This experiment is a study of the rate of a chemical reaction and the

Chemistry 112 Laboratory: Kinetics Page 83 Chemical Kinetics T his experiment is a study of the rate of a chemical reaction and the dependence of ...
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Chemistry 112 Laboratory: Kinetics

Page 83

Chemical Kinetics

T

his experiment is a study of the rate of a chemical reaction and the dependence of that rate on the temperature. You will first verify the reaction is first-order in the reactant of interest, a metal complex, and then you will study the temperature dependence of the rate. This latter study will enable you to calculate the activation energy, Ea, for the reaction. The compound to be studied is the beautiful green, cobalt-based complex called trans-dichlorobis(ethylenediamine)cobalt(+) ion. +

Cl H2C H2C

H2

N

H2

N Co N

H2

N

H2

CH2

The ethylenediamine molecule and the chloride ions are called “ligands.”

CH2

Cl

The nitrogen atoms of the ethylenediamine molecule all lie in a plane surrounding the Co3+ ion. The line connecting the two Cl- ions is perpendicular to this plane and passes through the Co3+ ion. The trans portion of the name indicates this arrangement—the Cl- ions are across the molecule from one another. The reaction you will study is the aquation of the green complex ion trans–[Co(en)2Cl2]+ to form the red complex ion [Co(en)2(H2O)Cl]2+. trans–[Co(en)2Cl2]+(aq) + H2O(liq) → [Co(en)2(H2O)Cl]2+(aq) + Cl-(aq) green

red

That is, a Cl- ion is replaced with a water molecule on the Co3+ ion. Reactions such as this have been studied extensively, and experiments suggest that the initial, slow step in the reaction is the breaking of the Co–Cl bond to give a five-coordinate intermediate. Slow: trans–[Co(en)2Cl2]+(aq) → [Co(en)2Cl]2+(aq) + Cl-(aq) Fast:

[Co(en)2Cl]2+(aq) + H2O(aq) → [Co(en)2(H2O)Cl]2+(aq)

This means the rate of the overall reaction is a measure of the rate of the slow, bond-breaking elementary step. The rate-determining, elementary reaction is first-order in the reactant trans–[Co(en)2Cl2]+. The rate law for this step is therefore Rate = k{trans–[Co(en)2Cl2]+}

December 2005

Information on rate laws and reaction mechanisms is found in Chapter 15 of Chemistry & Chemical Reactivity.

Chemistry 112 Laboratory: Kinetics

Page 84

(where k is the rate constant). This is a first order process whose rate depends on the concentration of the reactant in solution. A feature of all first-order reactions is that the logarithm of the fraction of original reactant remaining [ln(C/Co)] after some time has elapsed (t) is equal to the negative of the product of the rate constant (k) and the time.

ln C = - kt C0 ln (fraction remaining) = - (rate constant)•(time elapsed) C = concentration after time t has elapsed

C0 = original concentration at t = 0

k = first order rate constant

t = time

Be sure to notice this equation uses the natural logarithm. See page A-2 in the Appendix of Chemistry & Chemical Reactivity.

This equation informs us that the time to reach a given fraction remaining is the same no matter what the initial concentration (C0). The conclusion above is crucial to understanding this experiment. During the reaction there is a moment at which the mixture of green reactant and red product appears gray. This gray color is reached when the ratio of green reactant to red product has a given value. Or, rewriting the equation above in terms of this experiment,

C [green ion] = C0 [green ion] + [red ion] the gray color appears at a particular fraction remaining. You will use this feature of the reaction to verify that the reaction is first order in the green complex ion trans–[Co(en)2Cl2]+. The activation energy for a reaction, Ea, is the minimum energy the reactants must possess before reaction can occur, and we can determine this energy by studying the temperature dependence of the reaction rate. Specifically, the temperature dependence of the rate constant for a reaction is given by the equation

k = Ae-E a /RT where k = rate constant A = constant Ea = activation energy R = gas constant = 8.31 J/K•mol or 8.31 x 10-3 kJ/K•mol T = temperature (in kelvins) How can we interpret this equation? First, as T increases, the quotient Ea/RT becomes smaller (for a given Ea). But, importantly, e-x becomes larger as the quantity x gets smaller. Therefore, we conclude that the rate constant k increases as T increases. This is not a surprising conclusion, because it is reasonable that the reacting molecules acquire more and more energy as the temperature increases.

December 2005

This is the crucial idea in the experiment. In essence you are measuring the halflife of the reaction. For a first order reaction the halflife is independent of the initial concentration. See pp. 712-722 of Chemistry & Chemical Reactivity.

Chemistry 112 Laboratory: Kinetics

Page 85

In the aquation of trans–[Co(en)2Cl2]+, Ea is related to the energy of the Co-Cl bond. It is not the energy required to break the bond but rather it is the energy required to stretch the bond to the point at which it breaks completely rather than reforming. Rearranging the equation for the temperature dependence of k we have

ln k = -(

Ea 1 )( ) + ln A R T

This is an equation of a straight line of the type y = mx + b where y = ln k, m, the slope, is (-Ea/R), x is (1/T), and b is ln A. This means that if we plot ln k versus (1/T) we can obtain the activation energy from the slope of the plot. However, we do not need to measure k in this experiment to obtain Ea. Instead, go back to the first-order rate law on page 86.

ln

C = -kt C0

t = -(1/k)ln(C/C0) for a 1st order process. Thus, because the “gray point” in this experiment always occurs at the same C/C0 ratio, measuring time is a measure of the value of 1/k at a given temperature.

Rearranging, we have

⎛ 1⎞ C k = ⎜ ⎟ ln 0 ⎝ t⎠ C If we take the log of both sides, we obtain

lnk = ln

1 + ln t

⎡ C0 ⎤ ⎢⎣ ln C ⎥⎦

Now substitute this equation into the equation connecting ln k and Ea.

ln

1 E 1 ⎡ C ⎤ + ln ⎢ ln 0 ⎥ = -( a )( ) + ln A t R T ⎣ C⎦

Finally, rearrange the equation just above.

1 E 1 ln = -( a )( ) + ln A - ln t R T

⎡ C0 ⎤ ⎢⎣ ln C ⎥⎦

This equation informs us that if we plot ln (1/time) versus 1/T (where T is the kelvin temperature of the experiment), we will have a straight line plot with a negative slope of Ea/R.

December 2005

R = 8.31 x 10-3 kJ/K•mol

Chemistry 112 Laboratory: Kinetics

Page 86

Experimental Procedure Begin by preparing an ice-bath in which you cool a 10-mL graduated cylinder, a 25 x 150 mm test tube, two 18 x 150 mm test tubes, and a flask containing about 100 mL of distilled water. (You can use two, 600-mL beakers as ice baths.) The test tubes and graduated cylinder must be dry on the inside except for some slight condensation of moisture from the air. Fill a 250-mL beaker with water and begin heating it to prepare your constant temperature bath.

You may do this experiment with another student. Please do the calculations independently.

Determining the Ratio That Gives the Gray Color 1. Weight out about 0.3 g of trans-[Co(en)2Cl2]Cl and dissolve it in 35 mL of cold water in your 25 x 150 mm test tube. Mix the solution thoroughly and keep it cold in the ice bath to inhibit the aquation reaction. 2. To prepare a sample of the final product of the aquation, [Co(en)2Cl(H2O)]2+, add about 10 mL of the green solution, which you prepared in Step 1, to an 18 x 150 mm test tube and heat this in your boiling water bath for about 5 minutes. The solution should now be pink, indicating that the following reaction has occurred. trans–[Co(en)2Cl2]+(aq) + H2O(liq) → [Co(en)2(H2O)Cl]2+(aq) + Cl-(aq) green

red

Cool this pink solution in the ice bath. 3. Stand the 10-mL graduated cylinder in an ice bath, add about 4.5-5.0 mL of the green stock solution, and record the volume to the nearest 0.1 mL. Add the pink solution dropwise with thorough stirring until the solution assumes a gray color when viewed against a white background. Under TRIAL 1 on the Report Form, record the volume when the solution last appears faintly green, when the appearance is gray, and when the appearance is faintly pink. 4. Empty the graduated cylinder and add about 2.5 mL of the green stock solution about about 4.5 mL of cold water. Add pink solution and record the data as above under TRIAL 2. 5. Calculate the ratio indicated on your Report Form for the two trials. The values should be identical within the limits of your ability to detect the colors. This ratio represents the fraction of an originally pure trans– [Co(en)2Cl2]+ sample that will have become aquated (changed to the aqua complex trans–[Co(en)2(H2O)Cl]2+) when the gray color appears in your kinetic runs. Verification that the Aquation Reaction is First-Order 1. Add 5 mL of the green stock solution to a cold 18 x 150 mm test tube. To another test tube add 2.5 mL of the green stock solution and 2.5 mL of cold water. Mix well. Immerse these test tubes in the ice bath immediately. 2. Set up a constant temperature bath as demonstrated by your instructor and adjust the bath temperature to 55-60 ˚C. Simultaneously transfer the two cold test tubes made up in Step 1 to the bath and begin timing.

December 2005

You may remove the graduated cylinder from the ice bath periodically to observe the color.

Chemistry 112 Laboratory: Kinetics



The test tubes should be immersed far enough so the liquid level in the tubes is below the water bath level.



Starting at the same temperature and having the same solution volume, the solutions should warm at the same rate. (Do not worry if the bath temperature declines slightly.)

3. Observe the immersed test tubes against a white background, and record the time when the solutions turn gray. Because the rate of the reaction is first-order in the concentration of trans–[Co(en)2Cl2]+, the gray color should appear at the same time in each solution. If the reaction were second-order in the concentration of the ion, the less concentrated solution should take about twice as long for the gray color to appear. If the reaction were zero-order, the more concentrated solution should take twice as long for the gray color to appear.

Page 87

The statement in the paragraph at the left is important. As explained on page 2, you are essentially measuring the halflife of the reaction. For a first order reaction the half-life is independent of the initial concentration.

Measurement of Activation Energy 1. Prepare a new stock solution of trans–[Co(en)2Cl2]+ by dissolving about 0.3 g of the solid in 5 mL of cold water in a small beaker or test tube. Store the solution in an ice bath. 2 Add 5 mL of water to an 18 x 150 mm test tube. Clamp the tube so it is immersed in the warm temperature bath. From this point on do not remove the tube from the bath. 3. Hold the bath temperature steady at ±1 ˚C at some temperature in the 5075 ˚C range. A good procedure is to bring the bath to the desired temperature and then remove the bunsen flame. If the temperature drops outside the desired range, heat briefly with the flame and stir well. 4. When the bath temperature has been constant for a few minutes, rapidly add about 8 drops of the cold, green stock solution to the warm water in the test tube in the water bath. Stir quickly with a stirring rod and observe the time. 5. Observe the color of the reaction solution as in the previous part of the experiment and record the time it takes to turn gray. 6. Repeat Steps 4 and 5 at three other temperatures (at the very least) in the 50-75 ˚C range. (It is best to have 4-6 data points for this part of the experiment.)

December 2005

Estimate the times here to the nearest second.

Chemistry 112 Laboratory: Kinetics

Page 88

Calculation of Activation Energy 1. Use LeastSquares, Mr. Plot, or MS Excel to plot ln(1/time) versus 1/T for the 4-6 determinations you did. Be sure to notice that “time” is in seconds and T is in kelvins.

Slope = –E a/R

Best straight line through the data points

1/T (where T is the kelvin temperature 2. The slope of the plot gives the value of the activation energy, Ea, because slope = -Ea/R.

December 2005

See Example 15.10 in Chemistry & Chemical Reactivity for help on this part of the experiment.

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