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Determining the Thermodynamics of Ca(OH)2 Dissolution in Water Names: Day and time of lab: Wednesday 01:00pm-03:00pm Date Experiment Performed: February 18 2020

Determining the solubility of Ca(OH)2 in room temperature water Temperature of the “room temperature” solution (21oC) Trial 1

Trial 2

10.00

10.00

10.00

Initial reading on buret, mL

0.10

0.00

0.00

Final reading on buret, mL

17.5

17.2

17.3

Volume of HCl used in titration, mL

17.4

17.2

17.3

[OH –] in room temperature saturated solution, M

4.23 × 10−2

4.40 × 10−2

4.26 × 10−2

Molar solubility (s) of Ca(OH)2 at room temperature, M

2.12 × 10−2

2.2 × 10−2

2.13 × 10−2

Volume of filtered Ca(OH)2 used in titration, mL

Average molar solubility of Ca(OH)2 at room temperature, M

2.15 × 10−2

Ksp for Ca(OH)2 at room temperature

3.97 × 10−5

ΔGofor the dissolution of Ca(OH)2 at

24.8

Trial 3

room temperature, kJ.mol-1

1

Determining the Thermodynamics of Ca(OH)2 Dissolution in Water Determining the solubility of Ca(OH)2 in boiling water Temperature of the boiling solution (98.1oC) Trial 1 Volume of filtered Ca(OH)2 used in titration, mL Initial reading on buret, mL

Trial 2

Trial 3

10.00

10.00

10.00

0.00

13.55

22.91

Final reading on buret, mL

9.43

22.91

31.98

Volume of HCl used in titration, mL

9.43

9.36

9.07

[OH –] in boiling saturated solution, M

2.29 × 10−2

2.25 × 10−2

2.21 × 10−2

Molar solubility (s) of Ca(OH)2 in boiling water, M

1.15 × 10−2

1.13 × 10−2

1.11 × 10−2

Average molar solubility of Ca(OH)2 in boiling water, M

1.13 × 10−2

Ksp for Ca(OH)2 at boiling

5.70 × 10−6

ΔGofor the dissolution of Ca(OH)2 in the

37.3

boilingsolution, kJ.mol-1

Other thermodynamic quantities: ΔSo, kJ.mol-1.K-1 ΔS = − 0.162kJ mol·K

ΔHo, kJ.mol-1 ΔH = − 22.8kJ mol

Example Calculations: 1. Show how you found the value of [OH –] in the room temperature solution. [OH ] M= −Lmol 17.4mL HCL × = 0.0174L HCL

1L

1000mL −

x

0.0243M OH =

0.0174L HCL −

0.0243M OH =x × 0.0174L HCL x = 0.000423mol OH− − 1000mL OH 10.00mL × = 0.0100L 1L 0.000423mol

OH−

x= 0.0100L x= 4.23 × 10−2 M OH− 2. Show how you found the molar solubility of Ca(OH)2 in the room temperature solution. −2 4.23 x 10−210 2

Molar Solubility of Ca(OH) = 2.12 x M Ca(OH) 2 = 2 3. Show how you found the Ksp value for Ca(OH)2 at room temperature. − −2 = Ave of [OH ] 4.30 x 10 Ksp = [OH ] 21 − 3eq Ksp = (4.30 x 10 ) 21 −2 3eq Ksp = 3.97 x 10−5 4. Show how you found the ΔGo value for the dissolution of Ca(OH)2 at room temperature. ΔG = − RTlnK kJ

−5

ΔG = − (0.008314 )(294K)(ln 3.97 x 10 ) K·mol kJ

ΔG = 24.8

mol

5. Show how you found ΔHo and ΔSofor the dissolution of Ca(OH)2. ΔS =ΔG 1− ΔG 2 ΔT 2 − ΔT 1 molkJ kJ

(24.771 − 37.255 )

ΔS = (371.1K − 294K) mol

ΔS = − 0.162kJ mol·K

ΔH = ΔG + TΔS mol + kJ

0kJ

ΔH = (24.8 ) (294K)(− .162 ) ΔH = − 22.8kJ mol mol·K...