Hess Law Lab

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Objective To determine of heats from three exothermic reactions and investigate the Hess Law Summary Based on this experiment, we have study three related exothermic reactions involving sodium hydroxide. The first reaction (Part A), solid sodium hydroxide will dissociate into water. The heat produce by this reaction (∆H1) and this called as the heat of solution of solid NaOH. From the experiment we have managed to determined that ∆H1 = -41.84 KJ/mol In the second reaction (Part B), an aqueous solution of NaOH is allowed to react with an aqueous solution of HCl. This is a neutralization reaction between 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 enthalpy changes when one mole of H+ ions from an acid (HCl) reacts with one mole of OH- from an alkali (NaOH) to form one mole of water molecules under the stated conditions of the experiment. In the final reaction of the experiment (Part C), solid NaOH will react with an aqueous solution of HCl. This reaction is also 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 then neutralized by the acid solution. Thus the heat of the reaction (∆H3) is said be equal to (∆H1+∆H2). It is called the heat of solution of solid NaOH. From our calculation it known that ∆H3 is -58.6 kJ/mol. When ionic solid dissolves in water, heat was librated. Reaction 1: Dissolving solid sodium hydroxide in water. NaOH(s) ---> Na+(aq) + OH-(aq) + heat Reaction 2: Reaction of sodium hydroxide solution with dilute hydrochloric acid. Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) ---> Na+(aq) + Cl-(aq) + H2O Reaction 3:

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