# Thermochemistry Essay

2435 Words Nov 27th, 2010 10 Pages
Thermochemistry Lab
Purpose: This lab taught procedures for determining heat of capacity of a calorimeter and measuring enthalpy of change for three reactions. It also enforced methods of analyzing data obtained through experimentation and calculating enthalpy. These procedures are used in the branch of thermodynamics known as thermochemistry which is the study of energy changes that accompany chemical reactions. Concepts from this lab can be used to determine the potential energy of a chemical reaction. Much of the energy people depend on comes from chemical reactions. For example, energy can be obtained by burning fuel, metabolizing of food or discharging a batter.
Materials and Methods: All work was done with coffee cup
…show more content…
We weighed and recorded the mass of an Erlenmeyer flask. We added NaOH(s) to the Erlenmeyer flask until we had about 2.00g of NaOH(s). We measured 55.0mL of 1 M HCl into a 100mL graduated cylinder and diluted it to 100mL with deionized water. We measured and recorded the mass of the empty calorimeter; we added the HCl to the calorimeter and recorded the mass of the calorimeter with the HCl. We placed the magnetic stirrer and a temperature probe in the calorimeter, and then placed the lid on the calorimeter. We turned on the magnetic stirrer and set it to a slow stirring rate. We recorded temperature data for one minute before beginning the experiment. We added the NaOH(s) to the calorimeter and recorded temperature data for 3 minutes after mixing. This experiment was repeated one time.

Calculations:
Part 1: Determining heat capacity of the calorimeter:

| Mass/g | Temperature | Change in | | | Initial | Final | Temperature | Run 1 | Chilled | 50.963 | 9.6OC | 34.2OC | 24.6OC | | Hot | 50.117 | 54.4OC | 34.2OC | 20.2OC | Run 4 | Chilled | 48.414 | 8.3OC | 30.9OC | 22.6OC | | Hot | 49.918 | 54.0OC | 30.9OC | 23.1OC |

Heat Lost by Heated Water qH=(Cp)(MH)(TH) Run 1 qH=(4.184J/ OC-g)(50.112g)(20.2OC)=4235.31J
Run 4 qH=(4.184J/ OC-g)(49.918g)(23.1OC)=4824.59J

Heat Gained by Chilled Water qC=(Cp)(MC)(TH) Run 1 qC=(4.184J/ OC-g)(50.963g)(24.6OC)=5245.44J
Run

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