Expt 9 - Heats of Reaction and Solution - FA 2010

Expt 9 - Heats of Reaction and Solution - FA 2010 -...

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1      ADDITIONAL READING The concepts in this experiment are also discussed in sections 6.3 – 6.8 of Principles of Chemistry – A Molecular Approach , by Tro. ABSTRACT The purpose of this experiment is threefold: 1) to become familiar with the equations associated with Calorimetry, 2) to measure the heat of solution of a salt, and 3) to verify Hess’s Law of Constant Heat Summation. This experiment is divided into three parts and may be performed in pairs. In Part A you will determine the heat of solution, Δ H Soln , for the dissolving of a salt using Calorimetry. Temperatures will be measured and recorded using the Vernier system equipped with a temperature probe, and the data analyzed using features built into the instrument. The reaction is: MA(s) M + (aq) + A - (aq) In Part B you will determine the heat of reaction, Δ H Rxn , for magnesium metal and hydrochloric acid using Calorimetry and graphing techniques. The net ionic equation (NIE) for this reaction is: Mg(s) + 2 H + (aq) Mg 2+ (aq) + H 2 (g) In Part C you will determine the heat of neutralization, Δ H Neut , for a reaction between hydrochloric acid and sodium hydroxide using Calorimetry and graphing techniques. The NIE for this reaction is: H + (aq) + OH - (aq) H 2 O( ) BACKGROUND HEATS OF REACTION In this experiment you will measure the amount of heat that is generated (exothermic) or consumed (endothermic) in a chemical reaction. You may recall that all acid-base reactions are exothermic, meaning that they produce heat as a product. HNO 3 (aq) + NaOH(aq) NaNO 3 (aq) + H 2 O( ) + heat As you can see in the reaction above, one mole of nitric acid reacts with one mole of sodium hydroxide to produce the salt, sodium nitrate, water, and heat. In our experiment you will be able to measure the heat that is generated in this reaction using and simple calorimeter. In addition, you will use this information to calculate the molar heat of reaction, i.e. how much heat is generated per mole of reactant (kJ/mol). Consider the following example: 38.5 mL of 2.2 M HNO 3 , initially at 25.2 o C, reacts with 42.2 mL of 1.9 M NaOH, initially at T i = 25.2 o C. After graphing the data it was determined that the final temperature, T f , was 38.6 o C. From this information we can determine the heat generated and the molar heat of reaction. Recall from the conservation of energy (1 st Law of Thermodynamics) that the following must be true. q 1 + q 2 + q 3 = 0 or q acid + q base = – q water The energy released when the acid reacts with the base is transferred to the water during the reaction; therefore, we have the following: q acid + q base = q Rxn = – q water q Rxn = – (m water )(C water )( Δ T water ) The mass of the water, or more correctly, the mass of the solution, is the sum of the masses of the acid and base solutions. The densities of dilute acidic and basic solutions are approximately the same as that
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2 of water, 1.00 g/mL. We will also assume the specific heat of the mixture is the same as that of water i.e., C water = 4.184 J/g .o C.
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This note was uploaded on 04/19/2011 for the course CHEM 1033 taught by Professor Price during the Fall '10 term at Temple.

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Expt 9 - Heats of Reaction and Solution - FA 2010 -...

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