Not a straight line!!!! Therefore, not zero order. 2N2O5 4NO2(g) + O2(g) Not a straight line!!!! Therefore, not zero order.

Not a straight line!!!! Therefore, not 2nd order 2N2O5 4NO2(g) + O2(g) Not a straight line!!!! Therefore, not 2nd order

Find the [N2O5] present at 15 minutes into the reaction. 2N2O5 4NO2(g) + O2(g) Produces the best straight line…therefore the reaction is first order and the rate law is Rate = k [N2O5] ln [N2O5] = -2.3 [N2O5] = 0.10 k = 0.030 min-1 Find the [N2O5] present at 15 minutes into the reaction. [N2O5] = 0.106 M at 15 minutes ln [N2O5] = -kt + ln [N2O5]0

How long does it take to use up 10.0% of the N2O5? 2N2O5 4NO2(g) + O2(g) How long does it take to use up 10.0% of the N2O5? t = 3.5 min. k = 0.030 min-1 Find the [N2O5] present at 15 minutes into the reaction. [N2O5] = 0.106 M at 15 minutes ln [N2O5] = -kt + ln [N2O5]0 ln ([N2O5]/[N2O5]0)= -kt

N2O5(g)  2NO2(g) + ½O2(g) First order in N2O5 (known from previous example) Dinitrogen pentoxide decomposes when heated. If the rate constant for the decomposition of N2O5 is 6.2 x 10-4/min, what is the half-life? How long would it take for the concentration of the N2O5 to decrease to 25% of its initial value? …to 12.5% of its initial value Time needed for the original concentration to be cut in half. t½ = 1100 min. …to 25% of initial [ ] = 2200 min. (two half-lifes) …to 12.5% of initial [ ] = 3300 min. (three half-lifes)

Rate = k [ClO2]m[ I-]n = k' [ClO2]m 2ClO2(aq) + 2 I-(aq)  2ClO2-(aq) + I2(aq) If [ I-] is quite large with respect to [ClO2]… k' = k [ I-]n Rate = k [ClO2]m[ I-]n = k' [ClO2]m If you needed to determine the order of the reaction with respect to the ClO2 what would you need to know? Time [ClO2] 0.00 s 0.000477 M 1.00 s 0.000431 M 2.00 s 0.000391 M 3.00 s 0.000353 M [ClO2]-1 ln [ClO2] 2.10 x 103 M-1 -7.65 2.32 x 103 M-1 -7.75 2.55 x 103 M-1 -7.85 2.83 x 103 M-1 -7.95 Most linear… therefore first order with respect to the chlorine dioxide