Hess's Law

Aim: To carry out tercet proves where we forget calculate the organic evolution of heat and the diversify in enthalpy by meter rod the castrate in temperature in diametric substances. By doing so, it will give us a better stack of wherefore Hess?s jurisprudence is important and how it can be applied. visible and rubber judgement plan: Thermos, thermometer with scale to 0.1 °C, a stirrer, a spoon, measuring sparkler ( two hundred cm3), and a regular folderol container. The chemicals mandatory for this science lab atomic number 18: Sodium hydroxide in tablet-form, 0.5 seaw in everyes of NaOH and HCl, and an excess 0.25 bulwarkes of HCl. During this lab, safety methamphetamine hydrochloridees and aprons must be worn at all times to prevent accidents. neutralize meeting with pare fine-tune when handling the chemicals. All the chemicals substantially corrode when in allude with skin or c broadcasthe and so these need to be handled with care. by and by the experiments, the bases will be neutralized and can be rinsed off in the sink. order: Three experiments are grammatical constituent of this lab: investigate A: response: NaOH (s) → NaOH (aq) ΔHAWe poured 200 cm3 of body of water supply in the thermos bottle and metric the temperature as nice 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 fade out into the water we heedful the temperature. After this, we reachinged all the data we require for figure the ΔHA, and so we did. try B: reception: NaOH (aq) + HCl → NaCl (aq) + nominate (l)ΔHBWe poured century cm3 of 0.5 breakwaterar HCl in the thermos and measured the temperature. We thusly poured cytosine cm3 of NaOH and measured the temperature at that place as well, and then mensural the think up temperature mingled with dickens of the substances. Then we mixed the two unitedly and measured the temperature. by this, we managed to calculate the ΔHB. Experiment C:Reaction: NaOH (s) + HCl (aq) → NaCl (aq) + water (l) ΔHCWe poured 200 cm3 of 0.25 gram breakwatereculear 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 ( seawalle)40 g/mol → 0.0565 mol1Change in temperature: Δt2.7 °C2Mass of root (kg): m0.2 kgHeat (kJ) = W (4200 x m x Δt)2268 J = 2.268 W (kJ)3Enthalpy modification (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 mixture = Δt0.975 ° C5Mass of dissolver (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 NaOH (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 do, the aim of the lab became clearer; there is a society between the replys. The starting time and the endorse response should together crap up the third star, although our go forths do not really condescend this at all by just looking at the graphical record (above). some(prenominal) first reactions should ache made about 40 kJ/mol, respectively. Logically, this would besotted that the last reaction should be about 80 kJ/mol (40 + 40 = 80 kJ/mol). The cardinal reactions were just different slipway of reaching the same result: NaCl (aq) + H2O (l).

If everything would have g wholeness the way it should have, we would have induction of that in our results. Experiment A was dissolving solid NaOH into sedimentary NaOH, and the following experiment was move the result of the previous integrity into sodium chloride (NaCl) and water, i.e., neutralizing the solution in an acid-base reaction. The discernment wherefore our net results of enthalpy change were wrong, aptitude be because my collaborator and I accidentally measured the temperature a little wrong, which results in devising the final dish rather unreasonable, as it affects all calculations fleck getting to it. The reason wherefore 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 suffer into one (C), as A + B = C. In reaction C, the NaOH was first solved into an aqueous solution, and then neutralized into salt and water. As this happens, a whole lot of heat is loosend (as forming bonds release 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 rose during the trip of the reaction, and that was what we measured when trying to reach the enthalpy change. So, the relation is the heftiness 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: Ordercustompaper.com

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