12N 2 g 32O 2 g HNO 3 aq H f 494 Kcalmol The other two enthalpy values needed

12n 2 g 32o 2 g hno 3 aq h f 494 kcalmol the other

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1/2N 2 (g) + 3/2O 2 (g) HNO 3 (aq) H f = -49.4 Kcal/mol The other two enthalpy values needed will be determined experimentally. The ammonia and the appropriate acid solution (HA where A = Cl or NO 3 ) will be mixed in equal mole quantities to form an aqueous solution of an ammonium salt (NH 4 A). The reaction will be carried out in a calorimeter so that the heat change will be detected indirectly by a thermometer with minimal loss to the surroundings. You will need to calculate the enthalpy change for this neutralization reaction ( H neut ). In similar fashion, you will need to carry out the dissociation of a known amount of the solid NH 4 A and determine H soln . Remember that in calorimetry, the temperature measurement is indirect (heat transfer), so the heat absorbed/released by the immediate surroundings will be equal to the heat released/absorbed by the reaction (assuming no heat loss to the cup or exterior surroundings). Therefore q rxn = - (q surr ) = - (m surr c surr T surr ). To find H you will have to calculate your heat values on a mole basis. The final calculated result for the H f for the solid ammonium salt can be compared to that determined by the National Bureau of Standards (NBS): Solid Ammonium Salt Hf, Kcal NH 4 Cl -75.38 NH 4 NO 3 -87.27 Cary Academy Chemistry II W.G. Rushin 1
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Procedure: Note: Temperature-time readings will be collected so that T can be obtained graphically from a plot of temperature (y axis) vs. time (x axis). The points
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