Heat of neutralization Essay

Discussions1.Why theoretical value different from value obtained?It may set off loss to the surroundings.It may have parallax error during pickings the reading.2.Why is usually the cup utilise in this experiment made of polystyrene? To thwart hotness loss to the surroundings because it is a combust insulator.3.Why the heat of neutralisation chemical response has a negative sign?The reaction gives out heat that results in the increase of temperature of the products formed.I. DiscussionDuring this experiment, the pressure in the lab will be constant or essentially constant. When pressure is constant, changes in energy (q) grass be related to change in the enthalpy of reaction (Hrxn).1Energy changes accompany chemical reactions as original bonds are bewildered and new bonds are formed. Usually 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 a bsorbed by the reaction, the enthalpy is positive and the reaction is endothermic. In this experiment you will measure the enthalpy (H) of an battery- demigod/base neutralization. A neutralization reaction occurs when acid and base combine to form peeing as shown in the example below. Overall Reaction HCl (aq) + NaOH(aq) NaCl(aq) + H2O(l) TIE H+(aq) + Cl(aq) + Na+(aq) + OH(aq) Na+(aq) + Cl(aq) + H2O(l)orNIE H+(aq) + OH(aq) H2O(l) broadside in the neutralization reaction above the Na+and Clions, the watchman ions, remain unchanged.The only chemical reaction occurring is between the H+and OHions. neutralisation reactions haveheat as a product since energy is released when H+and OHform a H2O molecule.The heat liquify, q, of a process alike a chemical reaction can be studied 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 = qsurroundingsThe equat ion above says that the heat disoriented by the system is equal to, but of opposite sign from the heat gained by the surroundings. Thus if the heat change in the surroundings is mensural then 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 same physical properties as water, s pecifically 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 satisfying base) with hydrochloric acid (a strong acid) and acetic acid (a wanton acid) will be investigated. The base is present in slight redundance and, therefore, the acid is the limiting reagent and patch ups the number 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. supplying of Solutions1. Each lab bench will hold make the 0.5000 M acid solutions needed for this lab by diluting 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 each(prenominal) bench will share these solutions.2. Each pair of students will need at least 120 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 volumetrical flasks. Your instructor will demonstrate how to perform the dilution. (Although volumetric glassware is non 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 = Molarity 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. allow in your dilution calculations in you r lab book.