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CHEM.1230L: Chemistry I Lab



Whenever a chemical reaction takes place, there is a change in the energy of the system. For many reactions, this energy change takes the form of a transfer of heat energy either into (endothermic) or out of (exothermic) the system. In this experiment, you determine the heat flows for three processes: the reaction of an acid (HCl) with a base (NaOH); the reaction of an active metal (Mg) with an acid (HCl); and the dissolving of a salt (KBr).

The write up for this experiment is very long, and may seem confusing. The steps in the calculations are very systematic and straightforward, however, and are based in sequence on the data you have recorded.

Temperature changes (DT's) are determined from the graphs of your data. These temperature changes are needed at many places in the calculations, so it is important that you make your graphs first before attempting the calculations.

Watch the Pre-Lab video for this experiment*. (7 Mb file: Broadband connections only; Real Player required)

*this video will need to be re-linked. Link from original document does not work.

Experiment 9: Heats of Reaction & Hess's Law, Part A

Part A: Neutralization of HCl and NaOH
Data: Page 89, Part IA


Concentration of Reagents




0.950 M

0.981 M














4.00 (mix)

mix solutions

mix solutions



same (mixed)



same (mixed)



same (mixed)

Based on your data, you prepare a graph for Trial 1 (example shown on Page 53), and determine from your graph the change in temperature during the reaction (DT). For the data above, DT would be 7.1oC for my Trial 1. 

Calculations, Page 91, Part IIA
1. Temperature from graphs

This is the information you read off your graphs for the 4.00 minute point where the solutions were mixed. You need the extrapolated (from the graph) temperatures of the separate solutions as well as the extrapolated (from the graph) temperature of the mixture.

These extrapolations allow for the fact that the solutions before mixing were very gradually warming up with time, whereas the combination after mixing was very gradually cooling off with time. The extrapolations try to estimate what the temperatures would have been at 4.00 minutes when the solutions were actually mixed.



Temperature from Graphs

Trial 1

Trial 2 (no data shown)

NaOH and HCl at mixing


21.2 (no data shown above)

Reaction mixture at mixing


28.2 (no data shown above)


For Trial 1, DT is 7.1oC. For Trial 2, DT is 7.0oC. 

2. Heat Change of Mixed Solutions (Trial 1)

For a transfer of heat energy to or from a material, the quantity of heat transferred is dependent on the mass of material, the temperature change of the material, and the specific heat capacity of the material (a constant).

Since we measured out the solutions used in this part of the experiment by volume (using a graduated cylinder), we will have to convert from the volume of solution taken to the mass of the solution, using the density of the solution (1.02 g/mL) given on Page 54. The specific heat capacity is also given on Page 54 as 3.93 Joules/goC

For Trial 1, calculate the Heat Change for the Mixed Solutions, then click here to check your answer. 

3. Heat Change of Calorimeter

It is noted in the discussion for the experiment that no calorimeter is completely inert as regarding its ability to transfer heat to and from itself. Although plastic foam is a good insulator, it is not a perfect insulator. When the reaction occurs in the calorimeter, a small portion of the heat generated by the reaction is absorbed by the calorimeter.

The amount of heat absorbed by a calorimeter is a function of the calorimeter itself and is called the heat capacity of the calorimeter, Ccal. We have previously measured the heat capacities of plastic foam calorimeters such as the one you used in this experiment and have found the average value of such heat capacities to be 41.7 Joules/oC. This number means that anytime a reaction is performed in such a calorimeter, for each degree increase in the temperature of the contents of the calorimeter, the calorimeter itself absorbs 41.7 Joules. For Trial 1 of my data, in which the temperature of the reaction mixture increased by 7.1 oC, calculate the quantity of heat absorbed by the Calorimeter, then click here to check your result. 

4. Heat of Reaction

The heat of reaction represents the total quantity of heat energy transferred by the reaction. This energy occurred in two parts: the heat change that we were able to measure for the solutions, as well as the heat energy that was lost to warming up the calorimeter itself by 7.1 oC. The equation given in this part on Page 59 of the lab manual takes some explanation. The equation reads

Heat of reaction + Heat Change of Solutions + Ccal(DTcal) = 0

This is nothing more than a mathematical statement of the First Law of Thermodynamics: during any process, the total energy of the universe is constant. In other words, whatever changes in energy occurred in the chemicals themselves, in the solutions, and in the calorimeter must total to a net change of zero. Whatever amount of heat energy is given off by the actual chemicals when they react with each other is exactly equal to the total amount of heat energy transferred to the water in which the chemicals were dissolved and to the calorimeter. Rearranging the equation above to solve for the Heat of Reaction gives the following:

Heat of Reaction = 0 - [Heat Change of Solutions + Ccal(DTcal)]



This rearranged equation says that the Heat of Reaction is just the negative of the sum of the heat change of the solutions and the heat change of the calorimeter.

                                    Heat of Reaction = -[2.8 kJ + 0.30 kJ] = -[3.1 kJ] = -3.1 kJ

5. Moles of Limiting Reactant

HCl and NaOH react on a 1:1 stoichiometric basis

HCl + NaOH ® NaCl + H2O

We took 50.0 mL of 0.950 M NaOH and combined it with 50.0 mL of 0.981 M HCl. The reactant which is present in the lesser amount on a mole/stoichiometric basis is the limiting reactant. For our experiment, the NaOH is the limiting reactant. Some of the HCl is present in excess and does not react. For the data above, calculate the number of moles of NaOH used, then click here to check your result. 


6. Molar Heat of Neutralization (per mole of water formed)

The heat of reaction calculated in Part 4 above (-3.1 kJ) was for the specific amounts of materials used in the experiment. Heats of Reaction are most commonly tabulated in handbooks in terms of the Molar heat of reaction: the heat of reaction when 1 mole of a particular reactant or product is involved.

We want here to calculate the Molar Heat of Reaction per mole of water formed in the reaction. In Part 5 above, we calculated that the limiting reactant was the NaOH used. Since 0.0475 moles of NaOH reacted completely, from the balanced chemical equation, 0.0475 moles of water must have been produced. For the data above for Trial 1, calculate the Molar heat of reaction (per mole of water formed), and then click here to check your answer.