1. { { Cl- + H2O + CO2  HCO3- + HCl Reactants Products R P
Reaction Order With Respect To
Cl- + H2O + CO2  HCO3- + HCl
Reactants
Products R P
rate = k[Cl-]x[H2O]y[CO2]z

2. Cl- + H2O + CO2  HCO3- + HCl Reaction Order With Respect To Exp.
Cl- (M)
Rel. conc.
H2O (M)
CO2 (M)
Rate (Ms-1)
Rate change 1
0.01
0.1
1.5x10-3

3. Cl- + H2O + CO2  HCO3- + HCl Reaction Order With Respect To Exp.
Cl- (M)
Rel. conc.
H2O (M)
CO2 (M)
Rate (Ms-1)
Rate change 1
0.01
0.1
1.5x10-3

4. Cl- + H2O + CO2  HCO3- + HCl Reaction Order With Respect To
Exp.
Cl- (M)
Rel. conc.
H2O (M)
CO2 (M)
Rate (Ms-1)
Rate change 1
0.01
0.1
1.5x10-3 2
0.02
none
When the concentration of Cl- was doubled, no change was observed in the reaction rate, hence the reaction is zero order with respect to Cl-

5. Cl- + H2O + CO2  HCO3- + HCl Reaction Order With Respect To
Exp.
Cl- (M)
Rel. conc.
H2O (M)
CO2 (M)
Rate (Ms-1)
Rate change 1
0.01
0.1
1.5x10-3 2
0.02
none 3
0.2
3.0x10-3
doubles
When the concentration of H2O was doubled, the observed reaction rate doubled, hence the reaction is 1st order with respect to H2O

6. Cl- + H2O + CO2  HCO3- + HCl Reaction Order With Respect To Exp.
Cl- (M)
Rel. conc.
H2O (M)
CO2 (M)
Rate (Ms-1)
Rate change 1
0.01
0.1
1.5x10-3 2
0.02 3
0.2
3.0x10-3 4
6.0x10-3

7. Cl- + H2O + CO2  HCO3- + HCl Reaction Order With Respect To
Exp.
Cl- (M)
Rel. conc.
H2O (M)
CO2 (M)
Rate (Ms-1)
Rate change 1
0.01
0.1
1.5x10-3 2
0.02
none 3
0.2
3.0x10-3 4
1.2x10-2 4x
When the concentration of CO2 was doubled, the observed reaction rate quadrupled, hence the reaction is 2st order with respect to CO2

8. Add all the exponent to get the total reaction order,
Reaction Order With Respect To
Cl- + H2O + CO2  HCO3- + HCl
rate = k[Cl-]0[H2O]1[CO2]2
rate = k[H2O]1[CO2]2
Add all the exponent to get the total reaction order,
For this reaction the order is 3rd

9. rate = k[R]x Zero Order 1st Order 2nd Order rate = k[R]0 rate = k[R]1
Overall Reaction Order
rate = k[R]x
Zero Order
1st Order
2nd Order
rate = k[R]0
rate = k[R]1
rate = k[R]2
rate = k
ln[Rt] = -kt + ln[Ro]
1/[Rt] = kt + 1/[Ro]
[Rt] = -kt + [Ro]
t1/2 = 0.693/k

10. The reaction, 2 NOCl  2NO + Cl2 obeys the rate law,
Student Example
The reaction,
2 NOCl  2NO + Cl2
obeys the rate law,
rate = 0.20 M-1s-1 [NOCl]2
If the initial concentration of NOCl is 0.50 M, calculate the concentration after 10 minutes.

11. NO + O3  [activated complex]  NO2 + O2
Reaction Coordinate Diagram
NO + O3  [activated complex]  NO2 + O2
NO + O3
[NO-O3]*
NO2 + O2 Ea
Energy
Reaction Coordinate

12. Catalysts
A catalyst is a substance that increases the reaction rate but is not consumed in the reaction.
A catalyst provides an alternate reaction path with a lower activation energy.
Reactants
Products
Uncatalyzed pathway
Catalyzed pathway
Reaction 
Energy

13. H2O2(aq) + 2Br-(aq) + 2H+(aq)  Br2(aq) + 2H2O(l) Step 2:
Reaction Steps
Step 1:
H2O2(aq) + 2Br-(aq) + 2H+(aq)  Br2(aq) + 2H2O(l)
Step 2:
H2O2(aq) + Br2(aq)  2Br-(aq) + 2H+(aq) + O2(g)
Overall reaction:
2H2O2(aq)  2H2O (l) + O2(g)

14. Chapter 13.6 Mechanisms II: Microscopic Effects Objectives
Describe a chemical reaction as a sequence of elementary processes.
Write the rate law from an elementary step and determine its molecularity.
Predict the experimental rate law from the mechanism and differentiate among possible reaction mechanisms by examining experimental rate data.

15. An elementary step describes an actual molecular level event.
Reaction Mechanisms
A mechanism is a sequence of molecular-level steps that lead from reactants to products.
An elementary step describes an actual molecular level event.
The concentration dependence in the rate law for an elementary step is given by the coefficients in the equation.

16. Molecularity
The molecularity of an elementary step is the number of reactants species involved in that step.
Most elementary steps are either unimolecular (involving a single molecule) or bimolecular (involving two molecules).

17. is believed to occur by the following sequence of elementary steps:
The gas phase reaction
2NO + O2  2NO2
is believed to occur by the following
sequence of elementary steps:
2NO ⇌ N2O2 bimolecular
N2O2 + O2  2NO2 bimolecular
2NO + O2  2NO2 overall reaction
N2O2 is an intermediate, not found among the reactants or products.

18. Rate Laws for Elementary Steps
The rate of an elementary step is proportional to the concentration of each reactant species raised to the power of its coefficient in the equation:
Step 1: 2NO ⇌ N2O2
rate1 = k1[NO]2
Step 2: N2O2 + O2  2 NO2
rate2 = k2[N2O2][O2]

19. Rate Limiting Steps
The overall rate of a multistep reaction is determined by its slowest step, called the rate-limiting or rate-determining step.
The rates of fast steps which follow the rate-limiting step have no effect on the overall rate law.
The rates of fast steps that precede the rate-limiting step usually affect the concentrations of the reactant species in the rate-determining step.

20. 2NO + O2  2NO2 Step 1: 2NO ⇌ N2O2 rate1 = k1[NO]2
Rate Determining Step
2NO + O2  2NO2
Step 1: NO ⇌ N2O2
rate1 = k1[NO]2
Step 2: N2O2 + O2  2NO2
rate2 = k2[N2O2][O2]
If the first step is slow and the rate determining step, the rate law is:

21. 2NO + O2  2NO2 Step 1: 2NO ⇌ N2O2 rate1 = k1[NO]2
Rate Determining Step
2NO + O2  2NO2
Step 1: NO ⇌ N2O2
rate1 = k1[NO]2
Step 2: N2O2 + O2  2NO2
rate2 = k2[N2O2][O2]
However, if the first step is rapid and the second step is the rate limiting step, the rate law is:

22. The definition of equilibrium is rate1 = rate-1 hence:
Complex Reaction Mechanisms
Step 1: NO ⇌ N2O2
rate1 = k1[NO]2
rate-1 = k-1[N2O2]
The definition of equilibrium is rate1 = rate-1
hence:
k1[NO]2 = k-1[N2O2] or
[N2O2] = (k1/k-1)[NO]2

23. The reaction is second order in NO and first order in O2.
Complex Reaction Mechanisms
Substituting into the expression for step 2: rate = k2[N2O2][O2] rate = k [NO]2[O2] Combine all the rate constants:
rate = k [NO]2[O2]
The reaction is second order in NO and first order in O2. k1
k -1

24. Student Example
Write the overall chemical equation by combining these two elementary steps: 2NO2 ⇌ N2O4
N2O4 + CO  NO2 + NO + CO2
If the 1st step is fast and the 2nd step is slow, determine the rate equation that omits the intermediate.

25. Most enzymes follow the Michaelis-Menten mechanism.
Enzyme Catalysis
Most enzymes follow the Michaelis-Menten mechanism.
Step 1: the enzyme binds the substrate in a rapid, reversible reaction to form a complex. This step reaches equilibrium. k1
k-1
E + S ES

26. Enzyme Catalysis
Step 2: the product forms from the complex, and the enzyme is released. This step is the rate-limiting step.
rate = k2[ES] k2 ES
E + P

27. Under normal conditions, [S] is much greater than [E]o.
Enzyme Catalysis k1
k-1
E + S ES k2 ES
E + P
Under normal conditions, [S] is much greater than [E]o.
Nearly all the enzyme is bound to the substrate, so [ES] ~ [E], and rate = k2[E] = constant
The reaction rate is zero order in substrate, S.