Enthalpy Of Heat Essay

680 words - 3 pages

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 ...view middle of the document...

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)
1 50 50 0.05 0.05 -2134.86 -42697.2 -42974.6
2 50 50 0.05 0.05 -2176.72 -43534.4
3 50 50 0.05 0.05 -2134.86 -42692.2

Using the final ΔH for the reaction of KOH and HCL we can determine whether the following reaction was exothermic or endothermic. Since the ΔH of KCl is negative it can be concluded that the reaction was exothermic. The acid/base in the reaction are strong acid/base. Since KOH and HCl are strong they will completely dissociate and water will from as a product.
KOH(aq)+HCl(ar)KCl(aq)+H2O(l)
A sample calculation for the enthalpy of KOH and HCl is presented below.
Trial 1 (Table 3)
ΔT = Tfinal – T¬initial
ΔT =
ΔT = °C
-qwater = -mwaterswaterΔTwater
-qwater = (-100g)(4.186J/g°C)( °C)
-qwater = -2134.86J
ΔH = q/n
ΔH = -2134.86J/.05mol
ΔH = -42697.2J/mol

Discussion
If we used half the amount of water to dissolve our samples, the measurements and results for the enthalpy of the solution would be different. The ΔH and the q values would be divided in half. A problem that might arise if the amount of water used to dissolve our samples is that the samples might not dissolve completely.

Since nitric acid and hydrochloric acid are both strong acids completely ionize when dissolved in water, the results should be the same. The net ionic for the reaction of nitric acid and potassium hydroxide is H2++O2-H2O. The net ionic for the reaction of hydrochloric acid and potassium hydroxide is H2++O2-H2O.

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