Chemical kinetics is the study of reaction rates, which includes how fast a reaction occurs, how the conditions change the rate and what molecular interactions are involved in a reaction step.

We define the reaction rate as the amount of reactant used, or amount of product used per unit time, and always has the units M/s. As rates are dependent on concentration, rate laws are typically written like so:

For that imaginary reaction, we have defined the rate as the decrease in concentration of A (which is why it is negative) over change in time. The rate is proportional to the concentrations raised to some power, (which must be determined experimentally, it is *not* the stoichiometric value!) and k, the rate (proportionality) constant.

The exponent values are referred to as the order of the reaction with respect to whichever species it is associated with, (in the previous example, the rate is nth order with respect to the concentration of B). The overall order of the reaction is the sum of the exponents.

*Determining the rate law (initial rates method):*

To determine the rate law, you need to know the order of reaction with respect to each reactant, and find the rate law. Typically, the experiment measures the initial rate at different concentrations, so the order can be calculated using simultaneous equations. Once the orders have been determined, it is easy to find k. An example (taken from *General Chemistry*, Ebbing, 5^{th} Edition) is shown below:

Iodide ions are oxidised in acidic solution to form triiodide ion (I_{3}^{–}), by hydrogen peroxide.

H_{2}O_{2(aq)} + 3I^{–}_{(aq)} + 2H^{+}_{(aq)} → I_{3}^{–}_{(aq)} + 2H_{2}O_{(l)}

The reaction was run at a variety of concentrations and the rate recorded:

We give the rate law the form:

Rate = k[H_{2}O_{2}]^{m}[I^{–}]^{n}[H^{+}]^{p}

To calculate the order with respect to H_{2}O_{2}, we look at experiments 1 and 2 where the concentration of H_{2}O_{2} doubles, but the other two stay the same. We divide one by the other:

We substitute the values in and see that the I^{–}, H^{+} and k values cancel leaving:

Simplify:

2 = 2^{m} therefore m = 1.

This procedure is repeated for [I^{–}] and [H^{+}], giving the rate law

Rate = k[H_{2}O_{2}][I^{–}]

The values for one of the experiments can be substituted in to find k, giving us:

Rate = 1.2×10^{-2}[H_{2}O_{2}][I^{–}]

*Integrated rate laws:*

The initial rate law is all well and good, but it doesn’t tell you what the concentration of a species will be at a given time during the reaction. To find this, we integrate the rate law and get the following:

For a first order rate law, (eg. Rate = k[A]^{a}) we get:

ln[A]_{t} = -kt + ln[A]_{0}

and a half life, (the time taken for half of your reactant to react) of:

For a second order reaction, we get the following:

Collision theory and reaction mechanisms to come…