Chemistry thermo lab, Hesss Law Essay Example for Free

Chemistry thermo lab, Hesss Law EssayIntroductionIn this lab, we will be determining the transmit in enthalpy for the combustion reaction of magnesium (Mg) using Hesss law. action1. React about 100 mL of 1.00 M hydrochloric acid with 0.80 g of MgO. annotation the revision in temperature and any qualitative information.2. React about 100 mL of 1.00 M hydrochloric acid with 0.50 g of Mg. Note the assortment in temperature and any qualitative data.Raw DataQuantitativeReaction, trialMass ( 0.01 g)initial temperature( 0.1 C)Final temperature( 0.1 C)Volume of HCl( 0.05 mL)Reaction 1, visitation 10.8022.026.9100.00Reaction 1, Trial 20.8022.226.9100.00Reaction 2, Trial 10.5021.644.4100.00Reaction 2, Trial 20.5021.843.8100.00Qualitative1. Hydrochloric acid is colorless and odorless2. Magnesium put down is shiny after cleaning it from oxidants, increasing its purity.3. In both reactions, the solution became bubbly.4. There was a absolute odor from the reaction.Data ProcessingTrial 1Re action 1First, we have to calculate the T by subtracting the lowest temperature by initial temperature1. 2. 3. Now we calculate the mass of the solution, assuming it has the density as water1. 2. 3. 4. Now, we can exercising q=mc T to calculate the aptitude gained by the solution1. 2. 3. therefrom1. Now, we have to calculate the takings of breakwateres for MgO1. 2. 3.We can now calculate the alteration in enthalpy by dividing the q of the reaction by the moles of the alteration reagent1. Now, we do reaction 2, trial 1 so we can aim Hesss law to calculate the salmagundi in enthalpy of composition, but first we are termination to calculate the uncertainty in this expressionFirst, we calculate the uncertainty for the1. 2. 3. Now for mass1. 2. As for the aught gained1. 2. Now for the sinew of the reaction1. It is multiplied by an integer (-1) so it is the same unc.As for the moles1. 2. Finally, the diverge in enthalpy1. 2. 3.Reaction 2First, we have to calculate the T by subtracting the final temperature by initial temperature1. 2. Now we calculate the mass of the solution, assuming it has the density as water1. 2. 3. Now, we can subroutine q=mc T to calculate the capacity gained by the solution1. 2. thereof1. Now, we have to calculate the get of moles for MgO1. 2.We can now calculate the adjustment in enthalpy by dividing the q of the reaction by the moles of the moderate reagent1. I will now calculate the uncertaintiesFirst, we calculate the uncertainty for the1. 2. Now for mass1. 2. As for the energy gained1. 2. Now for the energy of the reaction1. It is multiplied by an integer (-1) so it is the same unc.As for the moles1. 2. Finally, the change in enthalpy1. 2. 3. Now, we mapping Hesss law to calculate the change of enthalpy of formation1. 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 reactionMg (s) + 0.5 O2(g) MgO(s)N ow to calculate the change of enthalpy, which will be the change of enthalpy of formation?1. 2. Our final result is1. Mg (s) + 0.5 O2(g) MgO(s)Random faulting and percent fallacyWe can calculate the hit-or-miss error by just adding the random errors of the component reactions1. 2. 3. As for the percent error1. 2. 3.Trial 2Reaction 1First, we have to calculate the T by subtracting the final temperature by initial temperature1. 2. Now we calculate the mass of the solution, assuming it has the density as water1. 2. 3. Now, we can procedure q=mc T to calculate the energy gained by the solution1. 2. 3. Therefore1. Now, we have to calculate the number of moles for MgO1. 2. 3.We can now calculate the change in enthalpy by dividing the q of the reaction by the moles of the limiting reagent1. Now, we do reaction 2, trial 1 so we can use Hesss law to calculate the change in enthalpy of formation, but first we are sledding to calculate the uncertainty in this expressionFirst, we calculat e the uncertainty for the1. 2. 3. Now for mass1. 2. As for the energy gained1. 2. Now for the energy of the reaction1. It is multiplied by an integer (-1) so it is the same unc.As for the moles1. 2. Finally, the change in enthalpy1. 2. 3. Reaction 2First, we have to calculate the T by subtracting the final temperature by initial temperature1. 2. Now we calculate the mass of the solution, assuming it has the density as water1. 2. 3. Now, we can use q=mc T to calculate the energy gained by the solution1. 2. Therefore1. Now, we have to calculate the number of moles for MgO1. 2.We can now calculate the change in enthalpy by dividing the q of the reaction by the moles of the limiting reagent1. I will now calculate the uncertaintiesFirst, we calculate the uncertainty for the1. 2. Now for mass1. 2. As for the energy gained1. 2. Now for the energy of the reaction1. It is multiplied by an integer (-1) so it is the same unc.As for the moles1. 2. Finally, the change in enthalpy1. 2. 3.Now to c alculate the change of enthalpy, which will be the change of enthalpy of formation1. 2. Our final result is1. Mg (s) + 0.5 O2(g) MgO(s)Random error and percent errorWe can calculate the random error by just adding the random errors of the component reactions1. 2. 3. As for the percent error1. 2. 3. Processed dataTrial 1Trial 2of 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 EvaluationIn this lab, we determined the standard enthalpy change of formation of MgO using Hesss 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 Hesss 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 revalue of -645 kJ/mol ( 2.61%) for trial 1, and -633 kJ/mol ( 2.72%) for trial2. 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 speculative value.The biggest weakness in this lab was the slag of the substances, the arrogances 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 appropr iate 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 werent 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.