# Enthalpy of Heat

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Determination of Heats of Solution and Reaction

Data and Results

The following table shows the results and data for the dissolution enthalpy of salts.

NaC2H¬3¬O2 Dissolution Enthalpy Data (Table 1)

Trial Salts Mass (g) Mass (mol) Tinitial (°C) Tfinal (°C) ΔT (°C) qwater (J) ΔHwater (J/mol) ΔHaverage (J/mol)

1 NaC2H3O2 4.088 0.05 21.7 24.7 3.0 -627.9 -12608.4 -12435.3

2 NaC2H3O2 4.103 0.05 21.6 24.5 2.9 -606.9 -12139.4

3 NaC2H3O¬2 4.105 0.05 22.9 24.9 3.0 -627.9 -12558.0

KClO4 Dissolution Enthalpy Data (Table 2)

Trial Salts Mass (g) Mass (mol) Tinitial (°C) Tfinal (°C) ΔT (°C)

qwater (J) ΔHwater (J/mol) ΔHaverage (J/mol)

1 KClO4 6.933 0.05 22.0 20.6 -1.4 293.02 58552.72 57184.23

2 KClO4 6.921 0.05 21.5 20.2 -1.3 272.09 54447.25

3 KClO4 6.933 0.05 21.8 20.4 -1.4 293.02 58552.72

Using the final ΔH for the reaction of NaC2H3O2 with H2O we can determine whether the reaction was exothermic or endothermic. Since the ΔH of NaC2H3O2 is negative, the reaction was exothermic. When NaC2H3¬O2 was added to H¬2O the NaC2H3O2 will become hydrated causing the NaC¬2H3O2 to break apart.

NaC2H3O2(s)+H2O(l)Na+(aq)+C2H3O2-(aq)

Using the final ΔH for the reaction of KClO4 with H2O we can determine whether the reaction was exothermic or endothermic. Since the ΔH of KClO4 is positive, the reaction was endothermic. When KClO4 is added to H2O the KClO4 will become hydrated causing the KClO4 to break apart.

KClO4 (s)+H2O(l)K+(aq)+ClO4-(aq)

A sample calculation for the enthalpy NaC2H¬3¬O2 is presented below.

Trial 1 (Table 1)

ΔT = T final – T initial

ΔT = 24.7 °C – 21.7 °C

ΔT = 3.0 °C

-qwater = -mwaterswaterΔTwater

-qwater = -50g*(4.186J/g°C)*(3.0)

-qwater = -627.9J

ΔH = q/n

ΔH = -627.9J/0.0498 mol NaC2H¬3¬O2

ΔH = 15062J/mol

The following table shows the neutralization enthalpy an acid/base.

KOH + HCl Neutralization Enthalpy Data (Table 3)

Trial KOH mL HCl mL KOH mol HCl mol Tinitial Tfinal ΔT (°C) qsystem (J) ΔHsystem (J/mol) ΔHaverage (J/mol)...