Lab: Solubility and ThermodynamicsAlexander Kazberouk

Period 5/6 Chemistry APMrs. 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 nitrate2. Add 15 mL of water and heat the test tube in a boiling water bath, stirring until all of the

potassium nitrate dissolved3. 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 appear8. 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 nitrate3. First time, heated to 89.9 degrees Celsius4. Test tube feels cool to touch as potassium nitrate dissolves5. 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˚CCALCULATIONS:

Moles of KNO3

20.00g * (1mol/(39.10g + 14.01g + 16.00g * 3)) = 0.1978mol

Kelvin Temperatures:67.5˚C + 273.2 = 340.7K; 1/T = 2.935 * 10-3K-1

55.2˚C + 273.2 = 328.4K; 1/T = 3.045 * 10-3K-1

46.3˚C + 273.2 = 319.5K; 1/T = 3.130 * 10-3K-1

40.6˚C + 273.2 = 313.8K; 1/T = 3.187 * 10-3K-1

36.2˚C + 273.2 = 309.4K; 1/T = 3.232 * 10-3K-1

32.5˚C + 273.2 = 305.7K; 1/T = 3.271 * 10-3K-1

Concentration at Equilibrium:At 340.7K: [KNO3] = 0.1978mol / 23.9mL * (1000mL/1L) = 8.28MAt 328.4K: [KNO3] = 0.1978mol / 29.9mL * (1000mL/1L) = 6.62MAt 319.5K: [KNO3] = 0.1978mol / 35.0mL * (1000mL/1L) = 5.65MAt 313.8K: [KNO3] = 0.1978mol / 38.9mL * (1000mL/1L) = 5.08MAt 309.4K: [KNO3] = 0.1978mol / 43.9mL * (1000mL/1L) = 4.51MAt 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 = 4.23At 328.4K: K = 6.622 = 43.8; lnK = 3.78 At 319.5K: K = 5.652 = 31.9; lnK = 3.46At 313.8K: K = 5.082 = 25.8; lnK = 3.25At 309.4K: K = 4.512 = 20.3; lnK = 3.01At 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/molAt 328.4K: ∆G = - (8.31 J/ mol K) * (328.4K) * (3.78) * (1kJ/1000J) = -10.3kJ/molAt 319.5K: ∆G = - (8.31 J/ mol K) * (319.5K) * (3.46) * (1kJ/1000J) = -9.19kJ/molAt 313.8K: ∆G = - (8.31 J/ mol K) * (313.8K) * (3.25) * (1kJ/1000J) = -8.48kJ/molAt 309.4K: ∆G = - (8.31 J/ mol K) * (309.4K) * (3.01) * (1kJ/1000J) = -7.74kJ/molAt 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-032 3.78 3.045E-033 3.46 3.130E-034 3.25 3.187E-035 3.01 3.232E-036 2.81 3.271E-03

Solubility Equilibrium Constant vs. Inverse Temperature (ln K vs. 1/T)

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

R2 = 0.996

00.5

11.5

22.5

33.5

44.5

2.900E-03 3.000E-03 3.100E-03 3.200E-03 3.300E-03

1/T (inverse Kelvin)

ln K

Enthalpy Change:∆G = -RTlnK = ∆H - T∆SlnK = -∆H/RT + T∆S/RT = -∆H/R * (1/T) + ∆S/RlnK = -4140(1/T) + 16.4-∆H/R = -4140∆H = (-4140) * - (8.31J/mol K) * (1kj/1000J) = +34.4kJ/mol

Entropy Change:∆G = ∆H - T∆ST∆S = ∆H - ∆G∆S = (∆H - ∆G)/TAt 340.7K: ∆S = (34.4kJ/mol - -12.0kJ/mol) / (340.7K) * (1000J/1kJ) = + 136J / mol K At 328.4K: ∆S = (34.4kJ/mol - -10.3kJ/mol) / (328.4K) * (1000J/1kJ) = + 136J / mol K At 319.5K: ∆S = (34.4kJ/mol - -9.19kJ/mol) / (319.5K) * (1000J/1kJ) = + 136J / mol K At 313.8K: ∆S = (34.4kJ/mol - -8.48kJ/mol) / (313.8K) * (1000J/1kJ) = + 137J / mol K At 309.4K: ∆S = (34.4kJ/mol - -7.74kJ/mol) / (309.4K) * (1000J/1kJ) = + 136J / mol K At 305.7K: ∆S = (34.4kJ/mol - -7.14kJ/mol) / (305.7K) * (1000J/1kJ) = + 136J / mol K

Entropy Change #2:∆S/R = 16.4∆S = 16.4 * 8.31 J/ mol K = 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 KNO3 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 could 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 R2 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.