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Lab Report. Objectives 1. To determine the enthalpy of neutralization of strong acid and strong base. 2. To determine the quantity and direction of the heat transfer in the dilution of a salt.

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Introduction

Transfer-Encoding: chunked ´╗┐SKU3023 : Chemistry IIEXPERIMENT 3 EXPERIMENT 3 : Enthalpy 3.1 Objectives 1. To determine the enthalpy of neutralization of strong acid and strong base. 2. To determine the quantity and direction of the heat transfer in the dilution of a salt. 3.2 Introduction A chemical or physical change involves heat (energy) is known as Thermochemistry. While reaction that release energy (heat) is called exothermic reaction, and the reaction that absorbs energy from its surroundings is called endothermic reaction. Calorimeter is a device uses to measure a quantity of heat transferred in the Thermochemistry reaction, which undergoes chemical or physical change. The heat transferred in a chemical reaction is defined quantitatively as an enthalpy (heat) of the reaction, at a constant pressure. The is negative if the reactions are exothermic and positive for endothermic reactions. The specific heat of a substance is known as the amount of heat required to raise its temperature for of one gram substance. The unit of specific heat is . The enthalpy of neutralization involves acid base reaction and for strong acid and base, the reaction is exothermic. is determined by a) Assuming the density and specific heat for acid and base solution are the same as water b) Measure the temperature change, The unit of is kJ/mol of acid and base reacted. ...read more.

Middle

= 5.002 58.443 = 0.0856 mole Mass of cup and water = 21.500 g Mass of Styrofoam cup = 2.659 g Mass of water = 21.500 ? 2.659 = 18.841 g Section E : Calculation for Heat of Salt Solution Item Trial 1 Trial 2 1. Temperature change 5 5 2. Water heat released (J) 393.78 392.38 3. Salt heat released (J) 104.54 104.60 4. Total enthalpy change () 498.32 J 496.98 J 5. Amount of OH- reacted (g) 1.55 1.55 6. Amount of H2O produced (g) 1.55 1.55 7. , NaCl salt + 5.79 + 5.78 8. Average + 5.785 PART B : Heat of Salt Solution 3.7 Questions 1. Show your calculations for Trial 1 (in Section C) HCl and NaOH Volume of HCl = 50 ml Temperature of HCl = 22 Volume of NaOH = 50 ml Temperature of NaOH = 21.5 Maximum temperature from graph = 28 Calculations : Average initial temperature of acid and base = = 21.8 Temperature change = 28 ? 21.8 = 6.2 Volume of final mixture = 50 ml HCl + 50 ml NaOH = 100 ml Mass of final mixture = Density Volume = (1.00 g/ml)(100 ml) = 100g Specific heat of the mixture (Given) = 4.18 Jg-1 mol NaOH = MV = = 0.050 mol Molar mass of NaOH = 22.990 + 15.999 + 1.008 = 38.997 Amount of NaOH is used = 0.050 mol ...read more.

Conclusion

The experimental value for the enthalpy change for neutralization is less than the theoretical value of standard enthalpy change for neutralization. This is due to the fact that some heat is lost to the surrounding and the heat absorbed by the Styrofoam cup is not included in the calculation. A few suggestions to improve the accuracy of the experiments are the thermometer had to be calibrated, which improves accuracy but is itself an imprecise technique. The thermometer and the calorimeter should wipe till it completely dry to prevent the errors in reading. The calorimeter also should run in closed area to prevent the heat easily lost to the surrounding. PART B : Heat of Salt Solution The experimental value for is different from the value of ). If the system is not well insulated, energy from the surroundings will be absorbed as well as energy from the water so that the fall of water temperature will not be as great as expected. 3.9 Conclusions From this experiment, we get the average ?Hn between HCl and NaOH is -49.74 kJ/mol and the average ?Hn between HNO3 and NaOH is -43.47 kJ/mol. The major weakness of this experiment is the lost of significant amount of heat to the surrounding. This causes deviation of the experimental value of ?Hn from the theoretical value (?Hn = -57.1 kJ/mol). must be positive (+) because energy has been absorbed, that is, the process is endothermic. 3. ...read more.

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