1. Practice Problem #1 Consider the process in which methyl isonitrile is converted to acetonitrile. What is the order of this reaction? You can’t tell just by looking at the equation. You need some experimental data! CH 3 NC CH 3 CN

2. Show Me The Data! This data was collected for this reaction at 198.9°C. CH 3 NCCH 3 CN What can we conclude about the order of the reaction based on this concentration vs time graph? Are we ready to calculate k? What should we do next? CH 3 NC M

3. What happens when we take the derivative ? What can we conclude about the order of the reaction based on this natural log vs time graph? Are we ready to calculate k? What should we do next? ln [CH 3 NC]

4. First-Order Equation ln [CH 3 NC] 1 st Order Equation: ln[A] = -kt +ln[A] 0 Solve for K: DON’T USE RISE/RUN *Use Equation* ln[A] = -kt + ln[A] 0 4.0 = -k(20,000 sec) + 5.0 k = 5.0 x 10 -5 sec -1

5. First Order Reaction Spin off problem: If the reaction started with 150 M CH 3 NC, how much would remain after 5,000 seconds? *Need to use the equation! ln[A] = -kt +ln[A] 0 *We solved for k already on last slide ln[X] = -(5.0 x 10 -5 )(5000 sec) + ln[150 M] 0 ln[X] = 4.760635294 How do we get rid of natural log?  “E” it! [X] = 116.8 M

6. Practice Problem #2 The decomposition of NO 2 at 300°C is described by the equation: NO 2 (g)  NO(g) + ½ O 2 (g) Time (s)[NO 2 ], M 0.00.01000 50.00.00787 100.00.00649 200.00.00481 300.00.00380 Let’s pretend that we graphed this data, and it made a curved line. What would we know?

7. Graphing ln [NO 2 ] vs. t yields: Time (s)[NO 2 ], Mln [NO 2 ] 0.00.01000-4.605 50.00.00787-4.845 100.00.00649-5.037 200.00.00481-5.337 300.00.00380-5.573 The plot is not a straight line, so the process is not first- order in [A]. What would we do next?

8. Second-Order Processes A graph of 1/[NO 2 ] vs. t shows this! Time (s)[NO 2 ], Mln [NO 2 ]1/[NO 2 ] 0.00.01000-4.605100 50.00.00787-4.845127 100.00.00649-5.037154 200.00.00481-5.337208 300.00.00380-5.573263 Solve for k: Use the 2 nd order equation 1/[A] = kt + 1/[A] 0 1/0.00787 M = k(50.0) + 1/0.01000 M 127.06 M -1 = k(50.0 sec) + 100 M -1 27.06 M -1 = k(50.0 sec) K = 0.5412 M -1 sec -1

9. Second-Order Processes Practice Problem #2: If the initial concentration remains 0.01000 M, determine the concentration of [NO 2 ] remaining after 75.0 seconds. Use 2 nd order equation: 1/[A] = kt + 1/[A] 0 1/[X] = 0.5412 M -1 sec -1 (75.0 sec) + 1/[0.01000 M] 1/[X] = 40.59 M -1 + 100 M -1 1/[X] = 140.59 M -1 How do we get rid of an inverse? X = 1/140.59 M -1 X =.00711 M