at: Enthalpy of neutralization of HCI by NaOH (kJ)- Recalling th 4m # = {t(rn)(A

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at: Enthalpy of neutralization of HCI by NaOH (kJ)- Recalling th 4m # = {t(rn)(AT)(c)jsekem" [(AT)( heat capacity of the ays AHin (for the c (AT)( heat capacity of the calorimeter)]), and (for the conditions in this procedure) ale tate the enthalpy aH associated with the neutralization of HCl and NaOH You may assume that the solution in the cal orimeter has the same heat capacity as that of pure water. Moles the moles of each. Determine the li of limiting reagent (mol) - Use the volume and concentration of each reactant to calculate ch, betermine the limiting reagent, based on reaction stoichiometry. . Molar enthalpy of neutralization of HCI (kJ/mol) Molar enthalpy determine the enthalpy, AH neutralization of HCI (ku/mol) Based on the moles of limiting reagent, rn, on a per mole basis. ren, on a . Average enthalpy of neutralization of HCI KJ/mol) - Compute the average value. . Percent error in enthalpy of neutralization of HCI - Based on the given value for the heat of neutralization of H,0' of -55.80 kJ/mol, calculate the percent error in your result. (Since HCl is a strong acid, the heat of neutralization of HCl and of H30+ is the same.) Reaction 2: Neutralization of a Weak Acid by a Strong Base - Repeat the calculations described above for Reaction 1, with the exception of percent error.

Explanation / Answer

Part I.

Trial 1

Heat absorbed by cold water = 49.9 x 4.18 (33.38 - 25.78) = 1585.2232 J

Heat lost by hot water = 49.9 x 4.18 (42.18 - 33.38) = 1835.5216 J

Heat absorbed by calorimeter = 250.30 J

Heat capacity of calorimeter = 250.30/(33.38 - 25.78) = 242.70 J/oC

Part III.

Dissolution equation

KNO3(s) --> K+(aq) + NO3-(aq)

Trial 1

Heat of solution by salt = (75.9 + 1.8034) x 4.18 (27.8 - 26.2) + 242.7/(27.8 - 26.2) = 0.671 kJ

Enthalpy of solution = 0.671 kJ

Enthalpy of solution by per gram salt = 0.671/1.8034 = 0.372 kJ

Heat of solution by per mole salt = 0.372 x 101.1032 = 37.64 kJ/mol

Part III

Trial 1

Neutralization HCl + NaOH

Total mass of solution = 50 + 36 = 86 g

Enthalpy of neutralization = 86 x 4.18 x 1 = 360 J = 0.360 kJ

moles of limiting reactant = 1 M x 0.036 L = 0.036 mol

enthalpy of solution = 0.360/0.036 = 10.0 kJ/mol

Neutralization of acetic acid

Total mass of solution = 50 + 31 = 81 g

Enthalpy of neutralization = 81 x 4.18 (28.3 - 27.7) = 203.15 J = 0.203 kJ

moles of limiting reactant = 1 M x 0.031 L = 0.031 mol

enthalpy of solution per mole of HCl = 0.203/0.031 = 6.55 kJ/mol

Similarly for Trial 2 calculations can be done.

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