Objective To determine of arouse ups from three exothermic fightions and examine the Hess Law Summary Based on this audition, we have strike three related exothermic receptions involving atomic number 11 hydroxide. The first reception (Part A), steadfast atomic number 11 hydroxide give uncouple into body of water. The heat piss by this answer (?H1) and this called as the heat of resolution of upstanding NaOH. From the test we have managed to intractable that ?H1 = -41.84 KJ/mol In the turn reaction (Part B), an aqueous solution of NaOH is allowed to react with an aqueous solution of HCl. This is a neutralization reaction in the midst of a strong acid and strong base. Therefore the heat of reaction (?H2) is called as the heat of neutralization of HCl and NaOH solutions. The ?H2 calculated from this experiment is -6.6944KJ/mol. This is because the atomic number 1 changes when 1 mol of H+ ions from an acid (HCl) reacts with one mole of OH- from an alkali (NaOH) to form one mole of water molecules low the stated conditions of the experiment. In the final reaction of the experiment (Part C), inviolable NaOH willing react with an aqueous solution of HCl. This reaction is in addition the combination of the first and two reactions. The solid NaOH will dissociate into its ions as it dissolves in the acid solution which is thence neutralized by the acid solution.
Thus the heat of the reaction (?H3) is verbalise be equal to (?H1+?H2). It is called the heat of solution of solid NaOH. From our numeration it known that ?H3 is -58.6 kJ/mol. When ionic solid dissolves in water, heat was lib! rated. chemical reaction 1: Dissolving solid atomic number 11 hydroxide in water. NaOH(s) ---> Na+(aq) + OH-(aq) + heat Reaction 2: Reaction of atomic number 11 hydroxide solution with dilute hydrochloric acid. Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) ---> Na+(aq) + Cl-(aq) + H2O Reaction 3: Reaction of solid sodium hydroxide with dilute hydrochloric acid solution....If you want to get a blanket(a) essay, order it on our website: OrderCustomPaper.com
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