- Level: International Baccalaureate
- Subject: Chemistry
- Word count: 11607
Chemistry thermo lab, Hess's Law.
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Introduction
ï»¿________________ ________________ ________________ ________________ ________________ Hess’s Law Lab ________________ Qusai Al Omari ________________ Introduction: In this lab, we will be determining the change in enthalpy for the combustion reaction of magnesium (Mg) using Hess’s law. Procedure: 1. React about 100 mL of 1.00 M hydrochloric acid with 0.80 g of MgO. Note the change in temperature and any qualitative data. 2. React about 100 mL of 1.00 M hydrochloric acid with 0.50 g of Mg. Note the change in temperature and any qualitative data. Raw Data: Quantitative: Reaction, trial Mass (± 0.01 g) Initial temperature (± 0.1â° C) Final temperature (± 0.1â° C) Volume of HCl (± 0.05 mL) Reaction 1, Trial 1 0.80 22.0 26.9 100.00 Reaction 1, Trial 2 0.80 22.2 26.9 100.00 Reaction 2, Trial 1 0.50 21.6 44.4 100.00 Reaction 2, Trial 2 0.50 21.8 43.8 100.00 Qualitative: 1. Hydrochloric acid is colorless and odorless 2. Magnesium tape is shiny after cleaning it from oxidants, increasing its purity. 3. In both reactions, the solution became bubbly. 4. There was a strong odor from the reaction. Data Processing: Trial 1: Reaction 1: First, we have to calculate the ΔT by subtracting the final temperature by initial temperature: 1. 2. 3. Now we calculate the mass of the solution, assuming it has the density as water: 1. 2. 3. 4. Now, we can use q=mc ΔT to calculate the energy gained by the solution: 1. 2. 3. ...read more.
Middle
1. 2. Our final result is: 1. Mg (s) + 0.5 O2(g) MgO(s) Random error and percent error: We can calculate the random error by just adding the random errors of the component reactions: 1. 2. 3. As for the percent error: 1. 2. 3. Trial 2: Reaction 1: First, we have to calculate the ?T by subtracting the final temperature by initial temperature: 1. 2. Now we calculate the mass of the solution, assuming it has the density as water: 1. 2. 3. Now, we can use q=mc ?T to calculate the energy gained by the solution: 1. 2. 3. Therefore: 1. Now, we have to calculate the number of moles for MgO: 1. 2. 3. We can now calculate the change in enthalpy by dividing the q of the reaction by the moles of the limiting reagent: 1. Now, we do reaction 2, trial 1 so we can use Hess?s law to calculate the change in enthalpy of formation, but first we are going to calculate the uncertainty in this expression: First, we calculate the uncertainty for the: 1. 2. 3. Now for mass: 1. 2. As for the energy gained: 1. 2. Now for the energy of the reaction: 1. It is multiplied by an integer (-1) so it is the same unc. As for the moles: 1. ...read more.
Conclusion
for trial 2. For trial 1, my value got a percent error of 7.14%, which is not that bad considering the weaknesses this lab had that will be discussed in the evaluation. However, in trial 2, I got a better percent error, which is 5.15%, we got a better value because we had a bigger ?H values thus when adding them (since one of them is positive and the other two is negative) we get a smaller value for the enthalpy change of formation thus bringing us closer to the theoretical value. The biggest weakness in this lab was the impurity of the substances, the assumptions that we made about the HCl solution, for example, we assumed that the specific heat capacity of the solution is the same as water, which is an assumption that is not a 100% accurate and affected our ?H values for both reactions and eventually our final ?Hf value. To fix this, In the different range of specific heat capacity values, 4.10 j/g k would have been more appropriate to get closer to our theoretical values, as you get a bigger qrxn values thus bigger ?H values. Another thing that I noticed is that the theoretical value that I got was the ?Standard? enthalpy change of formation. Standard meaning at standard conditions which are at 293 K and 101.3 kPa for pressure. These weren?t the conditions in the lab when I did the experiment. This might alter the experimental value closer to the theoretical value reducing the percent error. ...read more.
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