lab_10_heat-of-neutralization

lab_10_heat-of-neutralization - Bristol Community College...

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Experiment #10 Heats of Neutralization Bristol Community College CHM13: Fundamentals of Chemistry I Benjamin Tigano Shawn Morales CHM13.79L Date Performed: 11.15.2008 Date Submittied: 11.16.2008 Introduction All changes in nature are accompanied by changes in energy. Chemical reac- tions are accompanied by energy changes in the form of heat. The energy change of a reaction that occurs at constant pressure is termed the heat of reaction or the en- thalpy change. The symbol {delta}H ( the symbol “{delta}” means “change in”) is used to denote the energy change. If heat is evolved : the reaction is exothermic ({delta}H < 0, thus a negative value). If heat is absorbed : the reaction is endothermic ( {delta}H > 0 , thus a positive value). In this experiment you will measure the neutralization of a chemical reaction when an acid and a base react to form salt and water. The quantity of heat is measured experimentally by allowing the reaction to take place in a ther- mally insulated vessel called a calorimeter. The heat liberated in the neutralization re- action will cause an increase in temperature of the solution and is measured by a ther- mometer. If the calorimeter were prefect there would be no heat loss to the calorime- ter and the room. Thus we must measure the “heat capacity” of the calorimeter. This requires that we determine the “heat capacity of the calorimeter” by determining the amount of heat (in joules) required to raise its temperature 1 kelvin, which is the same as 1 oC ( both are in the centigrade scale). The change in enthalpy, {delta}H, is equal to the product of the temperature, {delta}T, times the heat capacity of the calo- rimeter and its contents. Heat, {delta}H = mCp{delta}T The heat capacity of the calo- rimeter is determined by measuring the temperature change that occurs when a spe- cific amount (in grams) of hot water is added to a specific amount (in grams) of cold water in the calorimeter. Heat lost by the warmer water = (Tf - T2 )* 100.0 g * 4.184 J/K-g = “ – value” (1) Where T2 equals the initial temperature of the 50.0 grams of warm water, and Tf is the final temperature after mixing the two waters together. T1 is the initial tempera- ture of the cold water, also remember the Specific Heat Capacity, Cp, of water is 4.184 J/K-g and the density of water is 1.00g/mL. Heat gained by the cold water = ( Tf - T1 )* 100.0 g * 4.184 J/K-g
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lab_10_heat-of-neutralization - Bristol Community College...

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