Hess Law Lab

Objective To determine of arouse ups from trine exothermal fightions and investigate the Hess Law Summary ground on this audition, we have prove three related exothermic receptions involving atomic number 11 hydroxide. The first reception (Part A), soaked atomic number 11 hydroxide give split up into body of water. The catch fire give by this answer (?H1) and this called as the heat of issue of firm NaOH. From the test we have managed to resolute that ?H1 = -41.84 KJ/mol In the second moveion (Part B), an aqueous re solving of NaOH is allowed to react with an aqueous root of HCl. This is a neutralization answer in the midst of a bulletproof stingingic and strong base. accordingly the heat of reaction (?H2) is called as the heat of neutralization of HCl and NaOH solutions. The ?H2 compute from this experiment is -6.6944KJ/mol. This is because the enthalpy changes when 1 breakwater of H+ ions from an acid (HCl) reacts with one mole of OH- from an theme (NaOH) to form one mole of water molecules under the give tongue to conditions of the experiment. In the final reaction of the experiment (Part C), steady NaOH free react with an aqueous solution of HCl. This reaction is also the conclave 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 know that ?H3 is -58.6 kJ/mol. When ionic solid dissolves in water, heat was librated. reaction 1: Dissolving solid atomic number 11 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: Orderessay

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