00 g/mL.
amt. of heat = mass x sp. heat x temp. change
J. absorbed = 100 g x 4.18 J/g˚C x (27.8 - 21.7) ˚C = 2550J
2nd Calculate amount of heat released in reaction
Joules released in reaction = Joules absorbed by solution + Joules absorbed by calorimeter
= 2550 J + [(27.8 —21.7) ˚C x 27 J/˚C]
= 2550 J + 165 J = 2715J
CAUTION! YOU MUST WEAR EYE PROTECTION AT ALL TIMES!!
KEEP ALL REACTION VESSELS WELL AWAY FROM YOUR FACE!!
PROCEDURES:
You will need a watch with a second hand or a stopwatch for the experiment.
Part A. - Calibration of thermometers
Step 1. Obtain two 110˚ C thermometers and label them “#1” and “#2.”
Step 2. Immerse both thermometers for two minutes in 150 mL of water contained in a 250 mL beaker.
Step 3. Carefully read the temperature on each thermometer to the nearest 0.1˚C. Record these values.
(NOTE: The difference in these two values will be used as a correction factor to correct readings
of #2 to the equivalent readings for #1. Thermometer #1 will always be used in the
calorimeter!)
Part B. - Heat Capacity of Calorimeter
Step 4. Assemble the calorimeter as shown in Figure 2.1. Use two 6-oz. styrofoam cups. The cardboard or
styrofoam lid should have a hole just large enough to allow the insertion of the thermometer. Use
a 1/4 inch section of rubber tubing around the thermometer to keep the bulb of the thermometer
about 1/2 inch above the bottom of the cup.
Step 5. Measure 50.0 mL of water in a graduated cylinder and pour this into the dry calorimeter. Put the
thermometer and lid in place and allow 5-10 minutes for the system to reach equilibrium.
Step 6. Measure 50.0 mL of water in a graduated cylinder and pour it into a dry 250 mL beaker.
Step 7. Heat the water in the beaker to give a 10-15˚C temperature rise on thermometer #2. Let this
sample stand for about 2 minutes to stabilize in temperature.
Step 8. Record the temperature of the calorimeter and cool water (from Step 5) to the nearest 0.1˚C.
(NOTE: Since you will be making measurements over a period of time, the following four steps,
9-12, must be done as a quick series. First read all four steps and plan your actions.)
Step 9. Record the temperature of the warm water in the beaker to the nearest 0.1˚C.
Step 10. NOTE: Use a watch to time the readings. One student should make readings while the other keeps
time and records the data.
Lab 2. Thermodynamics page 4

5. Step 11. Remove the lid of the calorimeter and quickly, add the warm water without splashing, and replace
the lid. Begin recording the time measurements as soon as the warm water is added. Gently swirl
the cups to mix the liquids. (NOTE: Do not break the thermometer.)
Step 12. Record the temperature of the water every 10 seconds on the time-temperature data sheet until a
maximum has been reached. Continue recording temperature readings every 30 seconds for 3
minutes. Record the data in a clearly labeled table.
Step 13. Plot the data (use a pencil) on a small portion of graph paper and determine what the temperature
should have been at time = zero seconds. Record this temperature. (NOTE: Since you will be
measuring the temperature change for four reactions, this type plot will be done for each reaction.
You can see that the calorimeter is losing heat to the surroundings!)
Step 14. Calculate the heat capacity of your calorimeter from your data and record this value.
Part C. Heats of Neutralization of Acids and Bases.
Step 15. Rinse and dry the calorimeter
Step 16. Measure 50.0 mL of the standardized HCl (3.00 ± 0.10M) in a dry graduated cylinder and pour
it into the calorimeter. Calculate the number of moles of acid in this volume. Record the exact
concentration and number of moles of acid. Replace the lid and thermometer #1.
Step 17. Measure 50.0 mL of the standardized NaOH (3.00 ±0.10M) in a dry graduated cylinder.
Calculate the number of moles of base in this volume. Record the exact concentration and number
of moles of base.
Step 18. Adjust the temperature of the base to the same temperature as the acid in the calorimeter. (NOTE:
You may use warmth of your hands or cool running water for this adjustment.) Record this
temperature as the initial temperature, Ti.
NOTE: Steps 19-20 must be performed quickly. Read all the steps and plan your actions!
Step 19. Remove the lid from the calorimeter, pour in the base solution without splashing, and replace the
lid and thermometer. Begin measuring time and temperature.
Step 20. Use the procedure in steps 11, 12, and 13 to determine and record the temperature at t = 0.
Step 21. Determine the change in temperature, ∆T, and record the value. Calculate the total heat evolved
for this quantity of reactants. (NOTE: Use Example 2.2 as a guide.)
Step 22. Calculate and record the ∆H4 per mole of acid neutralized.
Step 23. Repeat Steps 15-22 for each of these acid-base pairs (a) CH3COOH/NaOH, and (b)
CH3COOH/NH3. Calculate ∆H5, and ∆H6 per mole of acid neutralized in Equations 5 and 6 and
record your data and results.
Part D. Hess’s Law
Step 24. Use your value for the ∆H4, ∆H5, and ∆H6 and Hess’ Law to calculate ∆H7 .
HCl(aq) + NH3(aq)  NH4C1(aq) ∆H7 . =
Step 25. Use ∆Hf values from a text reference and calculate ∆H7 for the same reaction and compare these
Lab 2. Thermodynamics page 5

6. two values.
RESULTS: The following data and calculations must be included in your lab notebook.
DATA: Data should be clearly recorded in your lab notebook as you as you work gather the data.
Part A. - Calibration of the Thermometers
Step 3. Temperature of thermometer #1 in tap water:
Temperature of thermometer #2 in tap water:
Correction factor for thermometer #2 to make readings equal thermometer #1: (Indicate + or -)
Part B. - Heat Capacity of the Calorimeter (Calorimeter Constant)
Step 8. Temperature of calorimeter and cool H2O
Step 9. Temperature of warm H2O
Step 12. Table with the temperature of the mix in calorimeter every 10 seconds, then every 30 seconds
Step 13. Temperature of the mixture at time = 0 determined from the Temp/Time plot
Graphs of time compared to temperature must be inserted.
Part C. - Heat of Neutralization of Acids and Bases You should have three sets of this data.
Step 18. Temperature of the reactants before mixing
Step 19. Table with the temperature of the mix in calorimeter every 10 seconds, then every 30 seconds
Step 20. Temperature of the reaction mixture at Time 0 sec determined from the Temp/Time plot.
Graphs of time compared to temperature must be inserted.
Step 21. Temperature change for the reaction
CALCULATIONS: Calculations should be clearly labeled and show all steps.
Part B. - Heat Capacity of the Calorimeter (Calorimeter Constant)
Step 14. Calculations- show complete calculations with all appropriate units.
Amount of heat released by warm water
Amount of heat absorbed by cool water
Amount of heat absorbed by calorimeter
Heat capacity of calorimeter
Part C. - Heat of Neutralization of Acids and Bases
Reaction #4 Copy the molecular equation and write the other two equations for this reaction.
HC1(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
Balanced molecular equation and net ionic equation
Calculate the theoretical value ∆H4. for this reaction using the heats of formation from back of the
lab manual your text to calculate this value. Show each step in your calculations.
Step 16. Calculation of the number of moles in the volume of HCl solution
Step 17. Calculation of the number of moles in the volume of NaOH solution
Which reagent is the limiting reagent?
What is the number of moles of HCl neutralized?
Step 22. Calculation of the experimental value of value ∆H4
Amount of heat absorbed by the solution
Amount of heat absorbed by the calorimeter
Total amount of heat evolved by neutralization reaction
Amount of heat evolved per mole of HCl neutralized, ∆H4
Compare your experimental value with the calculated value.
Repeat all of the part C data, equations and calculations for reactions 5 and 6.
Reaction #5. CH3COOH(aq) + NaOH(aq)  NaCH3COO(aq) + H2O(l)
Reaction #6. CH3COOH(aq) + NH3(aq)  NH4CH3COO(aq)
Part D. - Using Hess Law
Lab 2. Thermodynamics page 6

7. Step 24. Hess’ Law Record your experimental values for each subreaction.
∆H neutralization
HC1(aq) + NaOH(aq)  NaC1(aq) + H20 ∆H4 = _________ kj/mol
CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H20 ∆H5 = ________ kJ/mol
CH3COOH(aq) + NH3(aq)  CH3COONH4 (aq) ∆H6 = ________ kJ/mol
Write these equations as total ionic equations and manipulate them and their enthalpy changes to calculate
the experimental value of the enthalpy change for the reaction given below (∆H7). Clearly show how you
manipulated the subreactions and calculated the value of ∆H.
HC1(aq) + NH3(aq)  NH4C1(aq) ∆H7 _________ kJ/mol
(from experimental data)
Step 25.
Calculate the theoretical value of ∆H using the heat of formation. Compare it to your experimental value.
Procedural Questions:
1. Why was it necessary to calibrate the two thermometers?
2. Why was it necessary to find the heat capacity of the calorimeter?
Concept Questions:
3. State Hess’s Law. Explain how this law allows you to calculate the value of ∆H for reaction #7.
4. Enthalpy.
a. State the first law of thermodynamics.
b. Compare internal energy (E) and enthalpy (H).
c. Define endothermic and exothermic reactions in terms of the direction of energy transfer and the
sign of ∆H. Sketch a graph to show the change in energy content.
d. Explain whether endothermic or exothermic reactions are favored based upon the dispersion of
energy. Describe these as either product favored or reactant favored.
5. Calorimetry.
a. Compare and contrast a bomb calorimeter with our stryofoam cup calorimeter.
b. Compare and contrast heat capacity and specific heat. How is each a useful quantity?
c. Describe how you could determine the specific heat of a metal by using the apparatus and techniques
in this experiment.
d. Predict and sketch the shape of the temperature versus time for an endothermic reaction. Show how
the temperature change is determined. Think: what will happen to the temperature of the solution?
6. Entropy.
a. State the second law of thermodynamics.
b. Relate the change in entropy that accompanies a reaction to whether it is product or reactant favored.
Use the concept of matter dispersal.
7. Gibbs Free Energy and Spontaneous Reactions.
a. Describe free energy
b. How is the calculated value of ∆G related to determining whether the reaction is spontaneous?
c. How does the value of ∆G depend upon temperature?
8. Strong vs Weak Acid
Lab 2. Thermodynamics page 7

8. a. Write the reversible reaction for the weak acid CH3COOH that occurs in aqueous solution. Write the
Ka for this equilibrium.
b. How does the behavior of CH3COOH in solution differ from the behavior of HCl?
c. Use bonding to explain this difference in behavior. Which is acid is representative of a weak acid
and which a strong acid?
d. Explain why the neutralization of weak acid is less exothermic than that of a strong acid.
e. How is calculation of ∆G related to equilibrium constant for a reversible reaction?
9. Strong vs Weak Base
a. How is sodium hydroxide a typical strong base? How does this base fit the Arrhenius model?
b. How does ammonia (NH3) behave as a weak base? Include a Lewis structure for ammonia molecule
and a discussion of the ions formed in solution using the Bronsted-Lowry model.
Additional Calculations:
10. How many joules are required to change the temperature of 80.0 g of water from 23.3˚C to 38.8˚C?
11. A piece of metal weighing 5.10 g at a temperature of 48.6˚C was placed into 20.00 mL of water in a
calorimeter at 22.1˚C, and the final equilibrium temperature was found to he 28.2˚C. What is the
specific heat of the metal?
12. If the specific heat of methanol is 2.51 J/gK, how many joules are necessary to raise the temperature of
50.0 g of methanol from 18˚C to 33˚C?
13. When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water,
the temperature rose from 23.9˚C to 32.2˚C. Calculate ∆H (in kJ/mol NaOH) for the solution process:
NaOH(s)  Na+(aq) + OH-(aq)
Assume it’s a perfect calorimeter and that the specific heat of the solution is the same as pure water.
14. Consider the reaction of 50.0 mL of 1.09 M CH3COOHwith 50.0 mL of 0.960 M NH3.
a. How many moles of CH3COOH and of NH3 are initially present?
b. How many moles of CH3COOH and of NH3 are actually neutralized?
15. Following the procedure described in Steps 4-13 of this experiment, this data was obtained when a
glass calorimeter was calibrated. Calculate the heat capacity (calorimeter constant) for this calorimeter.
Temperature of the calorimeter and 50.0 mL of cool water 20.6˚C
Temperature of 50 mL warm water 38.9˚C
Temperature of the calorimeter and mixture at time = 0 28.4˚C
16. The complete neutralization of 0.0482 moles of a monoprotic acid with excess NaOH evolves 2.62 kJ
of heat. What is the molar heat of neutralization of the acid with NaOH.
17. A student measured the molar heat of neutralization of a monoprotic acid, HA(1.02 M), with NaOH
(0.974 M) and obtained these data. Calculate the molar heat of neutralization of this acid with NaOH.
temperature of HA and NaOH before mixing 22.7 0˚C
volume of the acid, HA, used 50.0 mL
volume of the base, NaOH, used 50.0 mL
temperature of calorimeter and mixture at time = 0 29.1 ˚C
heat capacity of calorimeter 6.9 J/˚C
Lab 2. Thermodynamics page 8

9. AP 2004 (question 2)
3
2 Fe (s) + /2 O2 (g)  Fe2O3 (s) ∆Hf = -824 kJ/mol
Iron reacts with oxygen to produce iron (III) oxide, as represented by the equation above. A 75.0 g sample
of Fe (s) is mixed with 11.5 L of O2 (g) at 2.66 atm and 298 K.
a. Calculate the number of moles of each of the following before the reaction begins.
i. Fe (s)
ii. O2 (g)
b. Identify the limiting reactant when the mixture is heated to produce Fe2O3 (s).
Support your answer with calculations.
c. Calculate the number of moles of Fe2O3 (s) produced when the reaction proceeds to completion.
d. The standard free energy of formation, ∆Gf˚, of Fe2O3 (s) is –740. kJ/mol at 298 K.
i. Calculate the standard entropy of formation, ∆Sf˚, of Fe2O3 (s) at 298 K. Include units with your
answer. Use the ∆G equation to find this value not the entropy values from your table.
ii. Which is more responsible for the spontaneity of the reaction at 298 K, the standard enthalpy of
formation, ∆Hf˚, or the standard entropy of formation, ∆Sf˚, Justify your answer.
e. The reaction represented below also produces iron (III) oxide. The value of the ∆H˚ for the reaction
is –280. kJ/mol of Fe2O3 (s) formed.
2 FeO (s) + 1/2 O2 (g)  Fe2O3 (s)
Calculate the standard enthalpy of formation, ∆Hf˚, of FeO (s). Use the reactions above and Hess Law to
rearrange the reactions to get the heat of formation. Show the manipulations and calculation.
Lab 2. Thermodynamics page 9

10. There are two basic rules for calculating the enthalpy change for a reaction using Hess’s Law.
• Equations can be “multiplied” by multiplying each stoichiometric coefficient in the balanced chemical
equation by the same factor. The heat of reaction (∆H) is proportional to the amount of reactant. Thus, if an
equation is multiplied by a factor of two to increase the number of moles of product produced, then the heat
of reaction must also be multiplied by a factor of two.
• Equations can be ‘subtracted” by reversing the reactants and products in the balanced chemical equation.
The heat of reaction (∆H) for the reverse reaction is equal in magni tude but opposite in sign to that of the
forward reaction.
Lab 2. Thermodynamics page 10