Enthalpy of Neutralization Lab Report
Name: Partner:
Email:
Purpose : To verify Hess’s law by examining the solvation of NaOH and the reaction of NaOH and HCl both separately and together.
Procedure & apparatus : As given in the lab handout.
Data: Graphs attached at the end of the lab.
Reaction 1:
Volume of water: (mL)
Trial 1
Mass of NaOH solid:
(g)
T initial soln and calorimeter
( ° C)
T final soln and calorimeter
Mass of Soln
Moles of NaOH
∆ T soln and cal
q cal
(kJ)
q soln
q rxn
∆Hrxn
∆ Hrxn/mol (kJ/mole)
Reaction 2:
Volume of
HCl Soln:
(mL)
Mass of
NaOH (g)
T initial
Soln and cal
T final
∆ T
Soln and cal.
Mass of Solution
moles NaOH
Moles HCl
∆ Hrxn2
Molar Enthalpy (∆Hrxn 2) (kJ/mole)
Reaction 3:
Volume HCl (mL)
Volume NaOH (mL)
Concentration of HCl (M)
Concentration of NaOH (M)
T initial Soln and cal. ( ° C)
∆T Soln and cal. ( ° C)
Mass of Soln (g)
Mol NaOH
Mol HCl
q rxn 3
∆Hrxn3 (kJ)
Molar Enthalpy (∆Hrxn3) (kJ/mole)
Calculations : Include all your calculations on a separate sheet of paper.
Discussion and errors . My data is good (or bad) because
Discussion
My errors include..
1. E1
2. E2 3. E34. E4 Modifications :
mods Conclusion : . The ∆ H/mole for the solvation of NaOH (rxn1) is . The ∆ H/mole for the reaction of NaOH and HCl (rxn 3) is . The combined reaction (rxn 2) had an ∆ H/mole of kJ/mol. The sum of the parts (1&3) was close to the combined reaction (2) and showed only a % error.
1. Answer the prelab question 2 for the actual masses and Molarities used. Show all your work on a separate page.
2. The lab states that solid NaOH absorbs moisture from the air. What effect would weighing out “wet” NaOH have on your observed results?
post2 3. Do you have an isolated system? What effect would heat loss to the air have on your calculated ∆H value?
post3 4. What are the two simplifying assumptions that were made in the calculations? Do you think that they are reasonable?
post4