This is a very fast and easy section on kinetics, the study of reactions and speed. As always please do not use just the information provided here and always go to your textbook for information. These sections are only designed to assist in helping to learn.
The Rate Law
The rate law for a reaction is the dependence of the initial rate of a reaction on the concentration of the reactants. The rate of a reaction is just how fast the reactants disappear, or how fast the products appear. To determine the rate constant of a particular reaction, you must use experimental procedures like this:
A + 2 B + C --> D
Experiment |
Initial Concentrations of Reactants |
Initial rate of Formation of D |
||
[A] |
[B] |
[C] |
||
1 |
0.10 |
0.10 |
0.10 |
0.01 |
2 |
0.10 |
0.10 |
0.20 |
0.01 |
3 |
0.10 |
0.20 |
0.20 |
0.02 |
4 |
0.20 |
0.20 |
0.20 |
0.08 |
Now with this information and the equation Rate = k[A]x[B]y[C]z , where k is the Rate Constant, we can find the rate constant law by comparing similar experiments.
To find [A]x
The only two experiments that have a change in the concentration in [A] are experiments 3 and 4, so we will only be looking at those for this part. The easiest way to figure out x is to look at the numbers logically. The only thing that is changing in these two experiments are [A] (what we are looking for) and the Rate of formation of D. So to figure this out, just think to yourself, [A] is doubling (0.10 to 0.20) while the Rate of formation is being multiplied by four (0.02 to 0.08). So set up an equation for x: (2)x = 4, therefore x = 2, we have found our answer.
To Find [B]y
For this one, it is best to look at experiments 2 and 3, because [B] is doubling while the rest remain constant. Exactly the same as finding [A]x we look at the doubling of [B] (0.10 to 0.20) and the rate of Formation of D (0.01 to 0.02, also doubling). So now we set up an equation to solve for y, (2)y = 2, therefore y will equal 1.
To Find [C]z
Just like the previous ones, look for one where the others are constant and only [C] is changing. For [C] we will use experiments 1 and 2. Here [C] is doubling (0.10 to 0.20) while the Formation of D is remaining constant (0.01 to 0.01). Again, set up an equation to solve for z, (2)z = 1, so z is going to equal 0. This means that [C] will have no effect on the rate of this reaction.
Putting it all together to find K
We now know most of the rate law for this problem, Rate = k[A]2[B], there is no [C] because [C]0 is just 1. Now we just plug in numbers from any experiment to find k, so we will use experiment 1. The Rate is equal to the rate of formation of D so we use 0.01 = k[0.10]2[0.10] and solve for k. Upon solving we find that k is equal to 1 X10-3 M/sec.
Reaction Mechanisms
Many chemical reactions do not occur in one step, rather they are a combination of steps that occur at different speeds. For instance for the reaction 2 A + 2 B --> C + D could happen like this:
I.
A + A <--> X (Fast)
II.
X + B --> C + Y (Slow)
III.
Y + B --> D (Fast)
The important part to remember in this is that the slowest part is the rate determining step, so here is would be the second step.
Catalysts
The last part to mention about kinetics is that of catalysts. Catalysts are things that increase the rate of a chemical reaction without being consumed in the process, so they will not appear in the final balanced equation. A catalyst usually does this by bringing down the activation energy, so the reaction can go on much more quickly.