EXPERIMENT 9
ENTHALPY OF REACTION—HESS’S LAW
INTRODUCTION
Chemical changes are generally accompanied by energy changes; energy is absorbed or evolved, usually as
heat. Breaking chemical bonds in reactants requires energy, and energy is released as new bonds form in products.
Whether the combination of these steps absorbs or releases energy depends on the relative sizes of the energies
associated with breaking and forming bonds.
The amount of heat involved in a reaction depends not only on what the reaction is, but also on the temperature
at which the reaction occurs and whether the reaction occurs under conditions of constant pressure or constant
volume. In the laboratory, many reactions are conveniently carried out at constant pressure in beakers or flasks
that are open to the atmosphere. The amount of heat absorbed or released under this condition is the enthalpy
change, ∆H, for the reaction, where
∆H = Hproducts - Hreactants
(9-1)
Enthalpy, H, can be thought of as the heat content of a substance; this heat is stored as potential energy in the
form of bond, and other, energies. When atoms rearrange during a reaction, the heat content of the products is
usually different from the heat content of the reactants. This difference in heat content appears as heat absorbed
or released. This heat is generally indicated in Joules for the reaction as written. For example, in this experiment
you will examine an acid-base neutralization in aqueous solution:
H3O+(aq) + OH-(aq) → 2 H2O(l)
∆H = -55.8kJ
(9-2)
The enthalpy change for this reaction could be given in J/(mol of H3O+), or in J/(mol of OH−), or in J/(mol of
H2O). To avoid confusion it is customary to report ∆H for the reaction with the numbers of moles of reactants and
products simply as written. Thus, for reaction 9.2, in which 1 mole of H3O+ and 1 mole of OH− combine to form
2 moles of H2O, ∆H = -55.8 kJ, as shown.
Note that the enthalpy change, ∆H = Hproducts − Hreactants, is positive if heat is absorbed; that is, if Hproducts >
Hreactants, the reaction is endothermic. The enthalpy change is negative if heat is released. If Hproducts < Hreactants, the
reaction is exothermic. This is illustrated in the following diagrams:
reactants
products
H
H
heat evolved
heat absorbed
reactants
∆H > 0
products
∆H < 0
(endothermic)
Experiment 9
(exothermic)
9-1
In an exothermic reaction, the overall process causes the system to become more energetically stable; excess
energy is released as heat. The system changes from a higher energy state to a lower energy state. It seems
plausible intuitively that reactions would have a natural tendency to proceed spontaneously to a state of lower
energy; in fact, a negative enthalpy change is part of what causes some reactions to occur spontaneously.
Physical changes also have associated enthalpy changes. For the melting of water, for example:
H2O(s)
→
H2O(l)
∆H = + 6.1kJ
(9-3)
In this process, no chemical bonds are broken or formed. The energy input converts the potential energy from
that characteristic of the rigid solid-state organization of water molecules to that characteristic of the liquid state,
in which water molecules move over, under, and around one another, but are still held together in a relatively
small volume.
Similarly, enthalpy changes accompany the dissolving of solids or the dilution of solutions. For example,
when AlCl3(s) dissolves in water, the Al3+ and Cl− ions must be separated. In the solid state they are packed
together in a way that results in the greatest stability (cations surrounded by anions and vice versa). In liquid water
the water molecules are extensively H-bonded to one another. They must be separated to create cavities for the
Al3+ and Cl− ions to occupy. Both of these processes cost energy, i.e., they are endothermic. In contrast, energy is
released when Al3+ ions are hydrated, becoming closely surrounded by the negative ends of six polar water
molecules in solution. Similar interaction of Cl− ions with the positive ends of polar water molecules also releases
energy. The balance among the energies for all these processes determines whether the overall ∆Hsolution for AlCl3
will be positive or negative. (It turns out to be negative.)
Enthalpy is one member of an important class of thermodynamic functions known as state functions. A state
function has the unique characteristic that its value for a system does not depend upon the history of the system.
No matter how a system was formed, the value for any state function depends only on the present state of the
system. This has the extremely useful practical consequence that the change in any state function for a process
depends only on the initial and final states and thus is independent of the pathway by which the process takes the
system from the initial state to the final state. Another way of stating this for enthalpy is in the form of Hess’s
Law of Constant Heat Summation: If a reaction (or physical process) is carried out in a series of steps, ∆H for the
overall process is equal to the sum of the enthalpy changes for the individual steps. It is a consequence of Hess’s
Law that, as long as we begin with the reactants in one particular state (given temperature, etc.) and end with the
products in another particular state, the overall ∆Hrxn is the same, no matter what individual steps are carried out,
or in what order, to convert the reactants to the products. We will examine how well this principle holds for two
different chemical reactions in this experiment.
TECHNIQUE
Calorimetry is the measurement of the heat change for a reaction, and the device used to measure heat changes
is a calorimeter. In this experiment, the calorimeter consists of two nested Styrofoam cups with a MeasureNet
temperature probe inserted into the cup to monitor the temperature. If the reaction is exothermic, the heat released,
which will ultimately be transferred to the surroundings, is temporarily trapped by the insulating walls of the
Styrofoam cups and causes the temperature of the calorimeter contents to rise. For an endothermic reaction, the
temperature of the calorimeter contents falls. The Styrofoam cups temporarily prevent heat from entering the
system from the surroundings, and the calorimeter contents must supply the heat absorbed in the process. In either
case, the size of the temperature change is used to determine the quantity of heat released or absorbed.
Experiment 9
9-2
The amount of heat released or absorbed is related to the temperature change by the equation
q = m × c × ∆T = C × ∆T
where
(9-4)
q = amount of heat,
m = mass of material undergoing the temperature change,
c = specific heat of the material undergoing the temperature change,
∆T = temperature change = Tfinal − Tinitial, and
C = m × c = heat capacity of the material undergoing the temperature change.
In highly accurate calorimetric measurements, it is necessary to consider contributions to the heat capacity, C,
from the solution present, the calorimeter walls, and even the temperature probe itself (that is, everything which
undergoes the temperature change). While heat capacity data for many solutions can be taken from tables in
chemical handbooks, calorimeters and temperature probes differ, so their contributions to C must be determined
experimentally. (In practice the total heat capacity of calorimeter and contents is determined by measuring the
temperature change caused by adding a known amount of heat with an electrical heating device.) We will greatly
simplify the analysis by means of reasonable approximations:
1. We ignore the small contributions from the calorimeter walls and temperature probe.
2. We estimate the heat capacity of the calorimeter contents by considering only the amount of water
present and treating any solution added as though it were just water with a density of 1.000 g/mL. This
works fairly well since, although the separate values of m and c for a solution are different from m and c for
an equal volume of
pure water, the product
m × c changes little. So, Tf = 53.1°C
in this experiment we
will approximate the
specific heat of all
solutions as 4.184
J/(°C ⋅ g of water
present).
As EQUATION 9-4
indicates, q (the amount of
heat absorbed or evolved)
is proportional to ∆T. It is
often quite simple to
Ti = 22.6°C
measure ∆T. If no heat
escapes from or enters the
calorimeter from outside,
we need only to measure
Figure 9-1. Temperature vs. time plot for exothermic reaction.
Tinitial, mix the reactants,
and then measure Tfinal. If
the temperature in the calorimeter is close to room temperature and ∆T is small, this works well. If the temperature
in the calorimeter differs substantially from room temperature, however, (which always happens for large values
Experiment 9
9-3
of ∆T) the heat leakage must be taken into account. Such a case is shown in FIGURE 9-1, a plot of temperature
versus time for a highly exothermic reaction. The initial temperature, Ti, was monitored for 10 seconds, and found
to be 22.6°C. The reaction was initiated by mixing the reactants at t = 10 seconds. The temperature increased
rapidly at first and then slowly dropped due to heat leakage from the calorimeter. A straight line is drawn and
extrapolated to time t = 10 sec, giving a value of 53.1°C, which we take to be the final temperature, Tf, for the
reaction. This is the temperature that the calorimeter and its contents would have reached if complete mixing and
reaction had been instantaneous and if no heat leakage had occurred.
The extrapolation just
described is not necessary for
T1 = 23.51°C
small values of ∆T near room
T2 = 23.30°C
temperature, since there will
be very little heat leakage. This
situation is illustrated in
FIGURE 9-2, in which an
T4 = 22.02°C
endothermic process takes
place, followed by an
exothermic process. The entire
temperature range shown is
less than 5°C, so the
temperature is always close to
room temperature. The lack of
drift is evident in the data,
from the essentially constant
values before and after each of
the
processes.
(The
T3 = 19.55°C
temperature scale is greatly
expanded in FIGURE 9-2, so
even nearly constant values
Figure 9-2. Temperature vs. time plot for two-step experiment.
may appear to be changing a
little.) The interpretation of
FIGURE 9-2, particularly with regard to the labeled temperatures, T1 through T4, is discussed later in this
experiment.
We must be careful with the sign of ∆T when we use it to determine ∆H for the process. When ∆T is negative,
the value of q obtained from EQUATION 9-3 will also be negative. As discussed above, a negative ∆T is the result
of the occurrence of an endothermic process, for which ∆H is positive. Similarly, a positive ∆T results from an
exothermic process, for which ∆H is negative. Therefore, in general,
∆H process = −q
Experiment 9
(9-5)
9-4
OBJECTIVES
•
•
•
•
to determine the temperature changes for a series of chemical processes
to determine the enthalpy changes for the chemical processes
to compare the enthalpy changes with available literature values
to test the validity of Hess’s Law
EQUIPMENT NEEDED
burner
8-oz. Styrofoam cups (2)
100-mL graduated cylinders (2)
100-mL beaker
MeasureNet
temperature probe
CHEMICALS NEEDED
distilled water
1.00 M HCl; hydrochloric acid
NaOH(s); sodium hydroxide
NH4Cl(s); ammonium chloride
1.00 M NaOH; sodium hydroxide
PROCEDURE
The measurements in this experiment can be carried out in a relatively short time. However, the analysis and
interpretation of the results will probably take somewhat longer. Therefore, before you spend any time on the
analysis, be sure that you have data of adequate quality. Show the plots you obtain in the steps below to your
instructor for his/her comments and approval. There should be plenty of time (should your instructor advise it) to
simply repeat one or more of your measurements. In that case, discard the defective results and replace them with
the new.
A. Enthalpy Changes—NH4Cl and NaOH Reaction
In this part of the experiment you will compare two different pathways for carrying out the following overall
process:
NH4Cl(s, 2.675 g) + H2O(l, 50 mL) + NaOH(aq, 1.00M, 50 mL) → NH3(aq) + NaCl (aq)
(9-6)
Notice that 2.675 g of NH4Cl contains 0.0500 mol of NH4Cl, and that 50 mL of 1.00M NaOH solution contains
0.0500 mol of NaOH. Therefore, 0.0500 mol of NH3 and 0.0500 mol of NaCl will be formed when the reaction
occurs.
In the first pathway (call it SOLID FIRST), you will first dissolve the NH4Cl in the water, recording any
temperature changes that occur, and then add the NaOH solution to the NH4Cl solution, while continuing to record
any temperature changes occurring. (The resulting plot of temperature versus time should resemble FIGURE 9-2.)
In the second pathway (call it SOLID LAST), you will reverse the order (first adding the NaOH solution to the water
and then dissolving the NH4Cl in the resulting diluted NaOH solution). Note especially that both pathways have
the same starting and ending points. Thus this experiment will provide data allowing you to test the validity
of Hess’s Law.
Setting up the Temperature Probe
1. Press MAIN MENU on the workstation, select TEMPERATURE, then TEMP V TIME. The display should
confirm that you have selected temperature and list options now available to you. Press SETUP, then choose
Experiment 9
9-5
SET LIMITS FOR NEW ACQUISITION. Follow the instructions on the display to set the temperature (y)
limits at 10°C and 40°C and the time (maximum x) limit at 300 seconds. Leave the minimum time setting at
0. Press DISPLAY. The display should now show an empty graph with the values 10 and 40 along the y-axis.
Since we will be investigating changes in temperature, it will not be necessary to calibrate the temperature
probe, as long as it is reading 20-25°C while plugged into the workstation and exposed to the lab atmosphere.
A.1. SOLID FIRST Path
a.
Set up the calorimeter by nesting two Styrofoam cups together. Obtain ~100mL of 1.0 M
NaOH solution and ~200 mL of distilled water in separate beakers.
b.
Use separate graduated cylinders to measure 50.0 mL of distilled water and 50.0 mL of
1.0 M NaOH. Pour the distilled water into the calorimeter. Be sure to use distilled water from
the large containers in the lab, so that its initial temperature will be very close to room
temperature.
c.
Carefully weigh out as close to 2.675 g of NH4Cl as possible, within (0.005 g) using the
top-loading balance.
d.
Press START/STOP to begin collecting data. Place the temperature probe in the solution
in the graduated cylinder containing NaOH solution and begin stirring until the temperature
reading is fairly steady. Record its temperature. Rinse and wipe off the probe quickly and place
it in the calorimeter, where it will now indicate the temperature of the distilled water. Stir with
the temperature probe until the temperature becomes essentially constant, and record the
temperature. Note: Although you will be able to determine the initial temperatures of those liquids
from the plot you will print out, you should also record these values on the report sheet at the end
of the handout, since the values obtained from the MeasureNet workstation screen are more
precise.
e.
Quickly transfer the NH4Cl into the distilled water in the calorimeter and stir. Continue
until the temperature becomes essentially constant. Record the temperature.
f.
Quickly transfer the NaOH solution into the calorimeter and stir. Continue until the
temperature becomes essentially constant. Record the temperature.
g.
Press START/STOP to stop collecting data (or wait until it stops automatically at 300
seconds) and use FILE OPTIONS to print out copies of your data. Label this plot for later
analysis, indicating what is being measured in each portion of the plot and what has been added
to the calorimeter.
h.
Pour the calorimeter contents down the drain, then rinse and dry all equipment.
A.2. SOLID LAST Path
a.–d.
These are exactly the same as in the SOLID FIRST Path.
e.
Quickly add the NaOH solution to the water already in the calorimeter and stir. Continue
until the temperature becomes essentially constant, and then record the temperature.
f.
Quickly transfer the NH4Cl into the solution in the calorimeter and stir. Continue until
the temperature becomes essentially constant, and then record the temperature..
g.–h.
Same as in the SOLID FIRST Path.
Experiment 9
9-6
B. Enthalpy Changes—NaOH and HCl Reaction
In this part of the experiment you will compare two different pathways for carrying out the following process:
NaOH(s, 2.00 g) + H2O(l, 50 mL) + HCl(aq, 1.00M, 50 mL) →
(9-7)
H2O (l) + NaCl(aq)
Notice again that, as in Part B, the amounts have been chosen so that 0.0500 mol each of NaOH and HCl will
react to form 0.0500 mol each of H2O and NaCl.
In the first pathway (SOLID FIRST), you will first dissolve the NaOH in the water, recording the resulting
temperature, and then add the HCl solution to the NaOH solution, again recording the resulting temperature. In
the second pathway, you will reverse the order (SOLID LAST). As in Part A, both pathways have the same starting
and ending points, so this experiment will also provide data for a test of the validity of Hess’s Law.
B.1. SOLID FIRST Path
a.
Set up the calorimeter by nesting two Styrofoam cups together. Obtain ~100 mL of HCl
solution in a clean beaker.
b.
Use separate graduated cylinders to measure 50.0 mL of distilled water and 50.0 mL of
1.0 M HCl. Pour the distilled water into the calorimeter.
c.
Carefully weigh out as close to 2.000g of NaOH as possible, (within 0.005 g) using the
top-loading balance.
d.
Press START/STOP to begin collecting data. Place the temperature probe in the solution
in the graduated cylinder containing HCl solution and begin stirring until the temperature
reading is fairly steady. Record its temperature. Rinse and wipe off the probe quickly and place
it in the calorimeter, where it will now indicate the temperature of the distilled water. Stir with
the temperature probe until the temperature becomes essentially constant and record the
temperature.
e.
Quickly transfer the NaOH into the distilled water in the calorimeter and stir. Continue
until the temperature becomes essentially constant. Record the temperature.
f.
Quickly transfer the HCl solution into the calorimeter and stir. Continue until the
temperature becomes essentially constant. Record the temperature.
g.
Press START/STOP to stop collecting data (or wait until it stops automatically at 300
seconds) and use FILE OPTIONS to print out copies of your data. Label this plot for later
analysis, indicating what is being measured in each portion of the plot and what has been added
to the calorimeter.
h.
Pour the calorimeter contents down the drain, then rinse and dry all equipment.
B.2. SOLID LAST Path
a.–d.
These are exactly the same as in the SOLID FIRST Path.
e.
Quickly add the HCl solution to the water already in the calorimeter and stir. Continue
until the temperature becomes essentially constant.
f.
Quickly transfer the NaOH into the solution in the calorimeter and stir. Continue until the
temperature becomes essentially constant.
g.–h.
Same as in the SOLID FIRST Path.
Experiment 9
9-7
Waste disposal
Combine any leftover NaOH and HCl solutions and pour down the sink.
RESULTS
A. Enthalpy Changes—NH4Cl and NaOH Reaction
Use your printed graphs for the SOLID FIRST and SOLID LAST pathways for this overall process to determine
the enthalpy change for each step you carried out. In each case you need C and ∆T, which are combined in
EQUATION 9-4 to give q, the amount of heat released or absorbed in the step. EQUATION 9-5 then gives the
enthalpy change, ∆H.
Estimate the heat capacity C as described in the TECHNIQUE section. Thus, for example, the heat capacity for
the first step is 50.0 mL × 1.000 g/mL × 4.184 J/(°C g) = 209.2 J/°C. Use this value and your measured value for
∆T to calculate ∆H for this step. Using FIGURE 9-2 as an example, ∆T for the first step is T3 − T2 = −3.75°C.
Therefore ∆H = −q = −(−3.75°C × 209.2 J/°C) = +785 J.
Carry out the analogous calculation to find ∆H for the second step. Be careful here, since the initial
temperatures of the two solutions you are mixing are not the same. It is useful to imagine even this single step as
occurring in two simpler steps. First imagine the two liquids in thermal contact, that is, able to transfer heat, but
not to actually mix and react. Heat would flow from the warmer solution to the cooler solution until they were at
the same temperature. Since the heat capacities of the two solutions are nearly the same, this final temperature
would be the average of the two. (That is, it would be the average of the initial temperature of the NaOH solution
and the initial temperature of the NH4Cl solution, which, of course, is the final temperature of the first step).
From FIGURE 9-2, this average is (T1 + T3) / 2 = (23.51°C + 19.55°C) / 2 = 21.53°C. The change in temperature
caused by the actual reaction occurring upon mixing is the change from this average temperature to the final
temperature after mixing, T4 in FIGURE 9-2. Therefore ∆T = 22.02°C - 21.53°C = +0.49°C. To calculate ∆H for
the second step, remember that the amount of water present is doubled from what it was in the first step, since we
treat the solution added as though it were just water. Calculate the new heat capacity and use your temperature
data to find ∆H for the second step. In this example, ∆H = −0.49°C × 2 × 209.2 J/°C = −205 J.
Analyze your results for both pathways and enter them in the data sheets provided.
B. Enthalpy Changes—NaOH and HCl Reaction
Treat your data from this part of the experiment in exactly the same manner you treated the data in Part B.
Use the measured temperature changes to determine the enthalpy changes for the individual steps, and enter your
results in the data sheets provided.
C. Test of Hess’s Law
Transfer the enthalpy changes for the individual steps from your results in Parts B and C to the blanks provided
in the data sheets for Part D. Use Hess’s Law of Constant Heat Summation (see the INTRODUCTION) to calculate
the overall ∆H for each of the two pathways for each of the two overall processes studied. Calculate the difference
between the two overall ∆H values in each case and enter it into the data sheet. The size of this difference depends
mainly on two things: the validity of Hess’s Law, and the quality of your experimental technique. If Hess’s Law
is valid, the value should be zero, within experimental error.
Experiment 9
9-8
EXPERIMENT 9
REPORT SHEET
Name: _______________________________________
Date:__________
A. ENTHALPY CHANGES – NH4Cl AND NaOH REACTION
SOLID FIRST Pathway
SOLID LAST Pathway
Mass of NH4Cl
______________
_____________
Initial temperature (NaOH(aq))
______________
_____________
Initial temperature (water)
______________
_____________
(water) ______________
(average) _____________
Final temperature
______________
_____________
Temperature change
______________
_____________
Heat capacity
______________
_____________
q
______________
_____________
Enthalpy change
______________
_____________
FIRST STEP:
Initial temperature
SECOND STEP:
Initial temperature
(average) ______________ (diluted solution) ______________
Final temperature
______________
______________
Temperature change
______________
______________
Heat capacity
______________
______________
q
______________
______________
Enthalpy change
______________
______________
Experiment 9
9-9
Experiment 9
9-10
B. ENTHALPY CHANGES – NaOH AND HCl REACTION
SOLID FIRST Pathway
SOLID LAST Pathway
Mass of NaOH
______________
______________
Initial temperature (HCl(aq))
______________
______________
Initial temperature (water)
______________
______________
(water) ______________
(average) ______________
Final temperature
______________
______________
Temperature change
______________
______________
Heat capacity
______________
______________
q
______________
______________
Enthalpy change
______________
______________
FIRST STEP:
Initial temperature
SECOND STEP:
Initial temperature
(average) ______________ (diluted solution) ______________
Final temperature
______________
______________
Temperature change
______________
______________
Heat capacity
______________
______________
q
______________
______________
Enthalpy change
______________
______________
Experiment 9
9-11
EXPERIMENT 9
REPORT SHEET (CONT.)
Name: _______________________________________
Date:__________
C. TEST OF HESS’S LAW
NH4Cl + NaOH Reaction:
SOLID FIRST Pathway
SOLID LAST Pathway
∆H for first step
______________
______________
∆H for second step
______________
______________
Overall ∆H
______________
______________
Difference between the two overall ∆H values
_________________
NaOH + HCl Reaction:
SOLID FIRST Pathway
SOLID LAST Pathway
∆H for first step
______________
______________
∆H for second step
______________
______________
Overall ∆H
______________
______________
Difference between the two overall ∆H values
Experiment 9
_________________
9-12
Notes for Experiment 9
Procedure considerations and helpful hints:
•
1M HCl and 1M NaOH are caustic materials! Wear your goggles! Avoid spilling. If
you spill some on your clothes or skin, rinse with water as soon as possible. If you spill
a large amount on the floor or benchtop, notify your TA.
•
Solid NaOH is very caustic! Do not let it come into contact with your skin! Clean up
any NaOH spills around the balance area!
•
The calculations in this experiment are challenging! You should perform the
calculations for at least one pathway before you leave so that your TA can help make
sure that you are doing them correctly.
•
In order to answer one of the Discussion questions, you will need to look up the heat
of solution values for NaOH and NH4Cl. These values can be found in the CRC
handbook in your lab room. Note the units for these values given at the top of
the table. Alternatively, you can look them up online; one good source is
http://www.mindspring.com/~drwolfe/WWWolfe_dat_enthalpies.htm.
11/10
Experiment 9
9-13