CHEM 111 Kinetics Review Tom Teets Answer Key

CHEM 111 Kinetics Review Tom Teets Answer Key - Kinetics...

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Kinetics Review- ANSWER KEY Initial Rate Problem: For the reaction A + B products, the following initial rate data was collected at 298 K: Run [A] o , M [B] o , M Initial Rate, M/s 1 0.500 0.100 6.0 x 10 -5 2 0.500 0.200 6.0 x 10 -5 3 0.250 0.100 1.5 X 10 -5 a. Find the rate equation, including k with the proper units. E a for this reaction is 330J/mol. First we should write our rate equations, in the form rate = k[A] x [B] y , for all three runs. Run 1: rate = 6.0 x 10 -5 M/s = k(0.500M) x (0.100M) y Run 2: rate = 6.0 x 10 -5 M/s = k(0.500M) x (0.200M) y Run 3: rate = 1.5 x 10 -5 M/s = k(0.250M) x (0.100M) y To solve for x, we divide the rate equations for Run 1 and Run 3: 5 rate1 rate3 5 k 6.0x10 M /s 4 1.5x10 == = xy (0.500M) ( 0.100M) k (0.250M) (0.100M) x 24 x2 , ⇒= = To solve for y, we do the same for Run 1 and Run 2 5 rate1 rate2 5 k 6.0x10 1 6.0x10 = x (0.500M) y (0.100M) k x (0.500M) y y y0 1 1, 2 (0.200M) ⎛⎞ = ⎜⎟ ⎝⎠ = (Remember, it doesn’t matter what order you solve for x and y, or for that matter which rate goes in the numerator and denominator for your divisions) Now that we have x and y, to solve for k we just use one of the rate equations. For no particular reason, I used Run 1: 6.0 X 10 -5 M/s = k(0.500M) 2 (0.100M) 0 Using a little algebra, k= 2.4 x 10 -4 M -1 s -1 . Since I kept the units in the equation, the correct units for k automatically come out in the answer. Combining everything: rate=(2.4 x 10 -4 M -1 s -1 )[A] 2 (no need to put [B] 0 since it always is 1) b. The experiment is repeated at 373 K. What is the new value for Initial Rate 1? Any time we are asked to relate rate to temperature, we use the Arrhenius equation. Since the only thing we change between part (a) and (b) is the temperature, only the rate constant changes, and the ratio of the rate constants gives us the ratio of the rates. a 2 12 1 E k 11 ln kR T T =− From the problem, T 2 =373K, and T 1 =298K (the temperature of the original experiment). E a =330 J, and for R we must use 8.314 J/Kmol since we are dealing with energy in Joules. Putting these numbers into the equation and undoing the logs (taking the exponent), we get:
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22 5 5 21 11 kr a t e ( T ) 1.03 rate(T ) 1.03 rate(T ) 1.03(6. 6 01 0 M / s ) a .2 10 M e(T ) / t s == = × = × Notice that a significant temperature increase (75K) has little effect on the rate in this case.
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This note was uploaded on 02/08/2010 for the course CHEM 111 taught by Professor Kenney during the Spring '08 term at Case Western.

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CHEM 111 Kinetics Review Tom Teets Answer Key - Kinetics...

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