# Determination of Heat of Neutralization

Topics: Sodium hydroxide, Thermodynamics, Acid Pages: 8 (1000 words) Published: September 23, 2010
PRACTICAL 15:DETERMINATION OF HEAT OF NEUTRALIZATION

Data collection:

|Reaction |Initial Temperature/°C (±0.25) |Final Temperature of Mixture/°C (±0.25) | | |Acid |Base | | |HNO3 + NaOH |28.00 |28.25 |34.50 | |HNO3 + KOH |28.25 |28.25 |34.00 | |HCl + NaOH |28.25 |28.00 |34.75 | |HCl + KOH |28.25 |28.25 |34.00 | |H2SO4 + NaOH |28.00 |28.50 |36.50 | |H2SO4 + KOH |28.50 |28.00 |34.00 |

Volume of acid = 50 cm3 ± 0.25
Acid concentration = 1.0 M
Volume of base = 50cm3 ± 0.25
Base concentration = 1.0 M
Number of mole of solution used = [pic]
= [pic]
= 0.05 mole ± 0.5 %

Data Processing:

Calculation:

Temperature Change, [pic] = Final temperature – Average temperature of Acid and Base

|Reaction |Average Temperature of Acid and Base, °C|Final Temperature, °C |Change of Temperature, °C | | |(±0.5) |(±0.25) |(±0.75) | |HNO3 + NaOH |28.13 |34.50 |6.38 | |HNO3 + KOH |28.25 |34.00 |5.75 | |HCl + NaOH |28.13 |34.75 |6.63 | |HCl + KOH |28.25 |34.00 |5.75 | |H2SO4 + NaOH |28.25 |35.00 |6.75 | |H2SO4 + KOH |28.25 |34.00 |5.75 |

Heat Released, Q = [pic]

m - mass of solution
c - specific heat of water (4.2 Jmol-1°C-1)
[pic] - temperature change of solution

Enthalpy of neutralization, ∆H = Heat released
No of moles

[pic]Enthalpy change of each experiment :

1. HNO3(aq) + NaOH(aq) [pic]NaNO3(aq) + H2O(l)

Enthalpy of neutralization = [pic]
= 53.55 kJmol-1
Uncertainties:

m = 0.5 . × 100= 0.5 %
100

ΔT = 0.75 . × 100= 11.76 %
6.38

no of moles= 0.5 %

Total= 12.76 %

Enthalpy of neutralization= -53.55 kJmol-1± 12.76 %

2. HNO3(aq) + NaOH(aq) [pic]NaNO3(aq) + H2O(l)

Enthalpy of neutralization= [pic]
= 48.30 kJmol-1

Uncertainties:

m = 0.5 . × 100= 0.5 %
100

ΔT = 0.75 . × 100= 13.04 %
5.75

no of moles= 0.5 %

Total= 14.05 %

Enthalpy of neutralization= -48.30 kJmol-1 ± 14.05 %

3. HCl(aq) + NaOH(aq) [pic]NaCl(aq) + H2O(l)

Enthalpy of neutralization= [pic]
= 55.65 kJmol-1

Uncertainties:

m = 0.5 . × 100= 0.5 %
100

ΔT = 0.75 . × 100= 11.31%
6.63

no of moles= 0.5 %

Total= 12.31 %

Enthalpy of neutralization= -55.65 kJmol-1 ± 12.31 %

4. HCl(aq) + KOH(aq) [pic]KCl(aq) + H2O(l)

Enthalpy of neutralization= [pic]
= 48.30 kJmol-1

Uncertainties:

m = 0.5 . × 100= 0.5 %
100

ΔT = 0.75 . × 100= 13.04 %
5.75

no of moles= 0.5 %

Total= 14.04 %

Enthalpy of neutralization= -48.30 kJmol-1 ± 14.04 %

5. H2SO4(aq) + 2NaOH(aq) [pic] Na2SO4(aq) + 2H2O(l)...