Apparatus:
- Polystyrene cup
- Thermometer
- HCl
- NaOH pellets
-
H2O
- Digital weight reader
- Measuring cylinder
Method:
Measurement of ΔHθ1
Pour 50cm3 of 1M hydrochloric acid into the polystyrene cup and record its temperature as accurately as you possibly can. Weigh out 2g of sodium hydroxide pellets and quickly add these to the acid in your polystyrene cup. Stir and record the maximum temperature reached. Calculate the enthalpy change of this reaction in kJmol-1.
Measurement of ΔHθ2
Pour 50cm3 of water into an empty polystyrene cup and record its temperature as accurately as you possibly can. Weigh out 2g of sodium hydroxide pellets and quickly add these to the water in your polystyrene cup. Stir and record the maximum temperature reached. Calculate the enthalpy change of this process in kJmol-1.
Measurement of ΔHθ3
Pour 50cm3 of 1M sodium hydroxide into an empty polystyrene cup and record its temperature as accurately as you possibly can. Measure out 50cm3 of 1M hydrochloric acid into a measuring cylinder and record its temperature. Calculate the average initial temperature of the acid and the alkali. Add the acid to the alkali in your polystyrene cup and record the maximum temperature reached. Calculate the enthalpy change of this reaction inkJmol-1.
Data Collection:
Data Processing:
Measurement of ΔHθ1
NaOH(s) + HCl(aq) NaCl(aq) + H2O(l)
Initial Temperature = 26.5°C ±0.1°C
Maximum Temperature = 48.1°C ±0.1°C
ΔT = Maximum Temperature – Initial Temperature = 48.1°C – 26.5°C = 21.6°C ±0.2°C
% Error in ΔT =
The enthalpy change ΔHθ1 is given by multiplying the mass of HCl (m) by its specific heat capacity (Cp) and the change in temperature (ΔT). Since this reaction takes place in solution, which we assume to be mostly water. Hence the mass of the solution will be 50g and the specific heat capacity is given as 4.2J/g/°C. The calculation for the mass of the solution is as follows:
Mass = Density × Volume (Density given as 1g/cm3)
Mass = 1g/cm3 × 50cm3
Mass = 50g
Hence,
ΔHθ1= m × Cp × ΔT
ΔHθ1 = 50g × 4.2J/g/°C × 21.6°C
ΔHθ1 = 4536J ±0.926%
ΔHθ1 = 4.536kJ ±0.042kJ
Mass of NaOH Pellets: 2.20g ± 0.01g
% Error in mass =
2.20g of NaOH produces 4.536kJ ±0.042kJ
1 mole (40g) will produce: ± 1.381%
1 mole (40g) will produce 82.5kJmol-1 ± 1.14kJmol-1
Measurement of ΔHθ2
NaOH(s) NaOH(aq)
Initial Temperature = 27.2°C ±0.1°C
Maximum Temperature = 36.5°C ±0.1°C
ΔT = Maximum Temperature – Initial Temperature = 36.5°C – 27.2°C = 9.3°C ±0.2°C
% Error in ΔT =
The enthalpy change ΔHθ2 is given by multiplying the mass of NaOH (m) by its specific heat capacity (Cp) and the change in temperature (ΔT). Since this reaction takes place in solution, which we assume to be mostly water. Hence the mass of the solution will be 50g and the specific heat capacity is given as 4.2J/g/°C. The calculation for the mass of the solution is as follows:
Mass = Density × Volume (Density given as 1g/cm3)
Mass = 1g/cm3 × 50cm3
Mass = 50g
Hence,
ΔHθ2 = m × Cp × ΔT
ΔHθ2 = 50g × 4.2J/g/°C × 9.3°C
ΔHθ2 = 1953J ±2.15%
ΔHθ2 = 1.953kJ ±0.042kJ
Mass of NaOH Pellets: 2.02g ± 0.01g
% Error in mass =
2.02g of NaOH produces 1.953kJ ±0.042kJ
1 mole (40g) will produce: ± 2.645%
1 mole (40g) will produce 38.7kJmol-1 ± 1.02kJmol-1
Measurement of ΔHθ3
NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
Initial Temperature = 26.9.2°C ±0.1°C
Maximum Temperature = 36.8°C ±0.1°C
ΔT = Maximum Temperature – Initial Temperature = 36.8°C – 26.9°C = 9.9°C ±0.2°C
% Error in ΔT =
The enthalpy change ΔHθ3 is given by multiplying the mass of HCl (m) by its specific heat capacity (Cp) and the change in temperature (ΔT). Since this reaction takes place in solution, which we assume to be mostly water. Hence the mass of the solution will be 100g and the specific heat capacity is given as 4.2J/g/°C. The calculation for the mass of the solution is as follows:
Mass = Density × Volume (Density given as 1g/cm3)
Mass = 1g/cm3 × 50cm3
Mass = 50g
Hence,
ΔHθ3 = m × Cp × ΔT
ΔHθ3 = 50g × 4.2J/g/°C × 9.9°C
ΔHθ3 = 2079J ±2.02%
ΔHθ3 = 2.079kJ ±0.042kJ
Volume of NaOH: 0.05dm3 ± 0.0001dm3
% Error in volume =
0.05dm3 of NaOH produces 2.079kJ ±0.042kJ
1 mole (1dm3) will produce: ± 2.22%
1 mole (40g) will produce 41.6kJmol-1 ± 0.924kJmol-1
Conclusion and Evaluation:
According to Hess’s Law ΔHθ1 = ΔHθ2 + ΔHθ3
So using the results found above we see that:
ΔHθ1 = 82.5kJmol-1 ± 1.14kJmol-1
ΔHθ2 + ΔHθ3 = 38.7kJmol-1 + 41.6kJmol-1 = 80.3kJmol-1 ± 1.944kJmol-1
It can be seen that ΔHθ1 is almost equal to ΔHθ2 + ΔHθ3 but due to experimental errors such as heat being lost to the surroundings and not having exact readings they are not equal. Hess’s law has therefore been validated but the results would have been much more accurate if there was an insulating capsule around the polystyrene cup so as to prevent any heat from being lost to the environment and by having more accurate apparatus in terms of measuring cylinders and thermometers.