Determining Rate Law

Integrated Rate LawGoal: To determine the order and rate law from concentration and time data

Using the calculator to determine order

The test with the best linear fit is the order of the reaction:

order Time vs. Concentration1st order Time vs. ln(concentration)2nd order Time vs. (1/concentration)

Enter the data into your calculatorTime (s)[NO2] (mol/L)00.5001.20 x 1030.4449.00 x 1030.3814.50 x 1030.3409.00 x 1030.2501.80 x 1040.174For equation: NO2 + CO NO + CO2Below 225C only depends on [NO2]

Enter 4 columns of data:

1st column = time2nd column = concentration3rd column = ln (concentration)4th column = 1/concentration

Perform 3 linear regressionsThe linear regressions: order test Time vs. Concentration (1st column, 2nd column) 1st order Time vs. ln(concentration) (1st column, 3rd column)2nd order Time vs. (1/concentration) (1st column, 4th column)

Each time:1. Perform linear regression2. Write down the slope3. Write down the correlation coefficientWinner=Best correlation coefficient (closest to 1)

Writing the rate law and life equation:zero orderRate = k[A]oplot concentration [A] vs. time (t)k = +slope integrated rate law [A]t = -kt + [A]0half life t1/2 = [A]0/2k[A]Time (t)1st order Rate = k[A]1 plot of ln conc. vs. time is lineark = +slopeintegrated rate law ln[A]t = -kt + ln[A]0half life t1/2 = .693/kLn[A]Time (t)2nd order Rate = k[A]2 plot inverse of conc. vs. time is lineark = +slopeintegrated rate law 1/[A]t = kt + 1/[A]0half life t1/2 = 1/k[A]01/[A]Time (t)

Lets clarify some thingszero orderRate = k[A]oplot concentration [A] vs. time (t)k = +slopeintegrated rate law [A]t = -kt + [A]0half life t1/2 = [A]0/2k[A]Time (t)1st order Rate = k[A]1 plot of ln conc. vs. time is lineark = +slopeintegrated rate law ln[A]t = -kt + ln[A]0half life t1/2 = .693/kLn[A]Time (t)2nd order Rate = k[A]2 plot inverse of conc. vs. time is lineark = +slopeintegrated rate law 1/[A]t = kt + 1/[A]0half life t1/2 = 1/k[A]01/[A]Time (t)Order are exponents

Lets clarify some thingszero orderRate = k[A]oplot concentration [A] vs. time (t)k = +slopeintegrated rate law [A]t = -kt + [A]0half life t1/2 = [A]0/2k[A]Time (t)1st order Rate = k[A]1 plot of ln conc. vs. time is lineark = +slopeintegrated rate law ln[A]t = -kt + ln[A]0half life t1/2 = .693/kLn[A]Time (t)2nd order Rate = k[A]2 plot inverse of conc. vs. time is lineark = +slopeintegrated rate law 1/[A]t = kt + 1/[A]0half life t1/2 = 1/k[A]01/[A]Time (t)k from linear regression

Units = M(1-order) sY = mx + b

Lets clarify some thingszero orderRate = k[A]oplot concentration [A] vs. time (t)k = +slopeintegrated rate law [A]t = -kt + [A]0half life t1/2 = [A]0/2k[A]Time (t)1st order Rate = k[A]1 plot of ln conc. vs. time is lineark = +slopeintegrated rate law ln[A]t = -kt + ln[A]0half life t1/2 = .693/kLn[A]Time (t)2nd order Rate = k[A]2 plot inverse of conc. vs. time is lineark = +slopeintegrated rate law 1/[A]t = kt + 1/[A]0half life t1/2 = 1/k[A]01/[A]Time (t)[A]0= start concentration(From data table)

Lets clarify some thingszero orderRate = k[A]oplot concentration [A] vs. time (t)k = +slopeintegrated rate law [A]t = -kt + [A]0half life t1/2 = [A]0/2k[A]Time (t)1st order Rate = k[A]1 plot of ln conc. vs. time is lineark = +slopeintegrated rate law ln[A]t = -kt + ln[A]0half life t1/2 = .693/kLn[A]Time (t)2nd order Rate = k[A]2 plot inverse of conc. vs. time is lineark = +slopeintegrated rate law 1/[A]t = kt + 1/[A]0half life t1/2 = 1/k[A]01/[A]Time (t)t = time in seconds

Lets clarify some thingszero orderRate = k[A]oplot concentration [A] vs. time (t)k = +slopeintegrated rate law [A]t = -kt + [A]0half life t1/2 = [A]0/2k[A]Time (t)1st order Rate = k[A]1 plot of ln conc. vs. time is lineark = +slopeintegrated rate law ln[A]t = -kt + ln[A]0half life t1/2 = .693/kLn[A]Time (t)2nd order Rate = k[A]2 plot inverse of conc. vs. time is lineark = +slopeintegrated rate law 1/[A]t = kt + 1/[A]0half life t1/2 = 1/k[A]01/[A]Time (t)[A]t= concentration at time (t)

Lets clarify some thingszero orderRate = k[A]oplot concentration [A] vs. time (t)k = +slopeintegrated rate law [A]t = -kt + [A]0half life t1/2 = [A]0/2k[A]Time (t)1st order Rate = k[A]1 plot of ln conc. vs. time is lineark = +slopeintegrated rate law ln[A]t = -kt + ln[A]0half life t1/2 = .693/kLn[A]Time (t)2nd order Rate = k[A]2 plot inverse of conc. vs. time is lineark = +slopeintegrated rate law 1/[A]t = kt + 1/[A]0half life t1/2 = 1/k[A]01/[A]Time (t)t1/2= time when concentration halves

Lets try it:Time (s)[NO2] (mol/L)00.5001.20 x 1030.4449.00 x 1030.3814.50 x 1030.3409.00 x 1030.2501.80 x 1040.174Enter 4 columns of data:1st column = time2nd column = concentration3rd column = ln (concentration)4th column = 1/concentration

Lets try it:Do the linear regressions: order test Time vs. Concentration (1st , 2nd ) 1st order Time vs. ln(concentration) (1st, 3rd)2nd order Time vs. (1/concentration) (1st, 4th)

Each time:1. Perform linear regression2. Write down the slope3. Write down the correlation coefficient

What you should get:Do the linear regressions: order test Time vs. Concentration (1st , 2nd ) slope= -1.7 x 105corr coef. = -.9492941st order Time vs. ln(concentration) (1st, 3rd)slope= -5.8 x 105corr coef. = .9706862nd order Time vs. (1/concentration) (1st, 4th)slope= 2.1 x 104corr coef. = 0.999367

M = slope

r= correlation coefficientWinner=2ND ORDER!!!

Use formulas to solve for lots of stuff!!!2nd order

Rate = k[NO2]2 k = +slope = 2.1 x 104 1/Ms

k from linear regression

Units = M(1-order) = M(1-2) s s = M(-1) = 1 s Ms

Use formulas to solve for lots of stuff!!!2nd order

Rate = k[NO2]2 k = +slope = 2.1 x 104 1/Msintegrated rate law 1/[NO2]t = kt + 1/[NO2]0 1/[NO2]t = (2.1 x 104 1/Ms)t + 1/(0.500 M)half life t1/2 = 1/k[NO2]0 t1/2= 1/((2.1 x 1041/Ms)(0.500M)

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