1. The Progress of Chemical Reactions
Chemistry 122

2. Rate Laws
One of the variables influencing the rate, or speed of a reaction, is concentration
Due to effective collisions, the number of reactant particles eventually decrease and the reaction slows down
When a reactant’s concentration decreases, the products concentration increases
This makes the rate of a reaction always positive

3. A → B
When there is only one reactant (decomposition), its concentration can be written into the formula Rate = k[A]
Ex. H2O2 → H2 + O2
Rate = k[A]m
= k[H2O2]1
But what does it mean?

4. The significance of ‘k’
When ‘k’ is large, the rate of reaction is fast and products form quickly
Units include L/(mol∙s), L2/(mol2∙s), and s-1
k is unique for every reaction and must be supplied if we are to determine the rate of the reaction

5. Reaction order
Rate = k[A]; it is understood that the notation [A] means the same as [A]1
This exponent ‘1’ is referred to as its reaction order, not the coefficient
It is used to define how the rate is affected by the concentration of the reactant
When it is in the first order, or exponent = 1, the rate changes proportionally to the concentration

6. As the concentration changes, so does the reaction rate
First order reactions
Rate = k[A]m
As the concentration changes, so does the reaction rate
If the concentration decreases by ¼, the reaction rate slows down by ¼.
If the concentration doubles, so does the rate of reaction
Page 577

7. Second order reactions
In double replacement reactions, there are two reactants
Therefore, both concentrations have to be considered when determining the rate of reaction
The rate is calculated as follows:
Rate = k[A]m[B]n
The rate has to be determined through experimentation since it is not always the case that the exponents in the rate law correspond to the coefficients in a chemical equation

8. Finding the order of a reaction
2NO(g) + 2H2(g) → N2(g) + 2H2O(g)
This reaction occurs in more than one step
Rate = k[NO]2[H2]1
This has been determined experimentally, it is not a reflection of the balanced chemical equation
The reaction is described as second order in NO, first order in H2, and third order overall because the sum of the orders for the individual reactants is = 3.

9. In the real world of chemistry
Most reactions are more complex than one-step reactions
As a result, reactions cannot be predicted solely on the basis of a balanced equation; it must be determined experimentally
This is because the rate depends primarily on the slowest step in the reaction mechanism

10. Using a graph to determine the rate of reaction
When the rate of reaction produces a curve, you must draw a tangent to the curve in order to determine k
Take the slope of the line at two points along the curve and determine the rate (concentration/time)

11. Determining reaction order
The method of initial rates determines reaction order by comparing the initial rates of reaction carried out with varying reactant concentrations
Use the expression aA + bB → products as well as rate = k[A]m[B]n

12. Example Trial Initial [A] (M) Initial [B] (M) Initial Rate 1 0.100
Compare the concentration of Trial 1 and 2 to their respective rates.
Compare Trial 2 and 3 to their respective rates.
What do you notice?
Which reactant is first order? Which is second?
What is the order of the overall reaction?
Write the rate in the form of an equation.
Trial
Initial [A] (M)
Initial [B] (M)
Initial Rate 1
0.100
2.00 x 10-3 2
0.200
4.00 x 10-3 3
16.0 x 10-3
Example

13. Questions
1. Write the rate law for the reaction aA →bB if the reaction is third order in A. [B] is not part of the rate law. 2. Given the following experimental data, use the method of initial rates to determine the rate law for the reaction aA + bB → products. Hint: any number to the zero power equals one. For example, (0.22)0 = 1 and (55.6)0 = 1.
Trial
Initial [A] (M)
Initial [B] (M)
Initial Rate 1
0.100
2.00 x 10-3 2
0.200 3
4.00 x 10-3

14. Questions
3. Given the following experimental data, use the method of initial rates to determine the rate law for the reaction CH3CHO(g) → CH4(g) + CO(g)
Trial
Initial [CH3CHO] (M)
Initial Rate (mol/(L∙s) 1
2.00 x 10-3
2.70 x 10-11 2
4.00 x 10-3
10.8 x 10-11 3
8.00 x 10-3
43.2 x 10-11

15. Reaction mechanisms
If we were to know all the changes that occur as reactants turn into products, we would be able to graph the transformation
This is known as a reaction progress curve
The simplest curve is for an elementary reaction
Such reactions convert reactants to products in a single step
There is only one activated complex

16. Reaction mechanism
Most chemical reactions occur as a series of steps which produces many peaks and valleys
The series of reactions that take place constitute a reaction mechanism
The peaks are the individual activated complexes and the valleys are the energies of the intermediates
An intermediate is a temporary product; becoming a reactant in the next step of the overall reaction
Figure 18.28, p. 578

17. For the remainder of class
Guided reading for section 18.5
Questions 36, 37, p. 577
Important points to consider:
Rate in most cases is mol/L/s or mol∙s/L,
Anything is square brackets [A] is typically in molarity (mol/L).
Q. 38 – 42, p. 579.