ADDITIVITY OF HEATS OF REACTION
Mandie Caya and Courtney Roth Hour 1
Objective
1. To investigate energy changes which occur during a chemical reaction.
2. To determine the heat of reaction for three separate reactions.
3. To illustrate the additivity of reactions and their heats.
4. To illustrate the Law of Conservation of Energy.
Procedure
Given
Analysis
Table 1
Mass of NaOH pellets 2.0 g
Temperature of water (t1) 23.0 degrees C
Volume of water 200 mL
Mass of water (assume density=1 g/mL 200 g
Total mass of solution 202.0 g
Temperature of solution after mixing (t2) 25.5 degrees C
Heat absorbed by solution (assume specific heat 505.0 Q
capacity=1.0 cal/g deg.)
Moles of NaOH used .0502 moles
Heat evolved per mole of NaOH 10000 cal
Heat of solution for NaOH ( H1) -10000 kJ
Table 2
Temperature of HCl solution (t1) 23.0 degrees C
Volume of HCl solution 110.0 mL
Mass of HCl solution (assume D=1.0g/mL 110.0 g
Mass of NaOH pellets 2.0 g
Total mass of solution after reaction (Item 3 + Item 6) 112.0 g
Temperature of solution after reaction (t2) 30.0 degrees C
Temperature change (t2 – t1) 7.0 degrees C
Heat absorbed by solution (assume specific heat 780 Q
capacity=1.0 cal/g deg.)
Moles of NaOH .0502 moles
Heat evolved per mole of NaOH 16000 cal
Heat of reaction ( H2) -16000 kJ
Table 3
Temperature of acid and base solution (t1) 24.0 degrees C
Volume of acid solution 50.0 mL
Mass of acid solution (assume D=1.0 g/mL) 50.0 g
Volume of base solution 50.0 mL
Mass of base solution (assume D=1.0 g/mL) 50.0 mL
Total mass of NaCl solution after reaction (Item 3 + Item 100.0 g
5)
Temperature of NaCl solution (t2) 27.5 degrees C
Temperature change (t2 – t1) 3.5 degrees C
Heat absorbed by solution (assume specific heat 350 Q
capacity=1.0 cal/g deg)
Molar concentration of base solution .5 M
Moles of NaOH (.5 M x .0501) .025 moles
Heat evolved per mole of NaOH 14000 cal
Heat of neutralization ( H3) -14000 kJ
Calculations
For Table 1:
202.0 g x 1.0 x 2.5 = 505 Q
2.0 g NaOH 1 mole NaOH = .0502 moles NaOH
40.01 g NaOH
505 Q = 10059 = 10000 calories
.0502 moles
For Table 2:
112.0 g x 1.0 x 7.0 = 784 = 780 Q
2.0 g NaOH 1 mole NaOH = .0502 moles NaOH
40.01 g NaOH
780 Q = 15537 = 16,000 calories
.0502 moles NaOH
For Table 3:
100.0 g x 1.0 x 3.5 = 350 Q
350 Q = 14000 calories
.025 moles
Follow-Up Questions
1. 1: NaOH(s) Na+(aq) + OH-(aq) + Q1 H=-10000
2: NaOH(s) + H+(aq) + Cl-(aq) H2O + Na+(aq) + Cl-(aq) + Q2 H=-16000
Find:
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) H2O + Na+(aq) + Cl-(aq) + Q3
H=?
Reverse 1
Na+(aq) + OH-(aq) + Q1 NaOH(s) H= 10100
Add 1 and 2
Na+(aq) + OH-(aq) + Q1 +NaOH(s) + H+(aq) + Cl-(aq) NaOH(s) + H2O +
Na+(aq) + Cl-(aq) + Q2
H=Q3=10,000 + -16,000 = -6000
2. –10,000 + -14000 = -24,000
actual – experimental / actual x 100
-24,000 – (-16,000)/ -24,000 x 100 = 33.3 = 33 %
3.
NaOH (s)
H1 = -10,000
Na+(aq) + OH-(aq) + Cl-(aq) + H+ exothermic
H2 = -16,000
H2O + Na+(aq) + Cl-(aq) H3 = -14,000
4. The results illustrate the Law of Conservation of Energy because it shows
that when you add to substances the temperature will go up without you
provoking that result it just does it on its own. Energy cannot be created
or destroyed so the additivity of the energy is a constant.
5. (2 C – 1.9 C) / 2 C x 100 = 5.0 %
6. You may have read the thermometer to soon and the temperature may have
still been changing. You may have read the thermometer wrong. You
may have measured some of the substances wrong.
7. a) H+ + OH- H2O
b) HC2H3O2 H+ + C2H3O2
c) 2H+ + OH- + C2H3O2- HC2H3O2 + H2O
d) NaOH(aq) + HC2H3O2(aq) NaC2H3O2(aq) + H20
Na+(aq) + OH-(aq) + H+(aq) + C2H3O2(aq) Na+(aq) + C2H3O2(aq) + H20(l)
OH-(aq) + H+(aq) H20(l)
8. Part 1 – (-54.957 + -57.28) – (-425.6) = 313.363
Part 2 – (-285.830 + -57.28 + -40.023) – (-425.6 + 0 + -40.023) = 82.49
Part 3 – (-285.830 + -57.28 + -40.023) –
(-57.28 + -54.957 + 0 + -40.023) = -230.873