The procedure of the above experiment was repeated by mixing CH3COOH (50cm ,2 M, 0.1 mol) and NaOH (50cm ,2 M, 0.1 mol). The temperature rise was obtained and the heat of neutralization was determined. By using the enthalpy change of the first experiment, the first ionization of this experiment was determined.
The enthalpy of neutralization experiment was repeated using NaHCO3 (50cm ,
0.5 M, 0.1 mol) and NaOH (50cm ,0.5M, 0.1 mol). The temperature change was obtained and the enthalpy of the reaction was determined. The 2nd ionization energy was calculated.
Results:
- 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℃
- 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℃
-
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℃
The class mean of temperature change of the first part of experiment is 13.42℃, which is very close to this result (13.5℃).
The class mean of temperature change of the second part is 12.97℃, which is slightly higher than this result (12.8℃).
The class mean of temperature change of the third part is 2.39℃, which is much lower than this result(2.6℃)
Calculation:
-
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
No of mole of NaOH = 2.0175 x 0.05
= 0.1009 mol
No of mole of HCl = 1.9368 x 0.05
= 0.0968 mol
→ HCl is the limiting reagent.
Heat evolved = mc△T
= (0.05 + 0.05)(4.2)(13.5)
= 5.67 kJ
enthalpy change of neutralization = - 5.67/ 0.0968
= -58.55 kJ/mol
-
CH3COOH(aq) + NaOH(aq) → 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
-
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.
