Chemistry thermo lab, Hess’s Law Essay

IntroductionIn this lab, we will be determining the switch over in heat content for the burning answer of magnesium (Mg) using Hesss fairness.Procedure1. React approximately 100 mL of 1.00 M hydrochloric acid with 0.80 g of MgO. Note the interpolate in temperature and each qualitative data.2. React about 100 mL of 1.00 M hydrochloric acid with 0.50 g of Mg. Note the modify in temperature and any qualitative data.Raw DataQuantitativeReaction, streakMass ( 0.01 g) sign temperature( 0.1 C)Final temperature( 0.1 C)Volume of HCl( 0.05 mL)Reaction 1, streak 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 tape is showy after cleaning it from oxidants, increasing its purity.3. In both chemical chemical receptions, the solution became bubbly.4. in that location was a strong odor from the chemical reaction.Data ProcessingTrial 1Re action 1First, we have to await the T by subtracting the nett temperature by sign temperature1. 2. 3. at one time we play the plenteousness of the solution, assuming it has the engrossment as piss1. 2. 3. 4. Now, we passel uptake q=mc T to manoeuvre the sinew gained by the solution1. 2. 3. Therefore1. Now, we have to depend the government issue of jettyes for MgO1. 2. 3.We flush toilet directly auspicate the change in total heat by dividing the q of the reaction by the gram moleculees 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 going to calculate the scruple in this expressionFirst, we calculate 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 akin unc.As for the moles1. 2. Finally, the change in enthalpy1. 2. 3.Reaction 2First, we have to calc ulate the T by subtracting the final examination temperature by sign temperature1. 2. Now we calculate the mass of the solution, assuming it has the tightfistedness as urine1. 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 takings of moles for MgO1. 2.We can promptly 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, we use 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 reactionM g (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 is1. Mg (s) + 0.5 O2(g) MgO(s)random faulting and percent mistakeWe can calculate the random error by just adding the random errors of the destiny 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 use 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 going to calculate the uncertainty in this expressionFirst, we ca lculate 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.No w to calculate 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 model 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 rate o f -645 kJ/mol ( 2.61%) for trial 1, and -633 kJ/mol ( 2.72%) for trial2. For trial 1, my lever 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 value thus when adding them (since one of them is verificatory and the other two is negative) we get a smaller value for the enthalpy change of formation thus bringing us closer to the abstractive value.The biggest weakness in this lab was the impurity of the substances, the assertions that we made about the HCl solution, for example, we untrue 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 contrary range of specific heat capacity values, 4.10 j/g k would have been mor e 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 stock 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.