Hess Law Lab

Objective

To determine of arouse ups from three exothermic fightions and investigate the Hess Law

Summary

Based on this audition, we have study three related exothermic receptions involving atomic number 11 hydroxide. The first reception (Part A), steadfast atomic number 11 hydroxide give dissociate into body of water. The heat produce by this reaction (?H1) and this called as the heat of solution of solid NaOH. From the test we have managed to intractable 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 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 enthalpy changes when 1 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 willing 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 thence 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: 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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