1. Entry Task: Nov 15th Friday Turn in Na 2 S 2 O 3 Lab on my desk Discuss Rate and Order ws Pre-Lab Discussion on Crystal Violet lab MAYHAN

2. Clear off Desk Have out: Calculators Rates and Order ws MAYHAN

3. Answers MAYHAN

4. What are the four factors that can affect the rate of a chemical reaction? Temperature-  temp  Rate Surface area-  SA  Rate Concentration-  conc  Rate Add Catalyst-  catalyst  Rate Define chemical kinetics. The study of how fast (speed) of a reaction will take place. The change in molarity per second. MAYHAN

5. Consider the reaction A  2C The average rate of appearance of C is given by ∆[C]/∆t. How is the average rate of appearance of C relate to the average rate of disappearance of A? -2∆[A]/∆t MAYHAN

6. Consider the reaction: ___N 2 (g) + ___ H 2 (g)  ___ NH 3 (g) At the instant N 2 is reacting at a rate of 0.25 mol/L min at what rates are H 2 disappearing and NH 3 forming? ___N 2 (g) + ___ H 2 (g)  ___ NH 3 (g) Rate = − 1111  [N 2 ]  t = − 1313  [H 2 ]  t = 1212  [NH 3 ]  t 1111 0.25 = − 1313  [H 2 ]  t = 3131  0.75 M/min disappearing of H 2 − 32 MAYHAN

7. Consider the reaction: ___N 2 (g) + ___ H 2 (g)  ___ NH 3 (g) At the instant N 2 is reacting at a rate of 0.25 mol/L min at what rates are H 2 disappearing and NH 3 forming? ___N 2 (g) + ___ H 2 (g)  ___ NH 3 (g) Rate = − 1111  [N 2 ]  t = − 1313  [H 2 ]  t = 1212  [NH 3 ]  t 1111 0.25 = − 1212  [NH 3 ]  t = 2121  0.50 M/min appearance of NH 3 − 32 MAYHAN

8. The reaction between ozone and nitrogen dioxide has been studied at 231 K. ___NO 2 (g) + ___O 3 (g)  ___N 2 O 5 (s) + ___O 2 (g) Experiment shows the reaction is first order in O 3 and second order in NO 2. Write the rate law. How does tripling the concentration of NO 2 affect the reaction rate? How does halving the concentration of O 3 affect the reaction rate? 2 Rate= k [NO 2 ] 2 [O 3 ] [3] 2 = 9x rate of reaction [0.5] 1 = ½ rate of reaction MAYHAN

9. Data are given in the table at 660K for the reaction 2 NO(g) + O 2 (g)  2 NO 2 (g) a)Write the rate equation for the reaction. b) Calculate the rate constant. Rate= k [NO] 2 [O 2 ] 1.0 x 10 -4 M/s= k [0.020] 2 [0.010] 1.0 x 10 -4 M/s = k [0.020] 2 [0.010] = 25 M -2 s -1 MAYHAN

10. Data are given in the table at 660K for the reaction 2 NO(g) + O 2 (g)  2 NO 2 (g) c) Calculate the rate of reaction at the instant [NO] = 0.045 M and [O 2 ] = 0.025 M. Rate= 25 M -2 s -1 [0.045] 2 [0.025] 1.3 x 10 -3 M/s MAYHAN

11. Data are given in the table at 660K for the reaction 2 NO(g) + O 2 (g)  2 NO 2 (g) d) At the instant when O 2 is reacting at the rate 5.0 x 10 -4 mol/Ls, at what rate is NO reacting? Rate = − 1212  [NO]  t = − 1111  [O 2 ]  t = 1212  [NO 2 ]  t 1111 5.0 x 10 -4 = − 1212  [NO]  t = 2121 5.0 x 10 -4 1.0 x 10 -5 M/s = − MAYHAN

12. Given the following equations and experimental data, write the correct A 2 + B 2  2 AB a. Rate Law Expression b. Reaction Law c. Determine k, the Specific Rate Constant (including units) Rate = − 1111  [A]  t = − 1111  [B]  t = 1212  [AB]  t Rate= k [A] 2 [B] 0.01= k [0.001] 2 [0.001] 0.01 = k [0.001] 2 [0.001] = 1 x 10 7 M -2 s -1 MAYHAN

13. Given the following equations and experimental data, write the correct C + D  E a. Rate Law Expression b. Reaction Law c. Determine k, the Specific Rate Constant (including units) Rate = − 1111  [C]  t = − 1111  [D]  t = 1111  [E]  t Rate= k [D] 1 0.02= k [0.01] 1 0.02 = k [0.01] = 2 s -1 MAYHAN

14. For a 1 st order reaction, sketch out what the linear graph would look like and label the axis. - What would the integrated equation be for this graph? What would k be labeled as?_____________ ln[A] time -k slope ln[A] t = -kt + ln[A] 0 s -1 MAYHAN

15. For a 2 nd order reaction, sketch out what the linear graph would look like and label the axis. - What would the integrated equation be for this graph? What would k be labeled as?_____________ 1/[A] time +k slope 1/[A] 0 = kt + 1/[A] t M -1 s -1 MAYHAN

16. Data for the decomposition of N 2 O 5 in a particular solvent at 45°C are as follows: You’ll have to graph or EXCELL this data to determine order. Attach graphs to this paper. Time (min)Concln[Conc]1/[conc] 3.072.080.7323680.480769 8.771.670.5128240.598802 14.451.360.3074850.735294 31.280.72-0.32851.388889 Not zero order MAYHAN

17. Data for the decomposition of N 2 O 5 in a particular solvent at 45°C are as follows: You’ll have to graph or EXCELL this data to determine order. Attach graphs to this paper. Time (min)Concln[Conc]1/[conc] 3.072.080.7323680.480769 8.771.670.5128240.598802 14.451.360.3074850.735294 31.280.72-0.32851.388889 Might be 1 st order MAYHAN

18. Data for the decomposition of N 2 O 5 in a particular solvent at 45°C are as follows: You’ll have to graph or EXCELL this data to determine order. Attach graphs to this paper. Time (min)Concln[Conc]1/[conc] 3.072.080.7323680.480769 8.771.670.5128240.598802 14.451.360.3074850.735294 31.280.72-0.32851.388889 Not 2 nd order so 1 st order it is MAYHAN

19. Data for the decomposition of N 2 O 5 in a particular solvent at 45°C are as follows: You’ll have to graph or EXCELL this data to determine order. Attach graphs to this paper. What is the order of this reaction? Time (min)Concln[Conc]1/[conc] 3.072.080.7370.481 8.771.670.5130.599 14.451.360.3070.735 31.280.72-0.3291.39 1 st order Use the integrate rate equation to confirm the rate constant found in the graph. SHOW YOUR WORK!!! ln[0.72] t = -k(31.28-3.07= 28.21) + ln[2.08] ln[A] t = -kt = ln[A] 0 k=0.038s -1 -0.329 = -k (28.21) + -0.732 -0.329 -0.732 = -k (28.21) -1.061= -k (28.21) -1.061= -k 28.21 MAYHAN

20. Pre-Lab Discussion MAYHAN

21. In this experiment, you will observe the reaction between crystal violet and sodium hydroxide. One objective is to study the relationship between concentration of crystal violet and the time elapsed during the reaction. The equation for the reaction is shown here: A simplified (and less intimidating!) version of the equation is: CV + + OH – (aq)  CVOH (crystal violet) (hydroxide) MAYHAN

22. Absorbance (Concentration) will be used in place of concentration in plotting the following three graphs: *Concentration vs. time:A linear plot indicates a zero order reaction (k = –slope). *ln Concentration vs. time: A linear plot indicates a first order reaction (k = –slope). *1/Concentration vs. time: A linear plot indicates a second order reaction (k = slope). MAYHAN

23. The rate law for this reaction is in the form: rate = k[CV + ] m [OH – ] n, where k is the rate constant for the reaction, m is the order with respect to crystal violet (CV + ), and n is the order with respect to the hydroxide ion. As the reaction proceeds, a violet-colored reactant will be slowly changing to a colorless product. Using the green (565 nm) light source of a Colorimeter, you will monitor the absorbance of the crystal violet solution with time. We will assume that absorbance is proportional to the concentration of crystal violet (Beer’s law). MAYHAN