test1s-1 - . 3. We use the equation The first experimental...

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Up: Back to the Chemistry 2000 test index Chemistry 2000, Section A Spring 1996 Test 1 Solutions 1. Using the first two lines of the table, we see that the rate goes up by a factor of 5 when [A] goes up by a factor of 5. Using the last two lines, we see that the rate goes up by a factor of 4 ( ) when the [B] goes up by a factor of 2. Therefore 2. Different amounts of reactant disappear in 0.01s so this is not a zero-order reaction. Suppose that it's a first-order reaction. Then From the first two points, we calculate From the next two points, we get Since (within the accuracy of the data) we get the same rate constant in both cases, we conclude that this is a first-order reaction with
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Unformatted text preview: . 3. We use the equation The first experimental point tells us that From the second point, we have Subtract the two equations to eliminate : Therefore Bonus: Substitute into one of your two original equations and solve for : Use to get the out of the exponent. Recall that has the same units as k : 1. The equations are obtained by applying the law of mass action: 2. 3. Bonus: where is approximately constant. 4. If 99.9% of the plutonium has decayed, 0.1% of it remains. Put another way, we want to know how long it takes before . We can calculate k from the half-life: From the first-order equation, we have which is Marc Roussel Mon Sep 16 15:23:07 MDT 1996...
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This note was uploaded on 03/03/2012 for the course CHEM 2000 taught by Professor Roussel during the Fall '06 term at Lethbridge College.

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test1s-1 - . 3. We use the equation The first experimental...

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