CH 26x GFE Burand V6 - The Gibbs Free Energy Lab for general chemistry PDF

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The Gibbs Free Energy Lab for general chemistry...

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CH 26x Burand

Reaction Entropy and Gibbs Free Energy Michael W. Burand, Oregon State University Department of Chemistry

Introduction A spontaneous process is one that takes place without assistance from the surroundings. Chemical reactions that are not spontaneous under normal conditions can often be made to occur by changing the conditions such that the reaction is spontaneous under the new conditions. How can reaction spontaneity be determined? Both the enthalpy change (ΔH) and the entropy change (ΔS) during the reaction are important factors that relate to spontaneity. Exothermic reactions, where ΔH is negative, are usually spontaneous in nature, but not always (consider, for example, the spontaneous endothermic process of ice melting at room temperature). Likewise, the entropy change is also an imperfect predictor of spontaneity. The most reliable predictor of reaction spontaneity is the change in Gibbs free energy (ΔG) for the reaction. The following equation relates the Gibbs free energy change to the entropy and enthalpy changes for a reaction. Δ𝐺 = Δ𝐻 − 𝑇Δ𝑆 If ΔG is positive, the reaction is non-spontaneous; if ΔG is negative, the reaction is spontaneous. This behavior explains why being exothermic is not an ironclad approach to predicting spontaneity. Four different permutations of ΔH and ΔS are summarized in the following table; the third set clearly outlines that an exothermic reaction may indeed be non-spontaneous. Sign of ΔH − + − +

Sign of ΔS + − − +

Sign of ΔG − + − low T, + high T + low T, − high T

Result Always spontaneous Never spontaneous Spontaneous only at low T Spontaneous only at high T

If we know all the parameters on the right-hand side of the equation, we can calculate ΔG. Measuring temperature is not difficult and we have effective methods to determine ΔH. However, experimental measurement of ΔS is somewhere between impractical and impossible. When knowledge of the entropy is important, it is either found theoretically or experimentally determined in an indirect manner that involves the determination of ΔG first. Since we will be evaluating a chemical reaction at equilibrium, it will be best for us to alter the previous equation slightly and move the thermodynamic parameters to their standard state designations, as shown in the following equation. Δ𝐺° = Δ𝐻° − 𝑇Δ𝑆°

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CH 26x Burand

The Gibbs free energy change is also not possible to measure, and we will find it in an indirect manner by first determining the equilibrium constant for the reaction of interest and using the following equation. Δ𝐺° = −𝑅𝑇ln𝐾 You will recognize the K term in the preceding equation as the equilibrium constant and T as the absolute temperature. Despite our reaction having nothing whatsoever to do with gases, the R term is indeed the ideal gas constant. Check in your textbook and select the proper value of R that will yield ΔG° in units of joules. You will be familiar with finding K values experimentally from an earlier experiment, and thus now have a pathway to calculate ΔG° values. With this in mind, we are going to take a look at a reaction system where the equilibrium position is highly sensitive to temperature. Not all reactions respond in this manner. If we can see the equilibrium shift as temperature changes, we can calculate that ΔG° is also changing. For the most part, ΔH° and ΔS° are insensitive to temperature changes, and we will assume them to be constants within the range of temperatures we will study in this experiment. By doing so, we can measure the response of ΔG° as we change temperature, and then we can make a plot that will allow us to determine both ΔS° and ΔH° for this system. The chemical system we will study involves cobalt chloride in aqueous solution.1,2,3,4 Like many transition metals, cobalt dissolves in water to form a 2+ cation. The Co2+ is stabilized in water by forming six coordinate covalent bonds to yield the compound Co(H2O)6 2+. This compound is bright red in color. Co2+ can also form coordinate covalent bonds with the Cl− ion. As Cl− is added to the aqueous Co2+, the chloride will begin to displace the coordinated waters molecules, going through stepwise additions that are detailed by the following series of reactions. Co(H2O)6 2+(aq) + Cl−(aq) ⇌ CoCl(H2O)5 +(aq) + H2O(l) CoCl(H2O)5 +(aq) + Cl−(aq) ⇌ CoCl2(H2O)2(aq) + 3H2O(l) CoCl2(H2O)2(aq) + Cl−(aq) ⇌ CoCl3(H2O)−(aq) + H2O(l) 2− CoCl3(H2O)−(aq) + Cl−(aq) ⇌ CoCl4 (aq) + H2O(l)

We can write an equilibrium expression for any one of these reactions. Obviously, things will get fairly complicated if we have to deal with all four reactions at once. By controlling our reactant mixture, we can produce a situation where all of the intermediate chlorinated forms of cobalt (i.e. CoCl(H2O)5 +, CoCl2(H2O)2, and CoCl3(H2O)−) may be considered negligible in concentration. This allows us to write a single reaction formula, which is essentially the sum of the four previously-shown steps:

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Zeltmann, A. H.; Matwiyoff, N. A.; Morgan, L. L. J. Phys. Chem. 1968, 72, 121–127. Shakhashiri, B. Z. Chemical Demonstrations, Volume 1; The University of Wisconsin Press: Madison, 1983. 3 nd Levine, I. N. Physical Chemistry, 2 Edition; McGraw-Hill: New York, 1983. 4 Bjerrum, J.; Halonin, A. S.; Skibsted, L. H. Acta Chem. Scand. A 1975, 29, 326–332. 2

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CH 26x Burand

2+ − 2− Co(H2O)6 (aq) + 4Cl (aq) ⇌ CoCl4 (aq) + 6H2O(l)

Now that we are focused on this reaction, we can write an equilibrium expression for the reaction that is dependent on the concentration of four different substances. Even though there are plenty of solutes dissolved in our solution, it is still a reasonable assumption to presume that the solution is mostly water, and as you know, the concentration of a pure material in its standard state cancels to unity. The two cobalt species both show color in the visible range, and as such we can use a spectrophotometer and Beer’s law to find their concentrations. Recall that Beer’s law is given by: 𝐴 = εℓ𝑐 2+ 2− The Co(H2O)6 ion is a maroon red and the CoCl4 ion is blue. It might seem odd that such similar structures are so different in color, but this change upon alteration of the cobalt coordination can be explained by ligand field theory, which is beyond the scope of this course although an important topic in advanced inorganic chemistry. Normally we can use Beer’s law to find the concentration of a single solute, so the presence of two absorbing species in solution could present problems. Fortunately, the specific absorbances of the two cobalt species allow us to choose wavelengths where only one or the other compound absorbs. We will use absorbance at 500 nm to calculate the concentration of 2+ 2− Co(H2O)6 and absorbance at 690 nm to calculate the concentration of CoCl4 . The absorptivity constants needed to solve Beer’s law at each of these wavelengths are given in the following table.

Compound Co(H2O)6 2+ CoCl4 2−

Wavelength (nm) 500 690

ε (M−1cm−1) 4.613 577.2

The only remaining chemical species in the reaction is Cl−. We should be able to figure out the total amount of chloride present in our solution from the recipe of the solution that we made. We can determine the amount of chloride present in each of the two cobalt compounds at equilibrium and will expect the remainder to be present simply as free Cl−. Once we know all of these concentrations, further calculations will allow solving for the ΔS° and ΔH°.

Criteria for Completing the Project 2+ − 2− Considering the reaction: Co(H2O)6 (aq) + 4Cl (aq) ⇌ CoCl4 (aq) + 6H2O(l):

  

Determine the equilibrium constant, K, for the reaction for around five temperatures from the range of 10 °C–45 °C. From the aforementioned K and T data, determine a value of ΔG° for the reaction at each temperature. From the values of ΔG° and T, determine values of ΔH° and ΔS° for the reaction. Hint: Recall the following equation: Δ𝐺° = Δ𝐻° − 𝑇Δ𝑆° If we rearrange it as: 3

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CH 26x Burand Δ𝐺° = −Δ𝑆°(𝑇) + Δ𝐻°

It fits the equation of a line: 𝑦 = 𝑚𝑥 + 𝑏 Where y = ΔG°, x = T, the slope (m) = −ΔS°, and the y-intercept (b) = ΔH°.

Safety Considerations  

Consult the MSDS for every compound you will work with. 6.0 M HCl is a highly corrosive substance. Wear goggles and try to minimize inhalation of the vapors it produces. If you spill any HCl on your person, wash immediately with water and inform your TA. Cobalt chloride is an irritant to the skin, eyes, and respiratory tract. In addition, it has been shown to cause cancer when ingested by laboratory animals. Make sure not to ingest any of this material.

Waste 

Be sure to dispose of all cobalt waste properly. There will be disposal containers in the laboratory for your cuvette waste and any leftover solutions containing cobalt.

Useful Equipment, Techniques, and Concepts      

Using a volumetric flask Using a spectrophotometer Beer’s law Using water baths to control temperature Using Microsoft Excel Preparation hint: To make a CoCl2·6H2O/HCl solution, start by obtaining 0.43–0.48 g of the solid CoCl2·6H2O (actual mass known with precision). Using a piece of creased weighing paper to measure out the solid will make the subsequent transfer easier. Transfer this solid to a 50-mL volumetric flask. Add some 6.0 M HCl, and once all the solid is dissolved dilute to volume with more 6.0 M HCl. Be very careful with the HCl, and make sure to clean up any spills immediately.

Available Chemicals  

Solid CoCl2·6H2O HCl (6.0 M)

Planning Questions 1. What is the equilibrium expression for the reaction your group will study? 2. Briefly describe how you will find the concentrations of each species at a given temperature (and therefore be able to calculate the equilibrium constant, K). 3. Briefly describe how K and T data will allow you to calculate ΔG° at each temperature.

Summary Questions 1. Show how you determined the concentration of Co(H2O)6 need not show the calculation for each temperature.) 4

2+

at a particular temperature. (You

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CH 26x Burand

2− 2. Show how you determined the concentration of CoCl4 at a particular temperature. (You need not show the calculation for each temperature.) 3. Show how you determined the value of K and ΔG° at a particular temperature. (You need not show the calculation for each temperature.) 4. Construct a plot of ΔG° versus temperature that will allow you to calculate ΔH° and ΔS°, and report these values with the proper units. Include a copy of the plot with your summary. 5. Using your laboratory findings, would it be possible to use a water bath at atmospheric pressure to make ΔG° for this reaction equal to 25.0 kJ/mol? Discuss briefly. 6. You have experimentally measured ΔS° for this reaction. Does the sign of this quantity match what you would have expected from just examining the chemical reaction set being studied? Discuss briefly. 7. If an equilibrium reaction (such as the one your group studied) has a positive value for ΔG°, how can you explain that the forward reaction occurs at all? Doesn’t a positive value of ΔG° mean the reaction is non-spontaneous and will not occur? Hint: Consider the different between ΔG° and ΔG.

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