Chemical kinetics study reaction rates in a chemical reaction. When a chemical reaction takes place, a collision of sorts happens and the collision has to be strong enough to break bonds. There are several factors that affect the reaction rate such as concentration (according to the rate law), temperature and whether or not a catalyst is present.
A rate law describes the behavior of a chemical reaction. It's not possible to look at a reaction, & determine the rate law. You can, however, determine the rate law by observing the rate determining step of the reaction mechanism. Chemical reactions go through many steps before bonds are broken and new ones are formed. Everything that a chemical reaction goes through to break these bonds and form new ones is called the reaction mechanism 9which is usually not covered in general chemistry). The rate law of a reaction is determined experimentally by observing the appearance of the products or by noticing that some reactants disappear.
Let's look at this chemical reaction:
And its rate law:
In this rate law, the rate at which NO & O2 disappear is proportional to their concentrations and k is a proportional constant. The exponents are called "partial order". The sum of the partial orders will give us the overall order of our reaction. In the rate law shown above, NO has a partial order of 2 and is called 2nd order. O2's partial order is 1 and is called 1st order. Add them up and you get a 3rd order. It's important to understand that coefficients in the balanced equation are not the same as the exponents in the rate law.
For our next problem, we'll examine experimental data that we can use to find the rate law. We're going to find the rate by looking at the disappearnace of reactants.
We'll do this by compiling reaction rates in terms of the initial concentration of reactants:
In order to find the rate law, we need to look at the experimental data. Choose two of the trials to compare and choose ones where the concentration of a reactant is doubled and the other reactant stays the same. We're going to find the partial order of the reactant that has had a change in concentration.
We'll start by looking at trials 1 and 2. NO's concentration stayed constant but O2's concentration doubled. In order to find O2's partial order we need to consider the fact that its concentration doubled, which in turn doubled the rate of the reaction. We can set this up as an algebra problem:
Solving for X, we get 1, so O2 has a first order partial order.
Next we'll find NO's partial order. As before, pick two trials and choose a trial where O2's concentration stayed the same and NO's changed. We can see that these conditions are met in trials 1 and 3. O2's concentration stayed the same but NO's concentration tripled which means the rate of the reaction was increased nine times. We can set this up as an algebra problem:
Solving for X, we get 2, so NO has a second order partial order.
We now have enough information to write the law:
We need to solve for the rate constant. We'll randomly choose trial 1 and substitute in the rate and concentrations for NO & O2:
Which makes our rate law:
Determining Energy of Activation
Before you can have a chemical reaction, there has to be a collision of sufficient force and direction to break bonds and form new ones. The breaking of bonds takes energy and when new bonds are made, energy is released. This energy comes from the kinetic energy that the compound has due to its molecules moving about. At the beginning of this tutorial I mentioned that kinetic energy is influenced by temperature. Higher temperatures cause the molecules to move faster and produce more kinetic energy. This is called energy of activation. A collision must also have the proper orientation also called the collision factor. Each of these factors relate to one another mathematically by the Arrhenius equation:
If we graphed our data with log k on the y-axis and 1/T on the x-axis, we'd get a straight line. The slope of our line is Ea/2.303R, & the y-intercept corresponds to log A. By finding the slope, we can find the energy of activation. If we extend this line to the y-intercept, we can calculate the collision factor.
Since slope gives us the energy of activation, we can simplify the Arrhenius equation to find the energy of activation. We need to know the rate constants for two different temperatures. Remember that the slope of a line is found by using the formula:
Knowing this, we can write the equation:
The y-intercept (log A), is not a part of the slope, so we can ignore it. We can now use this modified equation to solve this problem:
The rate of reaction for a certain chemical process is viewed with two different temperatures.
Using this data, find the energy of activation in J/mol?
First, examine the units. The units for rate, (s-1), cancel. The energy of activation is supposed to be found in J/mol, so the gas constant, R, becomes 8.314 J/mol K. This gives us our units for temperature so all other temperatures need to be converted to K. Remember, add 273 to Celsius to find Kelvin:
Plug the rate constants & temperatures in the Arrhenius equation:
Solve for Ea. Remember to follow the order of operations.
