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To compare the first ionization enthalpy and the second ionization enthalpy of reaction

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Introduction

Title: Energetics Aim: To compare the first ionization enthalpy and the second ionization enthalpy of reaction Date: 7th November, 02 Introduction: There are various types of enthalpy changes of reaction, the enthalpy change of neutralization and ionization were determined through this experiment. The enthalpy change of neutralization was first determined by the reaction between a strong acid (HCl) and a strong alkali (NaOH). By this data, the first ionization of a weak acid(CH3COOH), and the second ionization of another weak acid (carbonic acid) were determined. Experimental: HCl (50cm , 2 M, 0.1mol) was put into a polystyrene cup and the temperature of the acid was recorded. NaOH (50cm , 2 M, 0.1 mol) was added to the acid quickly, and the highest temperature obtained was recorded. ...read more.

Middle

The temperature change was obtained and the enthalpy of the reaction was determined. The 2nd ionization energy was calculated. Results: 1. Determination of the heat of neutralization between HCl (a strong acid) and NaOH (a strong alkali) Initial temperature = 27.7? Final temperature = 41.2? Temperature change = 13.5? 2. Determination of the heat of neutralization between CH3COOH( a weak acid) and NaOH( a strong alkali) and the heat of ionization of CH3COOH Initial temperature = 27.6? Final temperature = 40.4? Temperature change = 12.8? 3. Determination of the heat of neutralization between NaHCO3(a weak acid) and NaOH( a strong alkali) and the heat of the 2nd ionization of carbonic acid Initial temperature = 27? Final temperature = 29.6? Temperature change = 2.6? ...read more.

Conclusion

--> CH3COONa(aq) + H2O(l) No of mole of CH3COOH = 2.0377 x 0.05 = 0.101885 mol No of mole of NaOH = 2.0175 x 0.05 = 0.1009 mol --> NaOH is the limiting reagent. Heat evolved = mc?T = (0.05 + 0.05)(4.2)(12.8) = 5.376 kJ enthalpy change of reaction = - 5.376 / 0.1009 = -53.29kJ /mol 3. NaHCO3(aq) + NaOH(aq) --> Na2CO3(aq) + H2O(l) No of mole of NaHCO3 = 0.5014 x 0.05 = 0.0251 mol No of mole of NaOH = 0.5041 x 0.05 = 0.0252 mol --> NaHCO3 is the limiting reagent heat evolved = (0.05 + 0.05)(4.2)(2.6) = 1.092 kJ enthalpy change of neutralization = -1.092 / 0.02507 = 43.56 kJ Discussion: The heats of neutralization calculated in the 1st part is different from 2nd part, the enthalpy change of the 2nd part is more exothermic. ...read more.

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