The Enthalpy of Neutralization

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IB Chemistry 2

Enthalpy of Neutralization

Data Collection


Experiment 1 – The reaction between hydrochloric acid and sodium hydroxide.

Table 1.1: Temperature of HCl and NaOH, separately and after mixing



Experiment 2 – The reaction between hydrochloric acid and potassium hydroxide.

Table 2.1: Temperature of HCl and KOH, separately and after mixing



Experiment 3 – The reaction between nitric acid and sodium hydroxide.

Table 3.1: Temperature of HNO3 and NaOH, separately and after mixing




Experiment 4.1 – The reaction between sulfuric acid and 2.00mol∙dm-3 sodium hydroxide.

Table 4.1: Temperature of H2SO4 and NaOH, separately and after mixing



Experiment 4.2 – The reaction between sulfuric acid and 4.00mol∙dm-3 sodium hydroxide.

Table 4.2: Temperature of H2SO4 and NaOH, separately and after mixing



Observations

Throughout the series of experiments there was really nothing drastic to observe, as all of our reagents were clear in color, there was no precipitate for any of the experiments, and the solution in all cases was clear as well. Notably however, was that the mixture of any two reagents gave off heat; meaning that the reaction in all cases was definitely exothermic. But again this was to be expected. Any change in temperature of the two aqueous reagents whatsoever was to be expected, as that is what we were looking to measure and observe in the first place.


Data Processing

Experiment 1 – Enthalpy of neutralization for hydrochloric acid and sodium hydroxide

NaOH(aq) + HCl(aq)→NaCl(aq) + H2O(l)

Finding the Moles of Water

Moles of hydrochloric acid:
[Volume x Concentration]

0.025dm3 x 2.00mol∙dm-3 = 0.050mol HCl

Moles of sodium hydroxide:
[Volume x Concentration]

0.025dm3 x 2.00mol∙dm-3 = 0.050mol NaOH

Moles of water:
[Equal to moles of limiting reagent]

0.050mol H2O

Calculating Enthalpy of Neutralization

Mass/volume of solution
[Mass = Volume, if we assume 1g = 1cm3]

25.00cm3 (± 0.5cm3) + 25.00cm3 (± 0.5cm3) = 50.00cm3 (± 1.0cm3) = 50.00g (± 1.0g)

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Calculation for measuring enthalpy changes
[Q = m∙c∙∆T]

Q = (50.00g ± 1.0g) x (4.18J∙g-1∙k-1) x (-13) = -2717J

Conversion of Uncertainty to Relative Error
[(Nearest unit of volume measurement observed ÷ desired volume) x 100 = Percentage error of volume]

(1÷50) x100= 2%

[(Nearest unit of temperature measurement observed ÷ recorded temperature)
x 100 = Percentage error of recorded temperature]
(1÷13) x 100= 7.7%

[Percentage value of volume error + percentage value of temperature error = relative error]

2%+7.7% = 9.7% relative error

Enthalpy change in joules with relative error percentage

Q= -2717J ± 9.7%

Conversion of Relative Error to Absolute Error

(2717÷100) ...

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