2. Integrated Rate Law Goal: To determine the order and rate law from concentration and time data

3. Using the calculator to determine order The test with the best linear fit is the order of the reaction: Ø order  Time vs. Concentration 1 st order  Time vs. ln(concentration) 2 nd order  Time vs. (1/concentration)

4. Enter the data into your calculator Time (s)[NO 2 ] (mol/L) 00.500 1.20 x 10 3 0.444 9.00 x 10 3 0.381 4.50 x 10 3 0.340 9.00 x 10 3 0.250 1.80 x 10 4 0.174 For equation: NO 2 + CO  NO + CO 2 Below 225°C only depends on [NO 2 ] Enter 4 columns of data: 1 st column = time 2 nd column = concentration 3 rd column = ln (concentration) 4 th column = 1/concentration

5. Perform 3 linear regressions The linear regressions: Ø order test  Time vs. Concentration  (1 st column, 2 nd column) 1 st order  Time vs. ln(concentration)  (1 st column, 3 rd column) 2 nd order  Time vs. (1/concentration)  (1 st column, 4 th column) Each time: 1. Perform linear regression 2. Write down the slope 3. Write down the correlation coefficient Winner= Best correlation coefficient (closest to ±1)

6. Writing the rate law and ½ life equation: zero orderRate = k[A] o plot concentration [A] vs. time (t) k = +slope integrated rate law  [A] t = -kt + [A] 0 half life  t 1/2 = [A] 0 /2k [A] Time (t) 1 st order Rate = k[A] 1 plot of ln conc. vs. time is linear k = + slope integrated rate law  ln[A] t = -kt + ln[A] 0 half life  t 1/2 =.693/k Ln[A] Time (t) 2 nd order Rate = k[A] 2 plot inverse of conc. vs. time is linear k = + slope integrated rate law  1/[A] t = kt + 1/[A] 0 half life  t 1/2 = 1/k[A] 0 1/[A] Time (t)

7. Let’s clarify some things zero orderRate = k[A] o plot concentration [A] vs. time (t) k = +slope integrated rate law  [A] t = -kt + [A] 0 half life  t 1/2 = [A] 0 /2k [A] Time (t) 1 st order Rate = k[A] 1 plot of ln conc. vs. time is linear k = + slope integrated rate law  ln[A] t = -kt + ln[A] 0 half life  t 1/2 =.693/k Ln[A] Time (t) 2 nd order Rate = k[A] 2 plot inverse of conc. vs. time is linear k = + slope integrated rate law  1/[A] t = kt + 1/[A] 0 half life  t 1/2 = 1/k[A] 0 1/[A] Time (t) Order are exponents

8. Let’s clarify some things zero orderRate = k[A] o plot concentration [A] vs. time (t) k = +slope integrated rate law  [A] t = -kt + [A] 0 half life  t 1/2 = [A] 0 /2k [A] Time (t) 1 st order Rate = k[A] 1 plot of ln conc. vs. time is linear k = + slope integrated rate law  ln[A] t = -kt + ln[A] 0 half life  t 1/2 =.693/k Ln[A] Time (t) 2 nd order Rate = k[A] 2 plot inverse of conc. vs. time is linear k = + slope integrated rate law  1/[A] t = kt + 1/[A] 0 half life  t 1/2 = 1/k[A] 0 1/[A] Time (t) k from linear regression Units = M (1-order) s Y = mx + b

9. Let’s clarify some things zero orderRate = k[A] o plot concentration [A] vs. time (t) k = +slope integrated rate law  [A] t = -kt + [A] 0 half life  t 1/2 = [A] 0 /2k [A] Time (t) 1 st order Rate = k[A] 1 plot of ln conc. vs. time is linear k = + slope integrated rate law  ln[A] t = -kt + ln[A] 0 half life  t 1/2 =.693/k Ln[A] Time (t) 2 nd order Rate = k[A] 2 plot inverse of conc. vs. time is linear k = + slope integrated rate law  1/[A] t = kt + 1/[A] 0 half life  t 1/2 = 1/k[A] 0 1/[A] Time (t) [A] 0 = start concentration (From data table)

10. Let’s clarify some things zero orderRate = k[A] o plot concentration [A] vs. time (t) k = +slope integrated rate law  [A] t = -kt + [A] 0 half life  t 1/2 = [A] 0 /2k [A] Time (t) 1 st order Rate = k[A] 1 plot of ln conc. vs. time is linear k = + slope integrated rate law  ln[A] t = -kt + ln[A] 0 half life  t 1/2 =.693/k Ln[A] Time (t) 2 nd order Rate = k[A] 2 plot inverse of conc. vs. time is linear k = + slope integrated rate law  1/[A] t = kt + 1/[A] 0 half life  t 1/2 = 1/k[A] 0 1/[A] Time (t) t = time in seconds

11. Let’s clarify some things zero orderRate = k[A] o plot concentration [A] vs. time (t) k = +slope integrated rate law  [A] t = -kt + [A] 0 half life  t 1/2 = [A] 0 /2k [A] Time (t) 1 st order Rate = k[A] 1 plot of ln conc. vs. time is linear k = + slope integrated rate law  ln[A] t = -kt + ln[A] 0 half life  t 1/2 =.693/k Ln[A] Time (t) 2 nd order Rate = k[A] 2 plot inverse of conc. vs. time is linear k = + slope integrated rate law  1/[A] t = kt + 1/[A] 0 half life  t 1/2 = 1/k[A] 0 1/[A] Time (t) [A] t = concentration at time (t)

12. Let’s clarify some things zero orderRate = k[A] o plot concentration [A] vs. time (t) k = +slope integrated rate law  [A] t = -kt + [A] 0 half life  t 1/2 = [A] 0 /2k [A] Time (t) 1 st order Rate = k[A] 1 plot of ln conc. vs. time is linear k = +slope integrated rate law  ln[A] t = -kt + ln[A] 0 half life  t 1/2 =.693/k Ln[A] Time (t) 2 nd order Rate = k[A] 2 plot inverse of conc. vs. time is linear k = +slope integrated rate law  1/[A] t = kt + 1/[A] 0 half life  t 1/2 = 1/k[A] 0 1/[A] Time (t) t 1/2 = time when concentration halves

13. Let’s try it: Time (s)[NO 2 ] (mol/L) 00.500 1.20 x 10 3 0.444 9.00 x 10 3 0.381 4.50 x 10 3 0.340 9.00 x 10 3 0.250 1.80 x 10 4 0.174 Enter 4 columns of data: 1 st column = time 2 nd column = concentration 3 rd column = ln (concentration) 4 th column = 1/concentration

14. Let’s try it: Do the linear regressions: Ø order test Time vs. Concentration  (1 st, 2 nd ) 1 st order  Time vs. ln(concentration)  (1 st, 3 rd ) 2 nd order  Time vs. (1/concentration)  (1 st, 4 th ) Each time: 1. Perform linear regression 2. Write down the slope 3. Write down the correlation coefficient

15. What you should get: Do the linear regressions: Ø order test Time vs. Concentration  (1 st, 2 nd ) slope= -1.7 x 10 5 corr coef. = -.949294 1 st order  Time vs. ln(concentration)  (1 st, 3 rd ) slope= -5.8 x 10 5 corr coef. =.970686 2 nd order  Time vs. (1/concentration)  (1 st, 4 th ) slope= 2.1 x 10 4 corr coef. = 0.999367 M = slope r= correlation coefficient Winner= 2 ND ORDER!!!

16. Use formulas to solve for lots of stuff!!! 2 nd order Rate = k[NO 2 ] 2 k = +slope = 2.1 x 10 4 1/Ms k from linear regression Units = M (1-order) = M (1-2) s s = M (-1) = 1 s Ms

17. Use formulas to solve for lots of stuff!!! 2 nd order Rate = k[NO 2 ] 2 k = +slope = 2.1 x 10 4 1/Ms integrated rate law  1/[NO 2 ] t = kt + 1/[NO 2 ] 0 1/[NO 2 ] t = (2.1 x 10 4 1/Ms)·t + 1/(0.500 M) half life  t 1/2 = 1/k[NO 2 ] 0 t 1/2 = 1/((2.1 x 10 4 1/Ms)·(0.500M)

18. NOW YOU KNOW!!! How to determine the order from concentration and time data How to write the rate law, integrated rate law and half-life equation Can use this to solve for time at any given concentration or concentration at any given time!