Enthalpy Of Neutralization Lab Report

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Enthalpy of neutralization
The purpose of this experiment is to determine the enthalpy change for the reaction between aqueous sodium hydroxide (NaOH) and aqueous hydrochloric acid (HCl).
A neutralization reaction is a chemical reaction where a base and an acid react with each other. The products will always be water and a salt. The balanced reaction equation for this experiment is the reaction below (Enthalpy of neutralization, 2018).
〖NaOH〗_((aq))+〖HCl〗_((aq))→〖NaCl〗_((aq))+H_2 O_((l))
In aqueous solutions the substances that are involved will experience dissociation, which changes the ionization state of the substances (Neutralization, 2018). When an acid is dissolved in water the covalent bond between the electronegative atom
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It is a subcategory of the standard enthalpy of reaction and defined as the energy released with the formation of 1 mole of water (Enthalpy of neutralization, 2018). Standard enthalpy of reaction is the enthalpy change that occurs in a system when matter is transformed by a chemical reaction. All products and reactants must be in their standard state (Standard enthalpy of reaction, 2017). The standard enthalpy change can be found by dividing the heat released during the reaction by the number of moles involved in the reaction (Enthalpy of neutralization,…show more content…
n=cV n=1.00 mol dm^(-3)×(25 dm^3±0.16%)/1000=0.025 mol±0.16 %
The enthalpy of neutralization is then calculated.
∆H=(-1356.5 J±3.104% )/(0.025 mol±0.16%)=-54260 J 〖mol〗^(-1)±3.3 %
∆H=-54260 J m〖ol〗^(-1)±3.3%÷1000=-54 kJ 〖mol〗^(-1)±3.3 %
Another way to calculate the enthalpy of neutralization is to assume the density of the solution to be equal to the density of water in order to assume the mass of the solution.
1.00 g 〖cm〗^(-3)×50 cm^3±0.96 %= 50.00 g ±0.96 %
The heat released can then be calculated using the assumed mass. q=50.00 g±0.96 % ×4.18 J g^(-1) ℃^(-1)×6.40℃±3.1 %=1337.6 J±4.06 %
∆H=(-1337.6 J±4.06 %)/(0.025 mol ±0.16 %)=-53504 J m〖ol〗^(-1)±4.22 %
∆H=-53504 J m〖ol〗^(-1)±4.22 %÷1000=-54 kJ m〖ol〗^(-1)±4.22 %
Conclusion and

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