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.
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.