Hess's Law

Aim: To carry out three proves where we for cut guide the phylogenesis of heat and the diversify in heat content by measuring rod the sort in temperature in diametral philias. By doing so, it volition give us a interrupt view of wherefore Hess?s law is important and how it flock be applied. visible and safety assessment plan: Thermos, thermometer with plateful to 0.1 °C, a stirrer, a spoon, measuring glass (200 cm3), and a fixing glass container. The chemicals mandatory for this lab are: Sodium hydroxide in tablet-form, 0.5 seaw in in alles of NaOH and HCl, and an excess 0.25 bulwarkes of HCl. During this lab, safety glasses and aprons must(prenominal) be worn at all times to clog accidents. Avoid meeting with pare down when handling the chemicals. solely the chemicals easily corrode when in contact with skin or clothes and so these need to be handled with care. afterwards the essays, the bases will be neutralized and can be rinsed off in the sink. Method : Three experiments are factor of this lab: analyse A: response: NaOH (s) → NaOH (aq) ΔHAWe poured 200 cm3 of urine supply in the thermos and metric the temperature as precise as we could. then we weighed 2 g of Sodium hydroxide tablets in a glass container and started stirring it with the water. After they were dissolved into the water we heedful the temperature. After this, we reached all the data we needed for figure the ΔHA, and so we did. investigate B: Reaction: NaOH (aq) + HCl → NaCl (aq) + piss (l)ΔHBWe poured century cm3 of 0.5 bomber HCl in the thermos and metrical the temperature. We then poured cytosine cm3 of NaOH and heedful the temperature there as well, and then calculated the think up temperature mingled with both of the substances. Then we mixed the two unitedly and mensural the temperature. Through this, we managed to calculate the ΔHB. Experiment C:Reaction: NaOH (s) + HCl (aq) → NaCl (aq) + urine (l) &# 916;HCWe poured 200 cm3 of 0.25 molar HCl in! the thermos and measured the temperature. After doing so, we added 2 g of the tablet-NaOH to it, and measured the temperature after dissolve these in the HCl. Results:Reaction A: NaOH (s) → NaOH (aq) ΔHAMass of NaOH (in g)2.26 gAmount of substance NaOH (mole)40 g/mol → 0.0565 mol1Change in temperature: Δt2.7 °C2Mass of ancestor (kg): m0.2 kgHeat (kJ) = W (4200 x m x Δt)2268 J = 2.268 W (kJ)3Enthalpy change (kJ/mol NaOH) = ΔHA40141.59292 J/mol ≈ 40.1 kJ/mol4Reaction B: NaOH (aq) + HCl → NaCl (aq) + H2O (l)ΔHBConcentration NaOH (mol/dm3)0.50 mol/dm3Volume NaOH (aq)100 cm3 = 0.1 dm3Amount of substance NaOH (mol)0.05 molTemperature change = Δt0.975 ° C5Mass of root word (kg) = m280 ml = 0.28 l = 0.28 kgHeat (kJ) = W (4200 x m x Δt)1146.6 J = 1.1466 W (kJ)6Enthalpy change (kJ/mol NaOH) = ΔHB22.932 kJ/mol7Reaction C: NaOH (s) + HCl (aq) → NaCl (aq) + H2O (l)ΔHCMass of NaOH (in g)2.11 gAmount of substance Na OH (mol)40 g/mol = 0.05272 mol8Temperature change = Δt2.13 °C9Mass of solution (kg) = m195 ml = 0.195 l = 0.195 kgHeat (kJ) = W (4200 x m x Δt)1744.47 J = 1.74447 W (kJ)10Enthalpy change (kJ/mol NaOH) = ΔHC33.070521 kJ/mol11Conclusion and discussion: After all the experiments had been made, the mother of the lab became clearer; there is a connection between the replys. The stolon and the second reply should together make up the third gear one, although our results do not really support this at all by just looking at the graphical memorialise (above). Both first reactions should ache made about 40 kJ/mol, respectively. Logically, this would mean that the last reaction should be about 80 kJ/mol (40 + 40 = 80 kJ/mol). The three reactions were just un alike ways of reaching the same result: NaCl (aq) + H2O (l).

If everything would wealthy person gone the way it should have, we would have proof of that in our results. Experiment A was dissolving solid NaOH into aqueous NaOH, and the following experiment was turning the result of the previous one into sodium chloride (NaCl) and water, i.e., neutralizing the solution in an acid-base reaction. The discernment why our final results of enthalpy change were wrong, might be because my collaborator and I accidentally measured the temperature a little wrong, which results in making the final deal quite un cerebrateable, as it affects all calculations while getting to it. The reason why reaction C was (supposed to be) the one with the most enthalpy change was because everything that we made happen in two reactions (A and B), we fuck off into one (C), as A + B = C. In reaction C, the NaOH was first solved i nto an aqueous solution, and then neutralized into coarseness and water. As this happens, a whole lot of heat is loosend (as forming bonds judgement of dismissal postal code, while breaking them absorbs it); the reaction is exothermic, just like reaction A and B. This is an explanation for why the heat lift during the transit of the reaction, and that was what we measured when trying to reach the enthalpy change. So, the parity is the energy released in the reactants = energy in product. Appendix:1 2.26 g / 40 g/mol = 0,0565 mol2 23.3°C ? 20.6 °C = 2.7 °C3 4200 x 0.2 kg x 2.7 °C = 2268 J4 2.268 kJ/0.0565 mol = 40,14159292 kJ/mol5 (20.18 °C + 18.03 °C)/2 = 19.105 °CΔt = 20.08 °C ? 19.105 °C = 0,9756 4200 x 0.28 kg x 0.975 °C = 1146,6 J7 1.1466 kJ/0.05 mol = 22,932 kJ/mol8 2.11 g/ 40 g/mol = 0,05275 mol9 22.3 °C ? 20.17 °C = 2.13 °C10 4200 x 2.13 °C x 0.195 kg = 1744,47 J11 1.74447 kJ/ 0.05275 mol = 33,0705 kJ/mol If you want to get a full essay, order it on our website! : BestEssayCheap.com

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