This page deals specifically with first- and second-order reaction kinetics, but you should know that other orders such as zeroth-order and 3rd order may also be involved in chemical kinetics.
d[H_{2}O_{2}] 1 d[H^{+}] 1 d[I^{-}] 1 d[H_{2}O] d[I_{3}^{-}] - ------- = - - ------ = - - ------- = - ------ = -------- dt 2 dt 3 dt 2 dt dtIn reality, if the concentration of H_{2}O_{2} is low, the changes in concentrations of H^{+} and H_{2}O are very difficult to detect because their quantities are so large in the solution. The merit in this equation is to show you that the rates of decreasing of reactant concentrations are governed by the stoichiometry. So do the rates of increasing of products concentrations.
There are many ways to express the reaction Rates, for example
d[H_{2}O_{2}] 1 d[H^{+}] 1 d[I^{-}] 1 d[H_{2}O] d[I_{3}^{-}] - ------- = - - ------ = - - ------- = - ------ = -------- = Rate dt 2 dt 3 dt 2 dt dtor
d[I^{-}] - ------- = Rate' dtObviously,
To generalize it, let the chemical reaction be represented by,
1 d[A] 1 d[B] 1 d[C] 1 d[D] rate = - --- ---- = - --- ---- = --- ---- = --- ---- a dt b dt c dt d dt
k A + other reactants -----> productswhere A is one of the reactants, and k is the rate constant.
For most experiments in chemical kinetics, the concentration of one reactant or product is monitored. If the concentrations of other reactants are high, they are not greatly changed. Thus, we have a pseudo decomposition reaction.
For the decomposition of A with a rate constant k,
d[A] - ---- = k [A]^{n} dtis called the differential rate law.
The differential expressions can be integrated to give an explicit relation of [A] with respect to time t. These explicit relations are called integrated rate laws.
Depends on the value of n, the integrated equations are different. If the reaction is first order with respect to [A], integration with respect to time, t, gives:
For a second order reaction, the integrated rate law is:
1 1 --- = ---- + k t - - - - (2) [A] [A]_{o}Derive the above equations yourself.
Determination of Rate Constants Using the Integrated Rate Laws
The usual approach to calculate the rate constant k makes use of the differential rate law. A series of experiments are performed with various initial concentrations, and their rates measured. The rate constants are calculated from the initial concentration and time of measurement. Often, you should construct a graph for their evaluation. However, results so obtained contain errors due to the approximation whereas values for k calculated using the integrated rate laws (1) and (2) are more accurate. Today, calculations can easily be performed with the aid of calculators and computers.
Thus, we emphasize that you use the integrated rate laws whenever possible. Whether you use the differential rate laws or the integrated laws, you have to evaluate the order first.
Equation (1) may be rewritten as
ln [A]_{o} - ln [A] k = ----------------- - - - - - - - (1') tIn a real experiment, you should plot - ln [A] vs. t and find a line to best fit your data. The slope of the line is k.
Similarly, equation (2) may be rewritten as:
1 1 --- - ---- [A] [A]_{o} k = -------------- - - - - - - - (2') tThus, plot of 1/[A] vs. t should yield a straight line, and the slope is k.
Either plot [A] vs. t or plot the appropriate linear relationship will reveal the order of the reaction. Both types are available in the simulation. Unfortunately, when we switched to the present format, the simulation cannot be implemented here.
A summary of the relationships for first and second order reactions are given below:
\ | Differential rate law | Integrated rate law | Linear plot | Half life |
---|---|---|---|---|
first order | - d[A]/dt = k [A] | [A] = [A]_{o} exp (-k t) | ln [A] vs t slope = - k | hl=ln(2)/k |
second order | - d[A]/dt = k [A]^{2} |
1 1 --- - ---- = k t [A] [A]_{o} |
1 --- vs t [A] slope = k | hl=1/k[A]_{o } |
Further applications of rate laws will be discussed in Integrated Rate Laws. Students must analyze the problems carefully, and appropriately apply the differential or integrated rate laws to derive the desirable results.
Discussion -
Since: - d[X]/dt = k; => [X] - [X]_{o} = - k t;
The plot of [X] and t has a linear relationship.
Discussion -
All radioactive decays follow first order kinetics.
Discussion -
k * half-life = ln(2) = 0.693
Discussion -
The first 12.5 y reduced 1.0 g of T to 1/2 g.
Another 12.5 y reduced 0.5 g of T to 1/4 g.
Calculate the time for 1.0 g T to reduce to 0.9 g.
Calculation of half-life from experiment is also a common exercise.
Discussion -
Amount left: 30, 15, 7.5, 3.75 Time: 8:00 8:45 10:15 13:15 t'increment: 45 90 180 (min)Calculate the time when 10 percent of the HI has decomposed.
Discussion -
Chain reactions may lead to explosion, but dust explosion is not a chain reaction.
Chemists explained the causes of a grain elevator explosion.
run [NO] [O_{2}] Rate 1 0.12 0.05 0.12 2 0.12 0.10 0.24 3 0.24 0.05 0.48What is the order of the reaction with respect to [NO]?
Discussion -
What is the order with respect to [O_{2}]?