kinetics – the speed (or rate) at which a reaction takes place 2 H 2 O 2 (aq) 2 H 2 O ( ) + O 2 (g) 2 ways to measure the rate of any reaction 1. measure the Increase in [Products] over time 2. measure the Decrease in [Reactants] over time How do you measure this ? measure a change in concentration per change in time

slope = yxyx Rate =   [ ]   t slope = Rate

* By convention, all rates are reported as positive values Rate =   [ Products ]   t when   [ Products ]   t slope = + # when   [ Reactants ]   t slope = - # Rate = -   [ Reactants ]   t

instantaneous rate – rate of rxn at a particular instant of time

initial rate – instantaneous rate at t = 0 (very beginning)

slope of tangent line at t = 0 is x M/sec Rate = -   [H 2 O 2 ]   t = - ( x M/sec) Rate = 2.0 x M/sec 2 H 2 O 2 (aq) 2 H 2 O ( ) + O 2 (g)

Can the rate of H 2 O 2 being consumed be related to the rate of products being generated ? 2 H 2 O 2 (aq) 2 H 2 O ( ) + O 2 (g) [H 2 O] increases at a rate of 2.0 x M/sec as [H 2 O 2 ] decreases at a rate of 2.0 x M/sec [O 2 ] increases at a rate of 1.0 x M/sec as [H 2 O 2 ] decreases at a rate of 2.0 x M/sec What is the rate that O 2 forms when the [H 2 O 2 ] is decreasing at a rate of 2.0 x M/sec ?

The rate of any rxn depends of four things: 1. physical state of the reactants (usually means surface area) 2. concentration of the reactants 3. temperature of the reactants 4. presence of a catalyst

Rate vs. [Reactants] 2 H 2 O 2 (aq) 2 H 2 O ( ) + O 2 (g) rate = k [H 2 O 2 ] x rate law expression takes the form: rate = rate of the reaction k = rate constant [H 2 O 2 ] = concentration of reactant x = order of the reactant (usually an integer)

what if the rxn has multiple reactants ? 2 HgCl 2 + C 2 O Hg 2 Cl CO Cl - Rate = k [HgCl 2 ] x [C 2 O ] y k = rate constant indicative of this rxn x = order wrt [HgCl 2 ] y = order wrt [C 2 O ] x and y must be determined experimentally

Trial[HgCl 2 ][C 2 O 4 2- ]initial rate, M/sec M0.30 M 7.65 x M0.30 M 1.53 x M0.10 M1.70 x to determine the numerical value for x and y, examine the data and find 2 trials where [reactant] of interest is varied, while all other reactants are held constant. Compare rate law expressions for these two trials to determine the order of that reactant Method of Initial Rates

Trial[HgCl 2 ][C 2 O 4 2- ]initial rate, M/sec M0.30 M 7.65 x M0.30 M 1.53 x M0.10 M1.70 x Use trials 1 and 2 to obtain the numerical value for the order wrt [HgCl 2 ] (which is x) examine trials 1 and 2 [HgCl 2 ] varies while the [C 2 O ] is held constant Rate = k [HgCl 2 ] x [C 2 O ] y x = 1 unitless

Trial[HgCl 2 ][C 2 O 4 2- ]initial rate, M/sec M0.30 M 7.65 x M0.30 M 1.53 x M0.10 M1.70 x Use trials 2 and 3 to obtain the numerical value for the order wrt [C 2 O ] (which is y) examine trials 2 and 3 Rate = k [HgCl 2 ] x [C 2 O ] y [C 2 O ] varies while the [HgCl 2 ] is held constant y = 2 unitless

Rate = k [HgCl 2 ] x [C 2 O ] y x = 1 y = 2 Rate = k [HgCl 2 ] 1 [C 2 O ] 2 x and y do NOT vary under any circumstances ! or Rate = k [HgCl 2 ] [C 2 O ] 2 rxn is 1 st order wrt HgCl 2 rxn is 2 nd order wrt C 2 O 4 2 -

Rate = k [HgCl 2 ] [C 2 O ] 2 Determine the numerical value for the rate constant, k = k Rate [HgCl 2 ] [C 2 O ] 2 1 M 2 ·sec 8.5 x k = k is constant and does NOT vary with [ ] units are very important k does vary with temperature

rate constant,

overall reaction order – sum of the individual reaction orders x + y = =3 reaction is overall 3 rd order

[ ] vs. time Δ aAaAbB + jJ if Rate = k[A] 1 rxn is overall 1 st order ln [A] o [A] t = kt [A] o = initial [A] at t = 0 [A] t = [A] at some time later, t k = rate constant t = time from t = 0

SO 2 Cl 2 decomposes as follows: SO 2 Cl 2 SO 2 + Cl 2 The rate constant, k = 2.2 x at 300  C. Suppose 0.30 M SO 2 Cl 2 is placed in a container at 300  C and allowed to decompose. What is the [SO 2 Cl 2 ] after 75 minutes ? 1 sec whenever the rate constant, k has units of 1/time ( or time - 1 ), the reaction is overall 1 st order ! 1 t

half-life – time required to convert exactly ½ reactants into products abb. for half-life is t 1/2 for rxn is overall 1 st order t 1/2 = k

Prozac is a prescription drug which is metabolically eliminated from the body by overall first order kinetics. Prozac has a body clearance half-life of x 10 5 sec. If a patient is given a 20.0 mg dose, how many hours until only 13.8 mg Prozac remains in the body ?

[ ] vs. time Δ aAaAuU + kK if Rate = k[A] 2 rxn is overall 2 nd order 1 [A] t = kt 1 [A] o t 1/2 = 1 k [A] o

ln [ ] vs. time is linear [ ] vs. time is NOT linear only occurs when the rxn is overall 1 st order slope = - k CH 3 NC CH 3 CN

NO 2 NO + ½ O 2 This rxn is NOT 1 st order only occurs when the rxn is overall 2 nd order vs. time is linear 1 [ ] slope = k

Temperature vs. Rate An increase in temperature results in an increase in the rate of a reaction collision theory – for a reaction to occur, reacting species must first collide 1. collision must possess enough energy to react 2. collision must have reactants in the proper orientation to react

activation energy – the minimum energy reactants must attain to react to form products EaEa ΔHΔH Reactants Products

how is activation energy, E a, determined ? In lab, keep [reactants] constant and vary the temperature. Calculate values of the rate constant, k at each temperature. Svante Arrhenius 1859 – 1927

Arrhenius equation k = (A) (e (−E a /RT) ) y = mx + b ln k = + ln A - E a RT take ln of both sides slope = - Ea R a plot of ln k vs. gives a straight line 1T1T

1T1T K -1 slope = - 19,000 K ln k vs. 1T1T K -1

slope = -EaR-EaR E a = - (slope) (R) slope = - 19,000 K E a = - ( - 19,000 K) (8.314 J/mole·K) E a = 157,966 J/mole E a = 158 kJ/mole E a must be positive

Arrhenius equation (again) = ln k1k2k1k2 1T11T1 1T21T2 EaREaR Trialk, 1/M·secTemp x  C x  C 2 HI H 2 + I 2 given the data above, determine the E a of the rxn

catalysts – increase the rate of rxn without being consumed 1. decrease the E a by providing a different pathway to products 2. provide the correct orientation for the reactants to react

uncatalyzed rxncatalyzed rxn

correct orientation Fritz Haber 1868 – 1934 Karl Bosch 1874 – 1940 N H 2 2 NH 3 Haber-Bosch process

catalyst is primarily Fe & Fe 2 O 3 H 2 and N 2 continuously added NH 3 has no affinity to absorb on surface H 2 and N 2 adsorb on the surface of the iron 3 H and N react on surface of the iron