preview

Calorimetry Lab Report Summary

Better Essays
Open Document

Summary Questions:
1. What value(s) of ΔHrxn did you determine? Compare to the literature value. The first trial that contained 10 mL of NaOH and 10 mL of CH3COOH produced a ΔHrxn of -13.11 kJ/m. The second trial, which contained 15 mL of NaOH and 5 mL of CH3COOH resulted in a ΔHrxn of -7.864 kJ/m. Based on the literature value of NaOH CH3COOH, -57.5 kJ/m, we were off by 44.39 kJ/m in the first trial, and 49.64 kJ/m in the second trial.

2. How did concentration and/or volume differences affect the heat change (q) for each trial? When the volumes of NaOH and CH3COOH were equal, the temperature increased by 5 degrees celsius. When we performed a second trial and added 15 mL of NaOH and 5 mL of CH3COOH, the temperature only showed an increase …show more content…

The first part of this lab’s objective was to find the calorimeter constant using DI water. We accomplished this by first checking the temperature and then adding 20 mL of cold DI water into our calorimeter. Next we collected 20 more mL of DI water in a 50 mL beaker and placed it onto a burner in order to heat it. We removed the beaker once the temperature of the water reached 60º celsius. After we removed the water, we poured it into the calorimeter with the cold water and took the temperature. The temperature ended up being 37º celsius. From this information we were able to calculate for qhot, qcold, qcal and Ccal. To be as accurate as possible we conducted this same test three more times and used the averages from all Ccal calculations as the final Ccal. After finding the Ccal of our calorimeter, we were able to integrate the acid and base pair CH3COOH and NaOH. In our first trial we added 10 mL of both CH3COOH and NaOH. Before mixing the two together in our calorimeter, we took the initial temperature. After mixing the two solutions together we took the final temperature. From this point we were able to calculate the qsoln, qcal, qrxn, and ΔHrxn. We conducted a second trial using 5 mL of CH3COOH and 15 mL of NaOH and performed calculations to determine ΔHrxn. …show more content…

We were also able to conclude that when you combine equal parts of an acid and base you can expect to see a higher temperature change than you would if you added more parts of a base than a acid. Our reasoning behind this conclusion is that when we added 10 mL of both solutions that temperature increased by 5ºc, but when we added 15 mL of NaOH and only 5 mL of CH3COOH there was only a temperature increase of 3ºc. In correlation, we calculated that when you have equal parts of acid and base, the ΔHrxn will be higher than when you use more parts base than

Get Access