1. Chemical Kinetics Chapter 14

2. The Rate Law Rate law – description of the effect of concentration on rate aA + bB cC + dD Rate = k [A] x [B] y reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall

3. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] Rate Laws Rate laws always determined experimentally Reaction order always defined in terms of reactant (not product) concentrations Order of a reactant is not related to the stoichiometric coefficients 1

4. Concentration and Initial Rates NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O(l) Double [NH 4 + ] with [NO 2 − ] constant:x = 1Rate doubles Assume: rate = k [NH 4 + ] x [NO 2 − ] y Double [NO 2 − ] with [NH 4 + ] constant:Rate doublesy = 1 Therefore rate law is rate = k [NH 4 +][NO 2 − ] Table 14.2

5. Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq) Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) 10.080.0342.2 x 10 -4 20.080.0171.1 x 10 -4 30.160.0172.2 x 10 -4 rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 k = rate [S 2 O 8 2- ][I - ] = 2.2 x 10 -4 M/s (0.08 M)(0.034 M) = 0.08/M s rate = k [S 2 O 8 2- ][I - ]

6. Relation Between Concentration and Time Rate law provides rate as a function of concentration Need relationship between concentration and time : i.e., How do we determine the concentration of a reactant at some specific time?

7. First-Order Reactions A product rate = −  [A] tt rate = k [A] k = rate [A] = 1/s or s -1 M/sM/s M =  [A] tt = k [A] − [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 ln[A] = ln[A] o - kt Integrated rate law Which now integrates to:

8. Consider the process in which methyl isonitrile is converted to acetonitrile CH 3 CN Fig 14.7(a) Data collected for this reaction at 198.9 °C. First-Order Reactions CH 3 NC

9.  Plot of ln P vs time, results in a straight line  Therefore,  Process is first-order  k is the negative of the slope: 5.1  10 -5 s −1 First-Order Reactions Fig 14.7

10. orange red-brown Decomposition of N 2 O 5 2 N 2 O 5 (in CCl 4 ) → 4 NO 2 (g) + O 2 (g) →

11. Decomposition of N 2 O 5 2 N 2 O 5 (in CCl 4 ) → 4 NO 2 (g) + O 2 (g) Plot of ln [N 2 O 5 ] vs time Linear plot indicates: 1st order

12. Second-Order Reactions A product rate = −  [A] tt rate = k [A] 2  [A] tt = k [A] 2 − [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 1 [A] = 1 [A] o + kt One type: A + B product Second type: rate = k [A][B] Initial rate law: Combining the two rate expressions: Which now integrates to: Integrated rate law

13. The decomposition of NO 2 at 300°C is described by the equation NO 2 (g)NO (g) + 1/2 O 2 (g) plot of ln[NO 2 ] vs time: Second-Order Reactions plot of 1/[NO 2 ] vs time: Fig 14.8 1 [A] = 1 [A] o + kt

14. Half-life = time required for one-half of a reactant to react t ½ = t when [A] = [A] 0 /2 Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0 1 st order t 1/2 = 0.693/k 2 nd order Fig 14.9 1 st order rxn t ½ = 1 k[A] o Half-Life

15. Summary of the Kinetics of Reactions OrderRate Law Concentration-Time Equation Half-Life 0 1 2 rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] o - kt 1 [A] = 1 [A] o + kt [A] = [A] o - kt t½t½ ln2 k = t ½ = [A] o 2k2k t ½ = 1 k[A] o