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Experiment-EnthalpyofDissolution.pdf
Catalyst Education 2020
Enthalpy of Dissolution and Neutralization
Objectives • Use simple calorimetry apparatus to determine the molar enthalpies of three different
reactions. • Understand that in a closed system, the heat of a reaction is equal in magnitude but opposite
in sign to the change in heat of the surroundings. • Directly determine the molar enthalpy of dissolution of NaOH(s) in water and use Hess’s
Law to indirectly determine the same thermodynamic value from two different reactions. Compare molar enthalpy results with literature values using a percent error calculation and compare the molar enthalpy of dissolution as determined by two different methods using a percent difference calculation.
• Use mass and molarity as needed to mathematically determine the limiting reactant for each reaction so enthalpy per mole can be determined.
Introduction This week the enthalpy due to both physical and chemical transformations will be
investigated. The first experiment will involve a physical transformation, the reorganization of
molecules in forming a solution. The resulting enthalpy in kJ/mol will be called the molar enthalpy
of dissolution, DHm,dissolution, since NaOH will be dissolving in water. The second experiment will
determine the molar enthalpy from the chemical neutralization of an acid with a base,
DHm,neutralization, again in kJ/mol.
Enthalpy of Dissolution When a solid is dissolved in a liquid, the solid is usually referred to as the solute and the
liquid is referred to as the solvent. In order for solvation (for the compound to dissolve) to occur,
the solvent molecules must rearrange to allow room for the solute. Similarly, the solute- solute
interactions must be broken apart and allow the solvent to surround individual solute molecules.
Finally, the interaction of the solute-solvent will be different than the solute- solute and solvent-
solvent interactions. All these changes in interactions involve energy, often detected in the form
of heat exchange. If the system gives off heat during these reactions, it is said to be exothermic,
and if the system takes in heat, it is said to be endothermic. If the system is exothermic, it will
transfer heat to the surroundings and the temperature of the surroundings will increase. If the
system is endothermic, it will absorb heat from its surroundings and the temperature of the
surroundings will decrease.
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An example of the dissolving process can be illustrated when solid sodium hydroxide,
NaOH(s), is placed in water (Eq. 1)
NaOH(s) !!"(⎯* Na#(aq) + OH$(aq)
Equation 1
This reaction is an exothermic process so an increase in temperature of the solution is
observed. A coffee cup calorimeter does a fairly good job of isolating the reaction. Assuming no
heat loss through the insulating walls of the calorimeter cup, then, because of conservation of
energy, the sum of the heat released when NaOH dissolves and the heat absorbed by the solution
in the container will be zero (Eq. 2).
q%&''()*+&(, − q'()*+&(, = 0
Equation 2
Rearranging the equation, you can see the heat released when NaOH dissolves (qdissolution)
is equal in magnitude but opposite in sign to the heat gained by the surroundings (qsolution), Eq. 3.
q%&''()*+&(, = −q'()*+&(,
Equation 3
Heat, q, is proportional to the total mass of a system and the change in temperature of a
system. The proportionality constant is the specific heat capacity, c. For our system, we can easily
measure the total mass of the solution, masssol, and the temperature change of the solution, DTsol =
Tfinal - Tinitial, and calculate qsolution using the following equation where csol is the specific heat
capacity of the solution:
qsolution = masssol ∙ csol ∙ ΔT'()
Equation 4
Using Equation 3, qdissolution is:
qdissolution = −(masssol ∙ csol ∙ ΔT'())
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Equation 5
Because heat of reaction, q, is equal to change in enthalpy, ΔH, for a reaction at constant
pressure and since we are doing this experiment on the lab benchtop where room pressure is
constant, then this experiment determines enthalpy, ΔH.
qdissolution = ∆H = −(masssol ∙ csol ∙ ΔT'())
Equation 6
The value obtained from Equation 5 or 6 is dependent on the amount of NaOH. If we divide
by the mole of NaOH, then we have a value independent of amount and can compare it with
literature values for the molar enthalpy of solvation of NaOH, ΔHm,dissolution, where the subscript,
m, indicates this is the enthalpy per one mole. Using the value for ΔH (Eq. 6) and dividing by
number of moles of NaOH, results in an equation for ΔHm,dissolution (Eq. 7).
∆H-,%&''()*+&(, = ∆H%&''()*+&(, 𝑛/0"!
= −(masssol ∙ csol ∙ ΔT'())
𝑛/0"!
Equation 7
Since csol is in units of joule/gramoCelsius (J/goC), the value of DHm,dissolution calculated using
this equation will have units of joule/mole (J/mol). Literature values of molar enthalpy are
frequently given in kJ/mol so convert to kJ/mol and enter this value on your report sheet to compare
with the literature value of - 44.5 kJ/mol.
When two chemical compounds react, they also can give off heat energy or absorb heat
energy. This energy is related to bonds being broken as well as bond formation. In the second part
of this experiment, we will investigate an acid-base reaction between aqueous sodium hydroxide,
NaOH(aq) and hydrochloric acid, HCl(aq) (Eq. 8a). The neutralization equation can be reduced to
the net ionic equation (Eq. 8b).
NaOH(aq) + HCl (aq) → H1O(l) + NaCl(aq)
Equation 8a
OH$(aq) + H#(aq) → H1O(l)
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Equation 8b
Notice that to determine the molar enthalpy of neutralization, ΔHm,neutralization, we must divide
by the mole of product formed, and the resulting value will be independent of amount and
comparable to literature values. Remember that the mole of product formed is determined by the
mole of limiting reactant. In this particular reaction, the mole to mole ratio of either reactant to
product is one to one so nproduct may represent either product.
∆H-,,2*+30)&40+&(, = − (masssol ∙ csol ∙ ΔT'())
𝑛/0"!
Equation 9
In this experiment we will be using 1.00 M solutions of NaOH and HCl, but the same molar
value for enthalpy should be measured regardless of the concentrations used. We also will be using
an excess of one of the reactants, and the identity of the limiting reactant will need to be
determined. Finally note that both the acid and base are already dissolved so no heat should be
produced from the dissolving process.
Using Hess’s Law In the third experiment, solid NaOH(s) will be placed into a solution of HCl(aq) (Eq. 10).
NaOH(s) + HCl (aq) → H1O(l) + NaCl(aq)
Equation 10
∆H-,'5'+2- = − (masssol ∙ csol ∙ ΔT'())
𝑛/0"!
Equation 11
This time, we will call the molar enthalpy, DHm,system, calculated using Eq. 11. If you think
about this reaction (Eq. 10), you will realize that it is a combination of the first two reactions that
you did to determine DHm,dissolution and DHm,neutralization.
In this third portion of the experiment, two processes are occurring. The first is the
dissociation of NaOH(s), and the second is the reaction of the dissolved NaOH(aq) with the
HCl(aq). There is a law that states, “If a reaction can be expressed as the sum of two or more
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reactions, then the enthalpy for the overall reaction is equal to the sum of the enthalpies.” This is
called Hess’s Law. Note that the values obtained from the first two parts of this experiment should
add up to give the value of the third part. Using Eq. 1 and Eq. 8a and recognizing that NaOH(aq)
= Na+(aq) + OH-(aq), we can derive Eq 12.
NaOH(s) !!"(⎯* Na#(aq) + OH$(aq) ∆H-,%&''()*+&(,
Equation 1
NaOH(aq) + HCl (aq) → H1O(l) + NaCl(aq) ∆H-,,2*+30)&40+&(,
Equation 8a
NaOH(aq) + HCl (aq) → H1O(l) + NaCl(aq) ∆H-,'5'+2-
Equation 10
∆H-,'5'+2- = ∆H-,%&''()*+&(, + ∆H-,,2*+30)&40+&(,
Equation 12
Rearranging Eq. 12, you can solve for DHm,dissolution (Eq. 13):
∆H-,%&''()*+&(, = ∆H-,'5'+2- − ∆H-,,2*+30)&40+&(,
Equation 13
This is a good demonstration of Hess’s Law to find the value of the enthalpy of a process.
You have determined DHm,dissolution directly in Part I of the experiment. You have determined
DHm,dissolution indirectly using Hess’s Law in Part III of the experiment. You may compare these
two values of DHm,dissolution to one another using a Percent Difference calculation and you may
compare these two values to the accepted literature value of -44.5 kJ/mol by a Percent Error
calculation. You may also compare the experimental DHm,neutralization to its accepted literature value
of -58 kJ/mol.
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Procedure
Part 1 1. Obtain a coffee cup calorimeter, which consists of a top and bottom made from two
different sizes of Styrofoam cups. Wash, rinse and dry the smaller cup. 2. Tare the mass of the smaller bottom cup on the balance. 3. Remove the cup from the balance to add between 100 to 120 g of deionized water. Make
sure the outside of the cup is dry and return it to the balance to record the mass of water added to the precision of the balance.
4. Place the lid on the system, insert the thermometer, and record the temperature to 0.1oC after 2 minutes of occasional swirling.
5. Measure 3-4 g of solid sodium hydroxide in a weigh boat and record the exact mass on your report form.
6. Remove the lid of the calorimeter, add the sodium hydroxide, and quickly replace the cover.
7. Continuously, stir with the thermometer to ensure mixing and record the highest temperature obtained to the nearest 0.1oC. (This may take 3 to 4 minutes, but time is not the factor that determines the end of the experiment. A temperature that stops increasing and stays steady for 30 seconds is the determining factor so watch the thermometer throughout the experiment.)
8. Check inside the cup to make sure the NaOH completely dissolved and, if not, the trial should be repeated.
9. Discard the solution into the proper waste container and repeat the process, this time using 5-6 g of sodium hydroxide and a fresh sample of deionized water.
Part 2 10. Tare your clean, dry coffee cup calorimeter. 11. Add approximately 60-65g of 1.0 M HCl solution to the calorimeter. 12. Record the mass to the precision of the balance. 13. In a clean graduated cylinder, measure approximately 50 mL of 1.0 M NaOH solution and
record the volume to the nearest 0.1 mL. 14. Measure the temperatures of both of these solutions allowing 2 minutes for the
temperatures to reach equilibrium and record the values to the nearest 0.1oC. (Use the same thermometer for both solutions, and the thermometer should not touch the glass or plastic walls of the graduated cylinder while equilibrating because of the materials’ poor insulating quality.)
15. Quickly, but carefully, pour the NaOH solution into the coffee cup calorimeter containing the HCl solution.
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16. Quickly, place the cover back on the system and stir with the thermometer to ensure mixing until a maximum temperature is reached, being sure to watch the temperature rise throughout the experiment.
17. Record the highest temperature obtained. 18. Discard the liquid into the proper waste container and repeat the process using fresh HCl
solution and approximately 55 mL of NaOH solution.
Calculations for Part 2
When calculating the moles of NaOH and HCl used in this portion of the experiment,
remember that 1.0 M NaOH is NOT solid, pure NaOH. It is mostly water with a little NaOH. You
will need to use values of molarity and volume to calculate moles. The same logic holds for 1.0 M
HCl.
Use the density of a dilute aqueous solution of 1.01 g/mL for calculations converting mass
of solution to volume of solution, or volume of solution to mass of solution. Since the solutions
are not pure water, they will have a specific heat capacity that is different from that of pure water.
The value of csol that you should use for all parts of the experiment is 4.06 J/g0C which is the
Specific Heat Capacity of a dilute aqueous solution. WATCH significant figures during
calculations.
Part 3 19. Into your clean, dry calorimeter, add approximately 60-65g of 1.0 M HCl solution and
record the exact mass of the solution. 20. Place the thermometer in the cup for 2 minutes to record the temperature of the solution. 21. In a weigh boat, measure approximately 1.5g of solid sodium hydroxide and record the
exact mass on your report form. 22. Remove the lid of your calorimeter, add the sodium hydroxide, and quickly replace the
cover. 23. Mix as before and record the highest temperature that is obtained. Make sure the NaOH
dissolved completely. 24. Discard the solution into the proper waste container and repeat the process using fresh
HCl solution and about 2g of solid sodium hydroxide. 25. The coffee cup calorimeters are reusable so after the last trial please wash both bottom
and top gently with soap and a brush and return.
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Enthalpy of Dissolution and Neutralization Report Sheet
Name: Partner:
Instructor: Section Code: Date:
Unless otherwise noted, 1 pt. for each blank
Part 1: DHm,dissolution Trial 1 Trial 2
Mass of Water ___________ ___________
Mass of NaOH ___________ ___________
Total Mass in cup ___________(calculated) ___________(calculated)
Mole of NaOH ___________(calculated) ___________(calculated)
Initial Temperature ___________ ___________
Final Temperature ___________ ___________
Change in Temperature ___________(calculated) ___________(calculated)
DHm,dissolution ___________ ___________(4 pts each)
Average DHm,dissolution ___________
Accepted DHm,dissolution ___________
Percent Error ___________
All calculations to determine DHm,dissolution for Trial 1 (4 pts).
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Part 2: DHm,neutralization Trial 1 Trial 2
Mass of 1.00 M HCl ___________ ___________
Volume of 1.00 M HCl ___________(calculated) ___________(calculated)
Volume of 1.00 M NaOH ___________ ___________
Mass of 1.00 M NaOH ___________(calculated) ___________(calculated)
Mole Product if HCl limiting reactant ___________(calculated) ___________(calculated)
Mole Product if NaOH limiting reactant ___________(calculated) ___________(calculated)
Limiting Reactant ___________ ___________
Total Mass in Calorimeter ___________(calculated) ___________(calculated)
Initial Temperature NaOH ___________ ___________
Initial Temperature HCl ___________ ___________
Average Initial Temperature ___________ ___________
Final Temperature ___________ ___________
Change in Temperature ___________ ___________
DHm,neutralization ___________ ___________(4 pts each)
Average DHm,neutralization ___________
Accepted DHm,neutralization ___________
Percent Error ___________
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All calculations to determine DHm,neutralization for Trial 1 (4 pts).
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Part 3: DHm,system Trial 1 Trial 2
Mass of 1.00 M HCl ___________ ___________
Volume of 1.00 M HCl ___________(calculated) ___________(calculated)
Mass of NaOH (s) ___________ ___________
Total Mass in Calorimeter ___________(calculated) ___________(calculated)
Mole Product if HCl limiting reactant ___________(calculated) ___________(calculated)
Mole Product if NaOH limiting reactant ___________(calculated) ___________(calculated)
Limiting Reactant ___________ ___________
Initial Temperature ___________ ___________
Final Temperature ___________ ___________
Change in Temperature ___________ ___________
DHm,system ___________ ___________(4 pts each)
Average DHm,system ___________
Use Hess’s Law to calculate DHm,dissolution indirectly (Average DHm,system – Average DHm,neutralization)(3 pts).
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Percent Difference: DHm,dissolution (Part 1 and value from box above)(2 pts)
Percent Error: DHm,dissolution (Part 1 and Accepted value)(1 pts)
