Lab, Solubility and Thermodynamics

Lab: Solubility and Thermodynamics Thermodynamics Alexander Kazberouk  Period 5/6 Chemistry AP Mrs. Murphree

TITLE: Solubility and Thermodynamics PURPOSE: The purpose of the lab was to determine the thermodynamics variables of ∆H, ∆S, and ∆G for  the dissolution reaction of potassium nitrate in water. The solubility of potassium nitrate in mol/L was measured over a range of various temperatures by finding out at what temperature crystallization began for solutions of different molarities. Then, the equilibrium constant was calculated and a graphical relationship between the natural logarithm of the equilibrium constant and the inverse of the temperature gave a linear plot that allowed the determination of Gibbs free energy and enthalpy changes associated with the reaction. From there, the entropy change associated with the reaction was determined. PROCEDURE: 1. Weigh out 20.0g of potassium nitrate and transfer it to a 25x200mL test tube. Do not ingest the  potassium nitrate 2. Add 15 mL of water and heat the test tube in a boiling water bath, stirring until all of the  potassium nitrate dissolved 3. Determine and record the volume of the potassium nitrate solution. Use a second large 25x200mm test tube and fill it with an equal volume of water. Then, measure that volume in a graduated cylinder  4. Remove the test tube with the potassium nitrate from the water bath and allow it to cool while slowly stirring the solution with a thermometer  5. Record the temperature when crystals first appear. It is assumed that at this temperature, the system is at equilibrium and it is possible to calculate the concentration of the ions 6. Add 5mL of water to the test tube and warm mixture in a hot water bath until all solid dissolves. Determine the solution volume as before and record it 7. Cool the solution slowly and record the temperature at which crystals first appear  8. Repeat the cycle of adding water/heating/cooling/recording temperature until crystals appear near  room temperature. Add 5mL of water each time

DATA & OBSERVATIONS: 1. 20.00g of potassium nitrate taken, white solid spheres (fairly large sized spheres) 2. Initially, add 10.01mL + 4.98mL of water to potassium nitrate 3. First time, heated to 89.9 degrees Celsius 4. Test tube feels cool to touch as potassium nitrate dissolves 5. Crystals form rapidly after some point when temperature drops, easy to see

Trial

Volume

Temperature

1

23.9mL

67.5˚C

2

29.9mL

55.2˚C

3

35.0mL

46.3˚C

4

38.9mL

40.6˚C

5

43.9mL

36.2˚C

6

48.5mL

32.5˚C

CALCULATIONS:

Moles of KNO3 20.00g * (1mol/(39.10g + 14.01g + 16.00g * 3)) = 0.1978mol Kelvin Temperatures: 67.5˚C + 273.2 = 340.7K ; 55.2˚C + 273.2 = 328.4K ; 46.3˚C + 273.2 = 319.5K ; 40.6˚C + 273.2 = 313.8K ; 36.2˚C + 273.2 = 309.4K ; 32.5˚C + 273.2 = 305.7K ;

1/T = 2.935 * 10-3K -1 1/T = 3.045 * 10-3K -1 1/T = 3.130 * 10-3K -1 1/T = 3.187 * 10-3K -1 1/T = 3.232 * 10-3K -1 1/T = 3.271 * 10-3K -1

Concentration at Equilibrium: At 340.7K: [KNO3] = 0.1978mol / 23.9mL * (1000mL/1L) = 8.28M At 328.4K: [KNO3] = 0.1978mol / 29.9mL * (1000mL/1L) = 6.62M At 319.5K: [KNO3] = 0.1978mol / 35.0mL * (1000mL/1L) = 5.65M At 313.8K: [KNO3] = 0.1978mol / 38.9mL * (1000mL/1L) = 5.08M At 309.4K: [KNO3] = 0.1978mol / 43.9mL * (1000mL/1L) = 4.51M At 305.7K: [KNO3] = 0.1978mol / 48.5mL * (1000mL/1L) = 4.08M Equilibrium Constant: K = [NO3-][K +] and [NO3-] = [K +] = concentration of KNO3 at equilibrium. K = [KNO3]2 At 340.7K: K = 8.282 = 68.6; lnK = 4.23 At 328.4K: K = 6.622 = 43.8; lnK = 3.78 At 319.5K: K = 5.652 = 31.9; lnK = 3.46 At 313.8K: K = 5.082 = 25.8; lnK = 3.25 At 309.4K: K = 4.512 = 20.3; lnK = 3.01 At 305.7K: K = 4.082 = 16.6; lnK = 2.81 Gibbs free energy: ∆G = -RTlnK = - (8.31 J/ mol K) * T * lnK * (1kJ/1000J) At 340.7K: ∆G = - (8.31 J/ mol K) * (340.7K) * (4.23) * (1kJ/1000J) = -12.0kJ/mol At 328.4K: ∆G = - (8.31 J/ mol K) * (328.4K) * (3.78) * (1kJ/1000J) = -10.3kJ/mol At 319.5K: ∆G = - (8.31 J/ mol K) * (319.5K) * (3.46) * (1kJ/1000J) = -9.19kJ/mol At 313.8K: ∆G = - (8.31 J/ mol K) * (313.8K) * (3.25) * (1kJ/1000J) = -8.48kJ/mol At 309.4K: ∆G = - (8.31 J/ mol K) * (309.4K) * (3.01) * (1kJ/1000J) = -7.74kJ/mol At 305.7K: ∆G = - (8.31 J/ mol K) * (305.7K) * (2.81) * (1kJ/1000J) = -7.14kJ/mol Trial

ln K

1/T

1

4.23

2.935E-03

2

3.78

3.045E-03

3

3.46

3.130E-03

4

3.25

3.187E-03

5

3.01

3.232E-03

6

2.81

3.271E-03

Solubility Equilibrium Constant vs. Inverse Temperature (ln K vs. 1/T) 4.5 4 3.5 3    K 2.5   n    l 2 1.5 1 0.5 0 2.900E-03

ln K = -4140(1/T) + 16.4 2

R = 0.996

3.000E-03

3.100E-03

3.200E-03

3.300E-03

1/T (inverse Kelvin)

Enthalpy Change: ∆G = -RTlnK = ∆H - T∆S lnK = -∆H/RT + T∆S/RT = -∆H/R * (1/T) + ∆S/R  lnK = -4140(1/T) + 16.4 -∆H/R = -4140 ∆H = (-4140) * - (8.31J/mol K) * (1kj/1000J) = +34.4kJ/mol Entropy Change: ∆G = ∆H - T∆S T∆S = ∆H - ∆G ∆S = (∆H - ∆G)/T At 340.7K: ∆S = (34.4kJ/mol - -12.0kJ/mol) / (340.7K) * At 328.4K: ∆S = (34.4kJ/mol - -10.3kJ/mol) / (328.4K) * At 319.5K: ∆S = (34.4kJ/mol - -9.19kJ/mol) / (319.5K) * At 313.8K: ∆S = (34.4kJ/mol - -8.48kJ/mol) / (313.8K) * At 309.4K: ∆S = (34.4kJ/mol - -7.74kJ/mol) / (309.4K) * At 305.7K: ∆S = (34.4kJ/mol - -7.14kJ/mol) / (305.7K) * Entropy Change #2: ∆S/R = 16.4 ∆S = 16.4 * 8.31 J/ mol K = 136J/mol K 

(1000J/1kJ) = +136J / mol K  (1000J/1kJ) = +136J / mol K  (1000J/1kJ) = +136J / mol K  (1000J/1kJ) = +137J / mol K  (1000J/1kJ) = +136J / mol K  (1000J/1kJ) = +136J / mol K 

RESULTS TABLE: Temperature (K)

Solubility (mol/L)

∆H (kJ/mol)

∆S (J/mol K)

∆G (kJ/mol)

340.7

8.28

34.4

136

-12.0

328.4

6.62

34.4

136

-10.3

319.5

5.65

34.4

136

-9.19

313.8

5.08

34.4

137

-8.48

309.4

4.51

34.4

136

-7.74

305.7

4.08

34.4

136

-7.14

CONCLUSION: The purpose of the laboratory exercise was to calculate the enthalpy change, the Gibbs free energy change, and the entropy change for the dissolution reaction of potassium nitrate in water. The solubility of potassium nitrate was measured in moles/liter for several temperatures. This was done by dissolving a known quantity of potassium nitrate in water, heating it until all dissolved, and letting it cool until crystals began forming soon after the equilibrium temperature was reached. The solution was then diluted to change its molarity, heated again if needed, cooled, and the new equilibrium temperature was determined. From this, the equilibrium constant at each temperature was gotten and the entropy and enthalpy changes were gotten from a graphical relationship between the natural logarithm of the equilibrium constant and the inverse of the temperature. The Gibbs free energy was gotten from a relationship between the equilibrium constant and temperature at equilibrium and the entropy values were verified through the Gibbs free energy equation. In order to get any of the values, it was first necessary to find the molar solubility of   potassium nitrate at various temperatures. A 20.00g amount of solid white sphere-shaped  potassium nitrate was taken and dissolved in water. The reaction was rather endothermic and the test tube felt cold. The solution was heated in a water bath until all of the potassium nitrate dissolved and the volume of the solution was determined by comparing it to a test tube with pure

water. The solution was then allowed to cool and the temperature at which crystals began forming was taken. Then, about 5mL of extra water were added to the solution, the solution was heated until all KNO3 dissolved again, the new volume was measured, and the temperature at which crystals formed was recorded. This time, the temperature where crystals formed was lower   because now there was more water and thus a lower molarity of the solution at equilibrium. All this was done four more times for a total of 6 trials and 6 different temperatures at equilibrium, ranging from 340.7K to 305.7K (see table). At this point, it was assumed that the temperature at which crystals become visible is equal to the equilibrium temperature and that the activities of  the ions played no role in all measurements. Knowing the volume of the solution at each of those trials and the original amount of potassium nitrate, it was possible to determine the molarity at each of the equilibrium temperatures and thus the molar solubility of KNO3 at each temperature (see table). The equilibrium constant for the dissolution of KNO 3 was equal to [K +][NO3-] and that because of 1:1 stoichiometry it was also equal to the molarity of KNO3 at equilibrium squared. Thus, the equilibrium constant at each equilibrium was gotten from the molar solubility. As temperature decreased, the equilibrium constant for the dissolution of potassium nitrate also decreased from 68.6 to 16.6 and reactants were more favored. This made sense, considering that the reaction was endothermic and thus needed heat to go towards the reactants. Knowing the equilibrium constant at each temperature, it was possible to determine Gibbs free energy by using the equation ∆G = -RTlnK where R was a constant, T was temperature, and K was the equilibrium constant. The Gibbs free energy for the reaction at all temperatures was negative because K was very large and thus the reaction proceeded far towards complete dissolution. The Gibbs free energy for the reaction ranged from -12.0kJ/mol at the highest temperature to -7.14kJ/mol near room temperature. Again, this made sense because at high temperatures, the equilibrium constant was larger for the endothermic reaction, there was a

larger amount of KNO3 dissolved in water, and the reaction was “more” spontaneous. From the two possible ways to calculate ∆G (∆G = -RTlnK = ∆H - T∆S), a linear relationship between enthalpy, entropy, the natural logarithm of the equilibrium constant, and the temperature was gotten, lnK = -∆H/R * (1/T) + ∆S/R. When the various equilibrium constants and their  respective constants were graphed, the relationship lnK = -4140(1/T) + 16.4 was gotten. From the slope of this line, -4140, the enthalpy change for the reaction was calculated to be +34.4kJ/mol. The enthalpy change did not depend on the temperature. It made sense that the enthalpy change was positive as the reaction was originally observed to be clearly endothermic.  Now, knowing the Gibbs free energy change, the temperature at each equilibrium, and the enthalpy change for the reaction it was possible to determine the entropy change for the reaction from ∆G = ∆H - T∆S. Although there was one minor variation, the entropy change for the reaction was 136J/mol K. The highly positive entropy change made sense because the reaction was spontaneous and endothermic. For the reaction to be both, it needed to have a positive entropy change. Additionally, the fact that a single molecule of potassium nitrate was broken up into smaller ions also fit in well with a positive entropy. Entropy co uld also be calculated from the y-intercept of the graph described above. The calculation there gave a matching value of  136J/mol K and it made sense that entropy was not dependent on temperature. Overall, the reaction was spontaneous, endothermic, had a positive enthalpy value, and a positive entropy value. As the temperature increased, the reaction got more spontaneous, the equilibrium constant increased, and Gibbs free energy got more negative. Without the actual values, error analysis is somewhat difficult. The R 2 value for the graph had a value a .996 out of a maximum of 1. Thus, the points matched a straight line very well. All the values for entropy were very close to each other and the values for the equilibrium constant and Gibbs free energy all matched the appropriate pattern. Minor errors such as misreading the

volume of the solutions and losing water to evaporation likely amounted to the experiment but did not play a significant role in the actual error. The interactions of ions and the ionic strengths of the solutions were ignored in the experiment, but it is unlikely that they had much of an influence. The main problem with the lab was that the temperature at which we observed crystal formation was lower than the temperature at which real equilibrium was established. This is  because visible crystal formation means that equilibrium has been passed and there is no way for  the human eye to see exactly when the very first crystals start forming. This delay resulted in a lower reported temperature and thus a less negative Gibbs free energy and a higher than actual entropy change. The enthalpy change did not depend on this much as it was the slope of a line, not its spatial position. Graphically, this meant that the line shifted to the right of what it should have been and at each equilibrium point, the temperature was actually lower and 1/T was actually higher. Thus, the experiment had a higher y-intercept and a higher entropy than there was in the actuality.