Qualitative Data:
- Beaker was warm after mixing the acid and base.
Processing:
2 NaOH(aq) + H2SO4(aq) = 2 H2O(l) + Na2SO4(aq)
c: 1.0mol/L 1.0mol/L
v: 50.0±0.2mL 30.0±0.2mL 80.0±0.4mL
T: 22.39±0.2°C 23.79±0.2°C 31.09±0.2°C
Temperature of Reactants: [(22.39±0.2°C)+( 23.79±0.2°C)]/2 = 23.09±0.2°C
Limiting Reagent: sodium hydroxide
Excess Reagent: sulphuric acid
Molar Enthalpy of neutralization for sodium hydroxide with sulphuric acid:
nΔrHm, NaOH = mcΔt
Δt = (31.09±0.2°C)-( 23.09±0.2°C) = 8.00±0.4°C
(0.050L ± 0.0002L NaOH(aq)) * 1mol NaOH(aq) * ΔrHm, NaOH = (0.080 ± 0.0004kg H2O(l)) * 4.19 kJ * (8.00± 0.4°C)
1L NaOH(aq) kg * °C
(0.050L ± 0.4% NaOH(aq)) * 1mol NaOH(aq) * ΔrHm, NaOH = (0.080 ± 0.50% kg H2O(l)) * 4.19 kJ * (8.00± 5.0%°C)
1L NaOH(aq) kg * °C
(0.050 ± 0.4% mol NaOH(aq)) * ΔrHm, NaOH = (2.6816... ±5.5% kJ)
ΔrHm, NH4Cl = _ (2.6816... ±5.5% kJ)
(0.050 ± 0.4% mol NaOH(aq))
ΔSolHm, NaOH = 53.632... ±5.9% kJ/mol
ΔSolHm, NaOH = 53.632... ±3.164288... kJ/mol
ΔSolHm, NaOH = 54 ±3.2 kJ/mol
Temperature increases = exothermic reaction = -(KJ/mol) so:
ΔSolHm, NaOH = -(54 ±3.2 kJ/mol)
Percent Error:
[(⏐57 kJ/mol -54 kJ/mol⏐) / 57 kJ/mol]*100% = 5.236...%
[(⏐57 kJ/mol -54 kJ/mol⏐) / 57 kJ/mol]*100% = 5.3%
Conclusion and Evaluation
Conclusion:
From the experiment I have calculated that the molar enthalpy of neutralization for sodium hydroxide wiht sulphuric acid is -(54 ±3.2 kJ/mol). The difference between the theoretical and experimental molar enthalpy of ammonium chloride is 3.0 kJ/mol meaning that the percent error is 0.676%.
Evaluation:
Weaknesses of the lab:
System of Experiment:
The calorimeter in which the dilution took place was not a closed system. This means that there are areas on the calorimeter which allowed the transfer of heat between the calorimeter and it's surroundings.
Specific Heat Capacity of Materials Used:
Since all the materials used also have specific heat capacity, it means that not just the product solution gained thermal energy in this experiment. This means that the temperature measured by the thermometer is off because it is also measuring the temperature of all other materials used.
Specific Heat Capacity of Water is Used:
In the calculations the specific heat capacity of water is used because a dilute aqueous solution is formed in the neutralization. In actuality the product of the neutralization has it's own specific heat capacity, this means that the calculations are off by a bit since the actual specific heat capacity was not used.
Improvements:
In order to close the system off even more, you can tape all the cracks that are visible on the calorimeter so that less energy is lost between the system and the environment. To fix the problem of thermal energy being lost by calorimeter material, just calculate the quantity of thermal energy each material loses and subtract it from you final temperature to compensate for the specific heat capacity of the other materials. This gives you the actual Q of the product, and since Q = mcΔt and nΔrHm = mcΔt, nΔrHm = Q. The last improvement that can be made to the experiment would be to use the actual specific heat capacity of the product rather than just 4.19 J/g * °C. All these improvements will help bring the calculated value closer to the theoretical value minimizing the percent error.