1. Kinetics

2. Introduction
Rate of Reaction – the rate at which reactants are consumed and products are formed.
Chemical Kinetics – the study of the rates of reactions.

3. Factors that affect Reaction Rates
Nature of the Reactants – surface area and color
Concentration
Temperature
Presence of a Catalyst

4. Rate = k[A]x[B]y
Zero order – the concentration of the reactant does not affect the rate of production of product. The concentration of reactant is doubled, the rate of production of product not affected.
First order – the concentration of the reactant does affect the rate of production of product. The concentration of reactant is doubled, the rate of production of product double.

5. Second order – the concentration of the reactant does affect the rate of production of product. The concentration of reactant is doubled, the rate of production of product quadruple.
Overall order = sum of all orders

6. Example 1
Expt # [CH3CHO] Rate (M/sec)

7. Example 2 Expt # [CO] [NO2] Rate (M/hr)
x x x 10-8
x x x 10-8
x x x 10-8
x x x 10-8
x x x 10-7

8. 2A + B  C + D Expt [A] [B] Rate (M/sec) 1 0.10 0.05 6.0 x 10-3

9. 2A + B + C  D + E Expt [A] [B] [C] Rate (M/min)

10. C2H4 + O3  2CH2O + ½ O2 Expt [O3] [C2H4] Rate (M/sec)
x x x 10-12
x x x 10-12
x x x 10-12

11. First Order Reactions ln([A]0/[A]) = akt A0 = Initial Concentration
A = Concentration at some point
a = Coefficient of A
k = Rate Constant
t = Time

12. Half Life of a First Order Rxn
t1/2 = ln2/ak

13. Compound A decomposes to form B and C in a reaction that is first order with respect to A and first order overall. At 25C, the specific rate constant for the reaction is 4.50 sec-1. What is the half life of A?
A  B + C

14. The reaction 2N2O5  2N2O4 + O2 obeys the rate law: rate = k[N2O5], in which the specific rate constant is sec-1 at a certain temperature. If an initial molarity is 4.2M, what would the molarity of it be after 1 minute?

15. How long would it take for it to decrease to 0.5M?

16. Second Order Rxn
1/[A] - 1/[A]0 = akt
t1/2 = 1/(ak[A]0)

17. Compound A reacts to form C and D in a reaction that was found to be second order overall. The rate constant is M-1 min-1. What is the half life of A when 4.10 x 10-2 M A is reacted.

18. The gas phase decomposition of NOBr is second order in [NOBr] with k = M-1sec-1 at 10C. We start with 4.00 x 10-3M NOBr in a flask at 10C. How many seconds does it take to reduce the concentration of NOBr to
2.5 x 10-3 M?
2NOBr  2 NO + Br2

19. Consider the reaction in the previous problem. If we start with 6
Consider the reaction in the previous problem. If we start with 6.2 M NOBr, what concentration of NOBr will remain after 5.00 minutes of reaction?

20. Rate Determining Step
A reaction can never proceed faster than its slowest step.
Intermediates – Substances that are produced by one reaction and later consumed by another reaction.
Catalyst – Substances that start a reaction and later produced by another reaction

21. Example 1 NO2 + NO2  N2O4 (fast) N2O4 + CO  NO + CO2 + NO2 (slow)
What are the intermediates?
What is the rate law for this reaction?

22. Example 2 NO + Br2  NOBr2 (slow) NOBr2 + NO  2NOBr (fast)
What are the intermediates?
What is the rate law for this reaction?

23. 2B  C + D Expt [B] Rate (M/sec) 1 0.40 6.0 x 10-3 2 0.80 1.2 x 10-2 3
1.20
1.8 x 10-2
What is the order with respect to B?
What is the rate law?
What is k?
If the initial concentration of B is 4.2, how long would it take for it to reach 0.5M?
Which of the following mechanisms correctly justifies your rate law? Justiify
B  C + E (Slow)
B + E  D (Fast)
2. A + B  C + E (Fast)
B + E  A + D (Slow)

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