Kinetics, or rates of chemical reactions, represent one of the most complex topics faced by high-school and college chemistry students. The rate of a chemical reaction describes how the concentrations of products and reactants changes with time. As a reaction proceeds, the rate tends to decrease because the chance of a collision between reactants becomes progressively lower. Chemists therefore tend to describe reactions by their “initial” rate, which refers to the rate of reaction during the first few seconds or minutes.
In general, chemists represent chemical reactions in the form
aA + bB ---> cD + dD,
where A and B represent reactants, C and D represent products, and a, b, c and d represent their respective coefficients in the balanced chemical equation. The rate equation for this reaction is then
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rate = (-1 ÷ a) d[A] ÷ dt = (-1 ÷ b) d[B] ÷ dt = (1 ÷ c) d[C] ÷ dt = (1 ÷ d) d[D] ÷ dt,
where square brackets denote the concentration of the reactant or product; a, b, c and d represent the coefficients from the balanced chemical equations; and t represents time.
Write a balanced chemical equation for the reaction under investigation. As an example, consider the reaction of hydrogen peroxide, H2O2, decomposing to water, H2O, and oxygen, O2:
H2O2(2) ---> H2O(2) + O2.
A “balanced” reactions contains the same number of each type of atom on both the left and right sides of the arrow. In this case, both sides contain four hydrogen atoms and two oxygen atoms.
Construct the rate equation based on the equation given in the Introduction. Continuing the example from step 1:
rate = -(1 ÷ 2) d[H2O2] ÷ dt = (1 ÷ 2) d[H2O] ÷ dt = (1 ÷ 1) d[O2] ÷ dt.
Substitute the concentration and time data into the equation from step 2 based on the information available in the problem or obtained during an experiment. For example, for the reaction described above, assume the following data was obtained:
time (s), [H2O2] (M) 0, 0.250 10, 0.226
This data indicates that after 10 seconds, the concentration of hydrogen peroxide decreased from 0.250 moles per liter to 0.226 moles per liter. The rate equation then becomes
rate = -(1 ÷ 2) d[H2O2] ÷ dt = -(1 ÷ 2) (0.226 - 0.250) ÷ 10 = 0.0012 M/s.
This value represents the initial rate of the reaction.