Chemistry thermo lab, Hesss Law Essay Example for Free

Chemistry thermo lab, Hesss Law Essay 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 ( ± 0.1à ¢Ã‚ ° C) Final temperature ( ± 0.1à ¢Ã‚ ° 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. 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. 2. Finally, the change in enthalpy: 1. 2. 3. Reaction 2: First, we have to calculate the ΔT 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 ΔT to calculate the energy gained by the solution: 1. 2. Therefore: 1. Now, we have to calculate the number of moles for MgO: 1. 2. We can now calculate the change in enthalpy by dividing the q of the reaction by the moles of the limiting reagent: 1. I will now calculate the uncertainties: First, we calculate the uncertainty for the: 1. 2. 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. 2. Finally, the change in enthalpy: 1. 2. 3. Now, we use Hess’s law to calculate the change of enthalpy of formation: 1. MgO(s) + 2HCl(aq) MgCl2(aq) + H2O(l) 2. Mg (s) + 2HCl(aq) MgCl2(aq) + H2 (g) 3. H2(g) + 0.5 O2(g) H2O(l) (given) By reversing reaction number 1, we can get our targeted reaction: Mg (s) + 0.5 O2(g) MgO(s) Now to calculate the change of enthalpy, which will be the change of enthalpy of formation? 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 ΔT 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 ΔT 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. 2. Finally, the change in enthalpy: 1. 2. 3. Reaction 2: First, we have to calculate the ΔT 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 ΔT to calculate the energy gained by the solution: 1. 2. Therefore: 1. Now, we have to calculate the number of moles for MgO: 1. 2. We can now calculate the change in enthalpy by dividing the q of the reaction by the moles of the limiting reagent: 1. I will now calculate the uncertainties: First, we calculate the uncertainty for the: 1. 2. 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. 2. Finally, the change in enthalpy: 1. 2. 3. Now to calculate the change of enthalpy, which will be the change of enthalpy of formation: 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. Processed data: Trial 1 Trial 2 of reaction 1 -104 kJ/mol ( ± 2.10%) -99 kJ/mol ( ± 2.19%) of reaction 2 -463 kJ/mol ( ± 0.509%) -446 kJ/mol ( ± 0.525%) of MgO -645 kJ/mol ( ± 2.61%) -633 kJ/mol ( ± 2.72%) Conclusion and Evaluation: In this lab, we determined the standard enthalpy change of formation of MgO using Hess’s law. First, we reacted HCl with MgO for the first reaction and got -104 kJ/mol ( ± 2.10%) for trial 1 and -99 kJ/mol ( ± 2.19%) for trial 2. As for reaction 2, where you react, I got -463 kJ/mol ( ± 0.509%) for trial 1 and -446 kJ/mol ( ± 0.525%) for trial 2. When we use Hess’s Law, we have to reverse reaction 1 to get the targeted equation, Mg (s) + 0.5 O2(g) MgO(s), and we get an enthalpy change value of -645 kJ/mol ( ± 2.61%) for trial 1, and -633 kJ/mol ( ± 2.72%) 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 ΔH 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 ΔH values for both reactions and eventually our final ΔHf 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 ΔH values. Another thing that I noticed is that the theoretical value that I got was the â€Å"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.