CHEM 110-Fall 2001-Debye Dissolving KNO3: Entropy or Energy Driven? BACKGROUND: The solubility of potassium nitrate, defined by the relative amounts of KNO3 and water required to make a saturated solution, is described by the chemical reaction: KNO3(s) --> K+(aq) + NO3-(aq) If s moles of KNO3 are required to make a saturated solution in 1 L of water, we can describe the solubility of KNO3 as s moles per liter of water. The volumes of solute and solvent are found not to be additive when making aqueous solutions of ionic compounds. Instead, the volume of the solution is very close to just the volume of water used to make the solution. The ions from the solute appear to fit into the empty spaces between the water molecules. Making use of this good approximation, the concentrations (in moles per liter of solution) of the K+ and NO3- ions in a saturated solution are equal to the solubility (in moles per liter of water) of KNO3: [K+]eq = s and [NO3-]eq = s The value of the solubility product varies with temperature as predicted by LeChatelier’s Principle. An endothermic (ΔHsoln>0) dissolution process would absorb heat from the surroundings heat + KNO3(s) --> K+(aq) + NO3-(aq) and the solubility of the salt would increase at higher temperatures. But an exothermic (ΔHsoln<0) dissolution solution process releases heat to the surroundings KNO3(s) --> K+(aq) + NO3-(aq) + heat and cause the solubility to decrease at higher temperatures. In this way, the sign of the heat of solution determines whether the solubility of KNO3 should increase or decrease with increasing temperature. Stated the other way around, the observed temperature variation of the solubility if of KNO3 indicates if the dissolution process of KNO3 in water is an endothermic or an exothermic process. In this experiment, you will determine the solubility of potassium nitrate at 5 different temperatures in the range 25–80 oC by dissolving a known amount of KNO3 in 20.0 mL of hot water and then letting the solution cool until the KNO3 just begins to crystallize and precipitate from the solution. At this particular temperature, you have a saturated solution of known composition, defining the solubility of the KNO3 at this temperature. In order to explain (or predict) the solubility of a substance in water, one must consider both energy effects and structure effects. The energy effects are summarized in the value of the heat of solution (ΔHsoln) and are related to the relative magnitudes of the forces between water molecules, between the two ions in solution and their surrounding water molecules (hydration spheres), and the forces between the ions in the crystal lattice of the undissolved salt. The structure effects are summarized in the value of the entropy of solution (ΔSsoln) and are related to the differences in the structural disorder between the separated solid KNO3 and water, and the solution resulting when the KNO3 is dissolved in the water. If the solution causes an overall increase in disorder, then ΔSsoln is positive. Remember that entropy expresses disorder and not order. By observing and recording how the solubility of KNO3 changes with temperature, you not only can calculate the solubility product and but also can separate out the relative importance of the energy (ΔHsoln) and structure (ΔSsoln) effects. The Gibbs free energy change is calculated at each temperature from the observed solubility ΔGsoln = -RT Ln s2 where T is in Kelvin, R is 8.314 J mol-1 K-1, and s is in moles/liter. Remember that ΔGsoln<0 favors lots of product (high solubility) and ΔGsoln>0 favors lots of reactant (low solubility). There is a simple relationship between the Gibbs free energy of solution, the heat of solution, and the entropy of solution which balances off these energy effects and structure effects in determining the value of ΔGsoln : ΔGsoln = ΔHsoln – T ΔSsoln The temperature (T in kelvin) in this basic relationship acts as a scaling factor in balancing the entropy change and the enthalpy change. Note that at very low temperatures (T->0), the term TΔSsoln becomes very small so that ΔGsoln = ΔHsoln (very low temperature) resulting in the general statement that reactions are energy driven at low temperatures. But at very high temperatures, – T ΔSsoln eventually always becomes much larger than ΔHsoln so that ΔGsoln = – T ΔSsoln (very high temperature) and we can make the generalization that chemical reactions are entropy driven at high temperatures. A careful look at the above equation shows that it is the equation of a straight line y = a + bx if you graph your values of (ΔGsoln) vs. T. The entropy of solution (ΔSsoln in J K-1 mol-1) then equals the negative slope of the line and the heat of solution (ΔHsoln in J mol-1) is equal to its y-intercept. EXPERIMENTAL: 1. Weigh 11.5-12.5 g of KNO3 to the nearest milligram directly into a clean, dry 50mL beaker. 2. Using a buret, add 20.0 mL deionized water to the beaker. Gently place a magnetic stir bar and a small boiling chip into the beaker. 3. Place the beaker on a stirrer/hot plate and position the temperature probe of a pH meter into the water so that its tip touches neither the beaker nor the stir bar. 4. Carefully turn on the magnetic stirrer to gently stir the mixture. CAUTION: If the stirring is too vigorous, some solution will splash out of the beaker and you will have to start over. 5. Set the heater control to Setting 4 or 5 and gently heat the contents of the beaker until all of the KNO3 is dissolved. Do not overheat. Below are guidelines for the temperatures at which the salt should dissolve. (If you overheat the solution, you will suffer through excessive cooling times.) Total KNO3 Heat to: 12 g 40 oC 16 50 20 60 24 70 28 75 6. Turn off the heat, place an insulating pad composed of seven paper towels doubly wrapped in aluminum foil between the hot plate and beaker, and allow the solution to cool. 7. Carefully watch the solution and, using your electronic thermometer, determine the temperature at which the solid KNO3 begins to crystallize out of the solution. 8. Repeat Steps 3 through 7 four times, each time adding an additional 3.5-4.5 g KNO3 (weighed to the nearest milligram) to your solution from the previous trial. REPORT: 1.For each temperature, calculate the solubility (s) of potassium nitrate as moles KNO3 per liter of water and the Gibbs free energy of solution (ΔGsoln). 2. Make a graph of the solubility vs. temperature (in oC) and fit the points with the best curve. This probably will not be a straight line. Report the equation of your fitted curve and interpret the general appearance of your graph to determine if the dissolving of KNO3 in water is an endothermic or an exothermic process. 3. Make a graph of ΔGsoln vs. T (in kelvin) and fit your points but now with a straight line. From its slope and y-intercept, obtain the entropy of solution (ΔSsoln) and the heat of solution (ΔHsoln). 4. Please answer the following questions and justify your answers: a. Does dissolving potassium nitrate in water cause a local increase or decrease in disorder? b. Is the solubility of potassium nitrate energy driven or entropy driven? Technical Notes: The graphical analysis used here to obtain ΔHsoln and ΔSsoln by using a straight line to fit the data holds well over the small temperature ranges investigated and gives reasonable results. The potassium nitrate is slow to dissolve unless finely powdered so students tend to overheat the solutions resulting in longer than necessary cooling times so it is advisable to grind the stock KNO3 before weighing. The crystallization temperatures are roughly with 5 oC of the indicated target heating times. A careful analysis of the experiment should consider the following: the values of ΔHsoln and ΔSsoln slowly change with temperature so that they are not truly constant as demanded by the linear fit; the volumes of these concentrated solutions are not strictly equal to the volume of water; the volumes of these solutions vary with temperature; the mean activity coefficient of KNO3 surely is not equal to one and varies in these concentrated solutions.
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