Heat of neutralization Essay

Discussions1.Why theoretical value diverse from value get under ones skined?It may waken loss to the surroundings.It may have parallax error during taking the reading.2.Why is usually the instill used in this experiment made of polystyrene? To prevent heat loss to the surroundings because it is a heat insulator.3.Why the heat of counteraction has a negative sign?The reaction gives out heat that results in the growing of temperature of the products formed.I. DiscussionDuring this experiment, the thrust in the testing ground give be constant or essentially constant. When pressure is constant, channelises in energy (q) can be related to change in the enthalpy of reaction (Hrxn).1Energy changes accompany chemical reactions as original bonds are broken and new bonds are formed. commonly the energy change takes the form of heat. If heat is released from the reaction, the change in energy or enthalpy (H) is negative and the reaction is exothermic. Conversely, if heat is absorbed by the reaction, the enthalpy is positive and the reaction is endothermic. In this experiment you leave alone measure the enthalpy (H) of an pane of glass/ ancestor neutralization. A neutralization reaction occurs when acid and base approve to form water as shown in the example below. Overall Reaction HCl (aq) + NaOH(aq) NaCl(aq) + water(l) TIE H+(aq) + Cl(aq) + Na+(aq) + OH(aq) Na+(aq) + Cl(aq) + H2O(l)orNIE H+(aq) + OH(aq) H2O(l)Notice in the neutralization reaction higher up the Na+and Clions, the spectator ions, remain unchanged.The only chemical reaction occurring is between the H+and OHions. neutralisation reaction reactions haveheat as a product since energy is released when H+and OHform a H2O molecule.The heat flow, q, of a service like a chemical reaction can be canvass by analyzing its heat exchange with its surroundings. The heat released by a system (a chemical reaction) is absorbed by its surroundings (often this is the solution).qrxn = qsystem = qsurroundi ngsThe equation above says that the heat lost by the system is equal to, but of resistance sign from the heat gained by the surroundings. Thus if the heat change in the surroundings is measured whence heat released by the chemical reaction can be calculated. Frequently, such a heat change measurement is done in an insulated container called a calorimeter. In a perfect calorimeter, all of the heat released by the chemical reaction would stay inside the calorimeter. Although our experimental setup utilizes a lessthan perfect calorimeter, a coffee-cup calorimeter, the data collected is close to that for a perfect calorimeter.The heat flow into the reaction surroundings (solution), qsurroundings, from the neutralization reaction can be calculated using the following equation where m is the mass of the calorimeter contents, T is the change in temperature, and Cs is the specific heat of the contents. We will assume that the solution in the calorimeter has the said(prenominal) physical properties as water, specifically that Cs = 4.184 J/gC. qsurroundings = m T CsIn contrast when volume is constant, changes in energy (q) can be related to changes in the internal energy, Erxn.2 In this experiment the neutralization of sodium hydroxide (a strong base) with hydrochloric acid (a strong acid) and acetic acid (a abstemious acid) will be investigated. The base is present in slight excess and, therefore, the acid is the limiting reagent and determines the weigh of moles of acid and base reacting. Therefore, the heat flow from the reaction isqrxn = Hrxn (mol acidII. Objectives1. To determine the enthalpy of neutralization of a strong base with a strong acid. 2. To determine the enthalpy of neutralization of a strong base with a weak acid. 3. To use Hess Law to determine the enthalpy of dissociation of the weak acid.III. Procedure This lab is done is pairs.A. Preparation of Solutions1. Each lab bench will need make the 0.5000 M acid solutions needed for this lab by diluti ng 1.000 M stock solutions of HCl and acetic acid. Read the label on the container to obtain the exact molarity of the acid solutions. It will be very close to 1.000M. Lab groups at individually bench will share these solutions.2. Each pair of students will need at least one hundred twenty mL of each acid. Accounting for errors and/or extra trials, a total volume of 500 mL of the 0.5000 M acids will be enough for 3 pairs of students.3. The equipment available for the dilution includes 250.00 and 500.00 mL volumetric flasks. Your instructor will demonstrate how to perform the dilution. (Although volumetric glassware is not commonly used to make quantitative dilutions, the process is appropriate for the solutions used in this lab.)Recall the formula for dilution calculationsM1V1 = M2V2For this dilutionM1 = 250.00 mLV1 = megabyte of the stock acid solution(known)M2, = Molarity of the dilute acid solution(solve for this should be close to 0.5000 M) V2 = 500.00 mL4. Include your dilut ion calculations in your lab book.