Lab Report 4 - Kinetics - Concentration Dependence of Reaction Rates
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Lab Report 4 - Kinetics - Concentration Dependence of Reaction Rates

Course Number: CHEM 1252L, Fall 2012

College/University: UNC Charlotte

Word Count: 1826

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Cooper Chung (Yu Cheng) 800736526 Name: Cooper Chung (Yu Cheng) Date: 02/22/12 Lab Partner: Andy Nguyen Kinetics: Concentration Dependence of Reaction Rates Purpose: To determine the rate of the reaction on [OH-] and to complete the differential rate law in terms of the initial reactant concentration with the order of the reaction. Procedure: Part 1: Order of Reaction with Respect to [phph2-] 1. Use a...

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Chung Cooper (Yu Cheng) 800736526 Name: Cooper Chung (Yu Cheng) Date: 02/22/12 Lab Partner: Andy Nguyen Kinetics: Concentration Dependence of Reaction Rates Purpose: To determine the rate of the reaction on [OH-] and to complete the differential rate law in terms of the initial reactant concentration with the order of the reaction. Procedure: Part 1: Order of Reaction with Respect to [phph2-] 1. Use a spectrophotometer and set it to transmittance mode. Set it to 0.0. 2. Pour about 10m L of the 0.3 M NaOH solution into a 50 mL beaker and use pipette to transfer exactly 4.0 mL of the solution to your cuvette. Close the sample holder cover and set the transmittance to 100.0. Record the initial transmittance value. 3. Remove the cuvette from the sample holder and add 1 drop of phenolphthalein. Record the absorbance every 5 seconds for 360 seconds. Once 360 seconds have elapsed, data is no longer needed to be recorded. Part 2: Order of Reaction with Respect to [OH-] 1. Repeat the experiment with a different concentration for [OH-]. Once you have calculated the m, the order with respect to [phph2-], n is able to be solved. For this run, rate will be kobs[phph2-]n, where k[OH-]. 2. Repeat the spectrophotometer if possible and necessary. Pour 5 mL of 0.3 M NaCl solution; this will stabilize the ions in the solution. 3. With a new cuvette, pipet 2.0 mL of the 0.3 M NaOH and 2.0 mL of the 0.3 M NaCl. Add 1 drop of phenolphthalein and record the data each 5 seconds for 360 seconds. Clean up later. Data/Calculations: Phph2- (Pink) + OH- Phph3- (Colored) Rate = - [Phph2-]/T = k[Phph2-]m[OH-]n Rate = k[Phph2-]m[OH-]n Part 1 Data Table Every 5 Seconds Absorbance Recorded Every 5 Seconds Absorbance Recorded Every 5 Seconds Absorbance Recorded Cooper Chung (Yu Cheng) 800736526 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1.00 0.945 0.940 0.930 0.92 0.915 0.905 0.895 0.885 0.874 0.865 0.855 0.845 0.835 0.825 0.815 0.806 0.796 0.788 0.778 0.768 0.746 0.736 0.728 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 0.718 0.708 0.698 0.688 0.678 0.668 0.658 0.648 0.638 0.630 0.620 0.610 0.600 0.592 0.582 0.574 0.564 0.556 0.548 0.538 0.530 0.522 0.514 0.506 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 0.498 0.490 0.482 0.474 0.466 0.458 0.450 0.444 0.438 0.430 0.424 0.418 0.412 0.404 0.398 0.392 0.386 0.378 0.374 0.368 0.362 0.356 0.350 0.345 Ln(Absorbance) Every 5 Seconds 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Ln(Absorbance ) Value 0 -0.05657 -0.06188 -0.07257 -0.08338 -0.08883 -0.09982 -0.11093 -0.12217 -0.13467 -0.14503 -0.15665 -0.16842 -0.18032 Every 5 Seconds 25 26 27 28 29 30 31 32 33 34 35 36 37 38 Ln(Absorbance) Every 5 Value Seconds -0.31745 -0.33129 -0.34531 -0.35954 -0.37397 -0.38861 -0.40347 -0.41855 -0.43386 -0.44942 -0.46204 -0.47804 -0.4943 -0.51083 49 50 51 52 53 54 55 56 57 58 59 60 61 62 Ln(Absorbance) Value -0.69716 -0.71335 -0.72981 -0.74655 -0.76357 -0.78089 -0.79851 -0.81193 -0.82554 -0.84397 -0.85802 -0.87227 -0.88673 -0.90634 Cooper Chung (Yu Cheng) 800736526 15 16 17 18 19 20 21 22 23 24 -0.19237 -0.20457 -0.21567 -0.22816 -0.23826 -0.25103 -0.26397 -0.29303 -0.30653 0 39 40 41 42 43 44 45 46 47 48 -0.52425 -0.54128 -0.55513 -0.5727 -0.58699 -0.60148 -0.6199 -0.63488 -0.65009 -0.66553 63 64 65 66 67 68 69 70 71 72 -0.9213 -0.93649 -0.95192 -0.97286 -0.9835 -0.99967 -1.01611 -1.03282 -1.04982 -1.06421 1/(Absorbance) Every 5 Seconds 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 1/(Absorbance) Value 1 1.058201 1.06383 1.075269 1.086957 1.092896 1.104972 1.117318 1.129944 1.144165 1.156069 1.169591 1.183432 1.197605 1.212121 1.226994 1.240695 1.256281 1.269036 1.285347 1.302083 1.340483 1.358696 1 Every 5 Seconds 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 1/(Absorbance) Value 1.373626 1.392758 1.412429 1.432665 1.453488 1.474926 1.497006 1.519757 1.54321 1.567398 1.587302 1.612903 1.639344 1.666667 1.689189 1.718213 1.74216 1.77305 1.798561 1.824818 1.858736 1.886792 1.915709 1.945525 Every 5 Seconds 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 1/(Absorbance) Value 2.008032 2.040816 2.074689 2.109705 2.145923 2.183406 2.222222 2.252252 2.283105 2.325581 2.358491 2.392344 2.427184 2.475248 2.512563 2.55102 2.590674 2.645503 2.673797 2.717391 2.762431 2.808989 2.857143 2.898551 Cooper Chung (Yu Cheng) 800736526 Part 2 Data Table Every 5 Seconds 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Absorbance Recorded 1.05 1.05 1.045 1.045 1.04 1.04 1.04 1.035 1.035 1.03 1.03 1.025 1.025 1.02 1.02 1.015 1.015 1.01 1.005 1.000 0.995 0.995 0.990 0.985 Every 5 Seconds 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 Absorbance Recorded 0.985 0.980 0.975 0.975 0.970 0.965 0.965 0.960 0.955 0.950 0.945 0.945 0.940 0.935 0.930 0.930 0.925 0.920 0.915 0.910 0.905 0.905 0.900 0.895 Every 5 Seconds 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 Every 5 Ln(Absorbance) Every 5 Absorbance Recorded 0.890 0.885 0.880 0.875 0.875 0.865 0.865 0.860 0.855 0.850 0.845 0.840 0.835 0.830 0.825 0.820 0.815 0.812 0.808 0.802 0.798 0.794 0.788 0.784 Ln(Absorbance) Every Chung 5 Ln(Absorbance Ln(Absorbance) Cooper (Yu Cheng) 800736526 Seconds 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 ) Value 0.04879 0.04879 0.04402 0.04402 0.03922 0.03922 0.03922 0.0344 0.0344 0.02956 0.02956 0.02469 0.02469 0.0198 0.0198 0.01489 0.01489 0.00995 0.00499 0 -0.00501 -0.00501 -0.01005 0.04879 Seconds 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 Value -0.01511 -0.01511 -0.0202 -0.02532 -0.02532 -0.03046 -0.03563 -0.03563 -0.04082 -0.04604 -0.05129 -0.05657 -0.05657 -0.06188 -0.06721 -0.07257 -0.07257 -0.07796 -0.08338 -0.08883 -0.09431 -0.09982 -0.09982 -0.10536 Seconds 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 Value -0.11653 -0.12217 -0.12783 -0.13353 -0.13353 -0.14503 -0.14503 -0.15082 -0.15665 -0.16252 -0.16842 -0.17435 -0.18032 -0.18633 -0.19237 -0.19845 -0.20457 -0.20825 -0.21319 -0.22065 -0.22565 -0.23067 -0.23826 -0.24335 1/(Absorbance) Every 5 Seconds 1 2 3 4 5 6 7 8 1/(Absorbance) Value 0.952381 0.952381 0.956938 0.956938 0.961538 0.961538 0.961538 0.966184 Every 5 Seconds 25 26 27 28 29 30 31 32 1/(Absorbance) Value 1.015228 1.020408 1.025641 1.025641 1.030928 1.036269 1.036269 1.041667 Every 5 Seconds 49 50 51 52 53 54 55 56 1/(Absorbance) Value 1.123596 1.129944 1.136364 1.142857 1.142857 1.156069 1.156069 1.162791 Cooper Chung (Yu Cheng) 800736526 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 0.966184 0.970874 0.970874 0.97561 0.97561 0.980392 0.980392 0.985222 0.985222 0.990099 0.995025 1 1.005025 1.005025 1.010101 1.015228 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 1.04712 1.052632 1.058201 1.058201 1.06383 1.069519 1.075269 1.075269 1.081081 1.086957 1.092896 1.098901 1.104972 1.104972 1.111111 1.117318 Graph of Part 1 [Phph2-]t = -kobs * t + [Phph2-]0 Ln([Phph2-]t) = -kobs * t + Ln[Phph2-]0 = kobs * t + Graph of Part 2 [Phph2-]t = -kobs * t + [Phph2-]0 Ln([Phph2-]t) = -kobs * t + Ln[Phph2-]0 = kobs * t + Analysis Part 1: Order of Reaction with Respect to [phph2-] 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 1.169591 1.176471 1.183432 1.190476 1.197605 1.204819 1.212121 1.219512 1.226994 1.231527 1.237624 1.246883 1.253133 1.259446 1.269036 1.27551 Cooper Chung (Yu Cheng) 800736526 1. 2. 3. 4. 5. Did not convert the % transmittance Provided Provided Provided Slope of [Phph2-]t = -kobs * t + [Phph2-]0 = = - 0.0125 Slope of Ln([Phph2-]t) = -kobs * t + Ln[Phph2-]0 = = -0.013467 Slope of = kobs * t + = = 0.0144165 The order is 2nd order due to the graph; therefore, m=2. kobs = kobs = kobs = 0.0144165 Part 2: Order of Reaction with Respect to [OH-] Tabulated Data: provided Linear Plot with fit line and equation: provided Slope of [Phph2-]t = -kobs * t + [Phph2-]0 = = - 0.002 Slope of Ln([Phph2-]t) = -kobs * t + Ln[Phph2-]0 = = -0.001923 Slope of = kobs * t + = = 0.0018493 The order is 2nd order due to the graph; therefore, n=2. Rate = k[Phph2-]m[OH-]n kobs = k [Phph2-]2[OH-]2 0.0144165 = k[OH-]2 =k Cooper Chung (Yu Cheng) 800736526 =k 0.1601833 = k Discussions/Conclusions: 1. Why did it not matter that you didnt know the initial concentration of the phenolphthalein solution? It did not matter because the purpose of phenolphthalein was to show change in color. Also, the original concentration of it was very little due to the amount that was added. 2. Explain the purpose of adding excess hydroxide compared to phenolphthalein. The purpose was to keep one variable constant. In other terms, if the concentration of hydroxide was too high to begin with; it would not matter in the equation anymore. Therefore, we can solve the other variable easily. 3. Explain the difference in approaches for Parts 1 and 2, including the equations relevant to each of the parts and what you are solving for in each part. In part1 of the solution, we are assuming [OH-] is constant and the value of m is 2 because of 2nd order reaction, so we can easily calculate the value of kobs. In part 2 of the solution, we already know the value of m and assume the value of n is 2 as well because of 2nd order reaction. With all the variables known ( kobs, m, n, [OH-], etc), k can be easily calculated. 4. What happened to the initial rates as you changed concentration of hydroxide? Justify, using any appropriate equations, your observations. The initial rate would change if the concentration of hydroxide is changed. Especially the n in this case is 2, because its 2nd order reaction. Therefore, the initial rate would most likely to be double if the concentration of hydroxide is doubled. Cooper Chung (Yu Cheng) 800736526 Conclusion Our lab had several mistakes to begin with. We did not measure out our volume of solution precisely and we did not setup the spec-20 correctly. So as result, our absorbance rate is somewhat higher than other groups rate. This led to difference in data, plot, and graph. But as result, the fact of those two solutions being 2nd order reaction is still unchangeable.

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