Hess Law in Heat Reactions of Sodium Solution and Hydrochloric Acid

Research Question: How can the heat of reaction of two separate equations be combined using Hess’s Law to determine the hear of reaction of sodium hydroxide solution with dilute hydrochloric acid?

Introduction: Hess’s Law has been the basis of thermochemistry for the past century. It states that the total enthalpy change of a reaction is independent of whether the reaction takes place in one or several steps. Using Hess’s Law, heats of reactions for individual steps can be algebraically combined the same as the reactions can be combined to form the overall reaction, and therefore the over all heat of reaction. This law is extremely important in many chemical evaluations, because often times it is hard, or impossible to determine the heat of reaction in one step. The chemical reaction must be broken down into simpler reactions, which can be measured directly.

The first and second reactions are dissociation reactions, and therefore the heat released is called the heat of dissociation. The third reaction, which is the overall reaction, is a neutralization reaction between an acid and a base, and therefore the overall heat released is called the heat of neutralization. Reactions 1 and 2 will then be combined to form reaction 3, their heats of dissociation being manipulated the same way in order to form the heat of neutralization of reaction 3.

Hypothesis: Looking at the three equations under consideration, my prediction is that if the opposite of the heat released from reaction 1 is combined with the heat released from reaction 2, then the overall heat of neutralization (reaction 3) of NaOH and HCl would be formed.

Variables: In this experiment, the variables I will be testing with to apply Hess’s Law will be solid sodium hydroxide, and diluted hydrochloric acid solution. These independent variables, when combined in the correct chemical, and eventual algebraic form, should be able to combine to determine the overall enthalpy of a larger reaction. This overall enthalpy acts as a dependent variable, while the individual enthalpies being manipulated act as independent variables. These variables will be tested in a styrofoam calorimeter in attempt to control the heat being released by each of the reactions.

Materials and Equipment:

• 2 styrofoam calorimeters
• 100 mL graduated cylinder
• Solid sodium hydroxide (NaOH)
• .25 M hydrochloric acid solution (HCl)
• Thermometers
• Digital Scale
• Glass stirring rod
• 200 mL beaker

Procedure:

Reaction 1:

1. Put 100 mL of room temperature distilled water into the beaker, and place it inside the calorimeter. Record the temperature of the water when it has reached a constant temperature.
2. Measure and record the mass of around 2.5 grams of solid sodium hydroxide, and place it into the water in the calorimeter.
3. Continuously stir and simultaneously watch the temperature of the solution until the solid is completely dissolved. Record the highest temperature reached during this process.
4. Discard the solution safely and rinse the beaker with water.
5. Repeat this process for three trials in order to obtain a mean value for the reaction’s enthalpy.

Reaction 2:

1. Repeat steps 1-4, but instead of 100 mL distilled water, use 100 mL 2.5 M hydrochloric acid.
2. Repeat step 6 for three trials in order to obtain a mean value for the reaction’s enthalpy.

For both reactions, the same equation to find the heat released from the reaction will be used: , where m is equal to the mass of the substance (g), C is the specific heat of water (J/gK), and is the change in temperature (K). The specific heat of water is assumed to be 4.184 J/gK.

Reaction 1:

Calculation of the average temperature change for reaction 1 using all three trails and their uncertainties:

Trial 1) Highest possible temperature change: 28.5 – 21.5 = 7 oC

Lowest possible temperature change: 27.5 – 22.5 = 5oC

Trial 2) Highest possible temperature change: 27.5 – 21.5 = 6 oC

Lowest possible temperature change: 26.5 – 22.5 = 4 oC

Trial 3) Highest possible temperature change: 27.5 – 21.5 = 6 oC

Lowest possible temperature change: 26.5 – 22.5 = 4 oC

Averages) Highest temperature change: oC

Lowest temperature change: oC

Average temperature change: oC

The temperature change will need to be in Kelvin for the calculation of heat released:

K = oC + 273.15, therefore, the average temperature change in K is:

278.45K

Calculation of heat released by reaction 1 using all three trials and their uncertainties:

Trial 1) Highest possible: J

Lowest possible: J

Trial 2) Highest possible: J

Lowest possible: J

Trial 3) Highest possible: J

Lowest possible: J

Averages) Highest: J

Lowest: J

Average: 2924.24 J

From these calculations, the average heat gained by the calorimeter (J) from reaction 1, the dissociation of NaOH(s) in water, is 2924.24 J. Therefore, the average heat released by the system is 2924.24 J and = -2924.24 J.

Reaction 2:

Calculation of the average temperature change for reaction 1 using all three trails and their uncertainties:

Trial 1) Highest possible temperature change: 31.5 – 21.5 = 10 oC

Lowest possible temperature change: 30.5 – 22.5 = 8 oC

Trial 2) Highest possible temperature change: 30.5 – 21.5 = 9 oC

Lowest possible temperature change: 29.5 – 22.5 = 7 oC

Trial 3) Highest possible temperature change: 30.5 – 21.5 = 9 oC

Lowest possible temperature change: 29.5 – 22.5 = 7 oC

Averages) Highest temperature change: oC

Lowest temperature change: oC

Average temperature change: oC

The temperature change will need to be in Kelvin for the calculation of heat released:

K = oC + 273.15, therefore, the average temperature change in K is:

281.45K

Calculation of heat released by reaction 1 using all three trials and their uncertainties:

Trial 1) Highest possible: J

Lowest possible: J

Trial 2) Highest possible: J

Lowest possible: J

Trial 3) Highest possible: J

Lowest possible: J

Averages) Highest: J

Lowest: J

Average: 2963.59J

From these calculations, the average heat gained by the calorimeter (J) from reaction 1, the dissociation of NaOH(s) in HCl(aq), is 2963.59J. Therefore, the average heat released by the system is 2963.59J and = -2963.59J.

Hess’s Law:

Now that the of both constituent reactions has been calculated, its must be manipulated using Hess’s Law to find the heat of neutralization of HCl(aq) and NaOH(aq).

1. NaOH(s) Na+(aq) + OH-(aq) = -2924.24 J
2. NaOH(s) + H+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + H2O = -2963.59J
3. Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + H2O = ?

If reaction number one is flipped to the reverse reaction, then the products and reactants of 1 and 2 will cancel eachother out algebraically to form the net ionic equation of reaction 3. The for reaction 1 is then multiplied by -1 because it is the reciprocal reaction.

1. Na+(aq) + OH-(aq) NaOH(s) = 2924.24 J
2. NaOH(s) + H+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + H2O = -2963.59J
3. Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) Na+(aq) + Cl-(aq) + H2O = ?

Now, for reaction 3 can be determined by adding together the from reactions 1 and 2. -35.39 J

The accepted value for the heat of neutralization between acids and bases is -44.51 J, so this can be used to calculate the percent error of the experiment.

19.3% error

Before the investigation, my hypothesis was that the first two reactions could be combined by adding the opposite of reaction 1 to reaction 2 in order to obtain reaction 3. I also assumed, based off of Hess’s Law, that the enthalpy changes for these respective reactions could be combined the same way to find the over all enthalpy change for the neutralization reaction of HCl(aq) and NaOH(aq). My hypothesis proved to be correct because when flipping reaction 1 and adding it to reaction 2, I obtained the products of reaction 3, and therefore the enthalpy change of reaction 3, without actually having to perform reaction 3 itself. There was, however, a 19.3% error in my experiment, as I obtained an enthalpy change of -39.53 J and the accepted value is

-44.51 J. I assumed this showed a simple lack of control on the escaping heat from the calorimeters that I used. The holes at the tops of the calorimeters that the thermometers and stirring rods went in were hand made, and could have allowed heat to escape, therefore skewing the experimental data. Because of the high precision, and mediocre accuracy, this error is assumed to be a systematic error, not random error.

One of the main limitations of this experiment is that I did not test reaction 3 itself, which could have helped prove that the error was systematic and not random. If I would have had a total value to compare my Hess’s Law value with, in addition to the accepted value then my experiment could have been more validated. If I were to perform the experiment again, I would react aqueous hydrochloric acid with aqueous sodium hydroxide in order to heighten this very validity. Another way that I would look to improve this experiment is in the calorimeters that I used. If I could obtain a calorimeter that already has a thermometer hole in it, rather than me having to make a hole, then I think the variables would have been more controlled, and the results less skewed.