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Hess’ Law
Purpose:
The purpose of this lab was to prove Hess’ Law.
Data:
Part I
Mass of NaOH
Moles of NaOH
Mass of Water
Initial Temperature of Water
Final Temperature of Water
Change in Temperature of Water
Heat absorbed by Water (cal)
Mass of Flask
Heat absorbed by Flask (cal)
Total Heat absorbed (cal)
Heat released by Reaction (cal)
Heat released per mole of Reaction (cal/mole)
Part II
Mass of NaOH
Moles of NaOH
Mass of Solution
Initial Temperature of Solution
Final Temperature of Mixture
Change in Temperature
Heat absorbed by Solution (cal)
Mass of Flask
Heat absorbed by Flask (cal)
Total Heat absorbed (cal)
Heat released by Reaction (cal)
Heat released per mole of NaOH (cal/mole)
Part III
Mass of NaOH
Moles of NaOH
Mass of Solution
Initial Temperature of Solution
Final Temperature of Solution
Change in Temperature of Solution
Heat absorbed by Solution (cal)
Mass of Flask
2.00 g
0.0500 moles
194.49 g
20.6 oC
23.6 oC
3.0 oC
583 cal
122.71 g
73.6 cal
657 cal
657 cal
13.1 kcal/mole
2.01 g
0.0503 moles
195.72 g
23.0 oC
29.4 oC
6.4 oC
1250 cal
122.71 g
157 cal
1410 cal
1410 cal
28.0 kcal/mole
2.04 g
0.0510 moles
199.05 g
23.0 oC
27.0 oC
4.0 oC
796 cal
122.71 g
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Heat absorbed by Flask (cal)
Total Heat absorbed (cal)
Heat released by Reaction (cal)
Heat released per mole of NaOH (cal/mole)
Part IV
Heats of Reaction for each part
∆HPart I
∆HPart II
∆HPart III
Proving Hess’ Law
Part I + Part III
Part II
Percent Error between Part II and Parts I and III
98 cal
894 cal
894 cal
17.5 kcal/mole
13.1 kcal/mole
28.0 kcal/mole
17.5 kcal/mole
30.6 kal/mole
28.0 kcal/mole
9.29 %
Calculations:
The following calculations are for Part I. Similar calculations were used for Parts II and III.
Moles of NaOH:
(Mass of NaOH) (1 mole / molar mass of NaOH)
(2.00 g) (1 mole / 40 g)
= 0.0500 moles
Change in Temperature of Water:
(Final Temp. of Water – Initial Temp. of Water)
23.6 oC – 20.6 oC
= 3.0 oC
Heat Absorbed by Water (cal):
Q=mc∆T
Q = (194.49 g) (1 cal/g·oC) (3.0 oC)
= 583 cal
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Heat Absorbed by Flask (cal):
Q=mc∆T
Q = (122.71 g) (0.2 cal/g·oC) (3.0 oC)
= 73.6 cal
Total Heat absorbed:
Heat Absorbed by Water + Heat Absorbed by Flask
583 cal + 73.6 cal
= 657 cal
Heat released per mole of Reaction (cal/mole):
Total Heat absorbed = x calories
Moles of NaOH
1 mole
657 cal
0.0500 moles
=
x calories
1 mole
657 = 0.0500x
x = 13100 cal/mole
x = 13.1 kcal/mole
The following are calculations for Part IV.
Part I + Part III:
∆HPart I + ∆HPart III
13.1 kcal/mole + 17.5 kcal/mole
= 30.6 kcal/mole
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Percent Error:
((((Part I + Part III) – Part II) / Part II) x 100
(((30.6 kcal/mole) – 28.0 kcal/mole) / 28.0 kcal/mole) x 100
= 9.29 %
Results:
The enthalpies of Part I and Part III were found to be 13.1 kcal/mole and 17.5 kcal/mole
respectively, which led to a total enthalpy of 30.6 kcal/mole. The enthalpy of Part II was found
to be 28.0 kcal/mole. This gave a 9.29 % error.
Conclusion:
Hess’ Law was proved using three parts. Part I dissolved NaOH in H2O to form an
aqueous solution of ions. (NaOH(s)  Na+(aq) + OH-(aq) + 13.1 kcal/mole). This gave an enthalpy
of 13.1 kcal/mole. Part II reacted NaOH with HCl to form water and an aqueous solution of
sodium chloride. (NaOH(s) + H+(aq) + Cl-(aq)  H2O + Na+(aq) + Cl-(aq) + 28.0 kcal/mole). This
gave an enthalpy of 28.0 kcal/mole. Part III reacted an aqueous solution of NaOH with HCl to
form water and an aqueous solution of NaCl. (Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq)  H2O + Na+(aq)
+ Cl-(aq) + 17.5 kcal/mole). This gave an enthalpy of 17.5 kcal/mole. By adding Parts I and III
(30.6 kcal/mole) and comparing it to Part II (28.0 kcal/mole), Hess’ Law was proven. This gives
a 9.29 % error between the two values.
The value for Parts I and III was higher that that of Part II. This high value can be
attributed to several factors, including factors such as the temperature being too high. The
temperature being too high can be a result of the heat not being evenly distributed throughout the
solution. The high temperature values affect results by making them have a higher value. Also,
high values can be attained through loss of solution, which may be from some solution being left
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over in a flask or beaker after being poured. This means that there is more mass being accounted
for than what really is there. This also affects results by giving them a higher value.