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When studying a chemical reaction, it is important to consider not only the chemical properties of the reactants, but also the conditions under which the reaction occurs, the mechanism by which it takes place, the rate at which it occurs, and the equilibrium toward which it proceeds. According to the law of mass action, the rate of a chemical reaction at a constant temperature depends only on the concentrations of the substances that influence the rate. The substances that influence the rate of reaction are usually one or more of the reactants, but can occasionally be a product. Another influence on the rate of reaction can be a catalyst that does not appear in the balanced overall chemical equation. The rate law can only be experimentally determined and can be used to predict the relationship between the rate of a reaction and the concentrations of reactants.
The relation between the rate of a reaction and the concentrations of reactants is expressed by its rate law. For example, the rate of the gasphase decomposition of dinitrogen pentoxide
2N_{2}O_{5} → 4NO_{2} + O_{2}
has been found to be directly proportional to the concentration of N_{2}O_{5}:
rate = k [N_{2}O_{5}]
Be very careful about confusing equilbrium constant expressions with those for rate laws. The expression for K_{eq} can always be written by inspecting the reaction equation, and it contains a term for each component (raised to the appropriate power) whose concentration changes during the reaction. For this reaction it is given by
In contrast, the expression for the rate law generally bears no necessary relation to the reaction equation, and must be determined experimentally.
More generally, for a reaction of the form
n_{A} A + n_{B} B + ... → products
the rate law will be
in which the exponents a and b are usually (but not always) integers and, we must emphasize once again, bear no relation to the coefficients n_{A} ,n_{B}.
Since the rate of a reaction has the dimensions of (concentration/time), the dimensions of the rate constant k will depend on the exponents of the concentration terms in the rate law. To make this work out properly, if we let p be the sum of the exponents of the concentration terms in the rate law
p = a + b + ...
then k will have the dimensions (concentration^{1–p}/time).
How fast a reaction occurs depends on the reaction mechanism, the stepbystep molecular pathway leading from the reaction to products. Chemical kinetics is concerned with how rates of chemical reactions are measured, how they can be predicted, and how reaction rate data is used to deduce probable reactions. The reaction rate or speed refers to something that happens in a unit of time. Consider that you are driving, and you want to know the distance from point A to point B, the distance from the points is the product of time x rate. Just think of it as the distance (concentration) is equal to the product of speed in M/sec and time(sec, min, and hour).
The rate itself is defined as the change in concentration of a reactant or product per unit of time. If A is a reactant and C a product, the rate might be expressed as:
In other words, the reaction rate is the change in concentration of reactant A or product C, over the change in time. This is an average rate of change and the minus sign is used to express the rate in terms of a reactant concentration. The reason for this is that by the conservation of mass, the rate of generation of product C must be equal to the rate of consumption of reactant A.
One may also wish to consider the instantaneous rate by taking the limit of the average rate as delta t approaches 0. This will give the instantaneous rate as:
Now the reaction rate is expressed as a derivative of the concentration of reactant A or product C, with respect to time, t.
Consider a reaction 2A + B > C, in which one mole of C is produced from every 2 moles of A and one mole of B. The rate of this reaction may be described in terms of either the disappearance of reactants over time, or the appearance of products over time:
rate = (decrease in concentration of reactions)/(time) = (increase in concentration of products)/time
Because the concentration of a reactant decreases during the reaction, a minus sign is placed before a rate that is expressed in terms of reactants. For the reaction above, the rate of reaction with respect to A is Δ[A]/Δt, with respect to B is Δ[B]/Δt, and with respect to C is Δ[C]/Δt. In this particular reaction, the three rates are not equal. According to the stoichiometry of the reaction, A is used up twice as fast as B, and A is consumed twice as fast as C is produced. To show a standard rate of reaction in which the rates with respect to all substances are equal, the rate for each substance should be divided by its stoichiometric coefficient.
Rate = (1/2)(Δ[A]/Δt) = Δ[B]/Δt = Δ[C]/Δt
Rate (as well as the Rate Law) is expressed in the units of molarity per second.
Measuring instantaneous rates is the most direct way of determining the rate law of a reaction, but is not always convenient, and it may not even be possible to do so with any precision.
The ordinary rate law (more precisely known as the instantaneous or differential rate law) tells us how the rate of a reaction depends on the concentrations of the reactants. But for many practical purposes, it is more important to know how the concentrations of reactants (and of products) change with time.
For example, if you are carrying out a reaction on an industrial scale, you would want to know how long it will take for, say, 95% of the reactants to be converted into products. This is the purpose of an integrated rate law.
For nearly all forward, irreversible reactions, the rate is proportional to the product of the concentrations of the reactants, each raised to some power. For the general reaction:
aA + bB → cC + dD
The rate is proportional to [A]^{m}[B]^{n }that is:
rate = k[A]^{m}[B]^{n}
This expression is the rate law for the general reaction above, where k is the rate constant. Multiplying the units of k by the concentration factors raised to the appropriate powers give the rate in units of concentration/time.
The dependence of the rate of reaction on the concentrations can often be expressed as a direct proportionality in which the concentrations may appear to be the zero, first, or second power. The power to which the concentration of a substance appears in the rate law is the order of the reaction with respect to that substance. In the reaction above the order of reaction is:
m + n
The order of the chemical equation can only be determined by experiment. In other words, one cannot determine what m and n are by just looking at a balanced chemical equation; m and n must be determined by the use of data. The overall order of a reaction is the sum of the orders with respect to the sum of the exponents. Furthermore, the order of a reaction is stated with respect to a named substance in the reaction. The exponents in the rate law are not equal to the stoichiometric coefficients unless the reaction actually occurs via a single step mechanism. However, the exponents are equal to the stoichiometric coefficients of the ratedetermining step. In general, the rate law can calculate the rate of reaction from known concentrations for reactants and derive an equation that expresses a reactant as a function of time.
The proportionality factor, k, called the rate constant is a constant at a fixed temperature. Nonetheless, the rate constant varies with temperature. There are dimensions to k and can be determined with simple dimensional analysis of the particular rate law. The units should be expressed when the kvalues are tabulated. The higher the k value, the faster the reaction proceeds.
The values of k, x, and y in the rate law equation (r =[A]^{m}[B]^{n}) must be determined experimentally for a given reaction at a given temperature. The rate is usually measured as a function of the initial concentrations of the reactants, A and B.
Example 
Example: Given the data below, find the rate law for the following reaction at 300K.
A + B → C + D
Trial  [A]_{initial} (M)  [B]_{initia}l (M)  r_{initial} (M/sec) 
1  1  1  2 
2  1  2  8.1 
3  2  2  15.9 
Solution: First, look for two trials in which the concentrations of all but one of the substances are held constant.
a. In trials 1 and 2, the concentration of A is kept constant while the concentration of B is doubled. The rate increases by a factor of approximately 4. Write down the rate expression of the two trials.
Trial 1: r_{1} = k[A]^{x}[B]^{y} = k(1.00)^{x}(1.00)^{y}
Trial 2: r_{2}_{ }= k[A]^{x}[B]^{y} = k(1.00)^{x}(2.00)^{y}
Divide the second equation by the first which yields:
4 = (2.00)^{y}
y = 2
b. In trials 2 and 3, the concentration of B is kept constant while the concentration of A is doubled; the rate is increased by a factor of approximately 2. The rate expressions of the two trails are:
Trial 2: r_{2} = k[A]^{x}[B]^{y} = k(1.00)^{x}(2.00)^{y}
Trial 3: r_{3} = k[A]^{x}[B]^{y} = k(2.00)^{x}(1.00)^{y}
Divide the second equation by the third which yields:
2 = (2.00)^{x}
x = 1
So r = k[A][B]^{2}
The order of the reaction with respect to A is 1 and with respect to B is 2; the overall reaction order is:
1 + 2 = 3
To calculate k, substitute the values from any one of the above trials into the rate law:
2.0 M/sec = k(1.00 M)(1.00M)^{2}
k = 2.0 M^{2} sec^{1 }
Therefore the rate law is r =2.0[A][B]^{2}
Chemical reactions are often classified on the basis of kinetics as zeroorder, firstorder, secondorder, mixed order, or higherorder reactions. The general reaction aA + bB → cC + dD will be used in the discussion next.
First lets note what each of these orders means in terms of initial rate of reaction effect:
A zeroorder reaction has a constant rate, which is independent of the reactant's concentrations. Thus the rate law is:
rate = k = constant
where k has the units of M(sec^{1}). In other words, a zeroorder reaction has a rate law in which the sum of the exponents is equal to zero. An increase in temperature or a decrease in in temperature is the only factor that can change the rate of a zeroorder reaction. In addition, a reaction is zero order if concentration data are plotted versus time and the result is a straight line. The slope of this resulting line is the negative of the zero order rate constant k.
At times, chemists and researchers are also concerned with the relationship between the concentration of a reactant and time. Such expression is called the integrated rate law in which the equation expresses the concentration of a reactant as a function of time (remember, each order of reaction has its own unique integrated rate law). The integrated rate law of a zeroorder reaction is:
[A_{t}] = kt + [A_{0}] (See page on zeroorder reactions to see how this is derived)
Notice, however, that this model cannot be entirely accurate since this equation predicts negative concentrations at sufficiently large times. In other words, if one were to graph the concentration of A as a function of time, at some point, the line will cross below 0. This is of course, physically impossible since concentrations cannot be negative. Nevertheless, this model is a sufficient model for ranges of time where concentration is predicted as greater than zero.
The half life (t_{1}_{/2}) of a reaction is the time needed for the concentration of the radioactive substance to decrease to onehalf of its original value. The halflife of a zeroorder reaction can be derived as follows:
Given a reaction involving reactant A and from the definition of a halflife, we know that t_{1}_{/2 }is the time it takes for half of the initial concentration of reactant A to react. So we can now substitute new conditions into the integrated rate law form to obtain:
We now solve for t_{1}_{/2 }to obtain the following:
A firstorder reaction has a rate proportional to the concentration of one reactant.
rate = k[A] or rate = k[B]
Firstorder rate constants have units of sec^{1}. In other words, a firstorder reaction has a rate law in which the sum of the exponents is equal to 1.
The integrated rate law of a firstorder reactions is:
ln[A]_{t} = kt + ln[A]_{0}
or
ln([A]_{t}/[A]_{0}) = kt
or
Moreover, a firstorder reaction can be determined by plotting a graph of ln[A] vs. time t and a straight line is produced with a negative slope of k.
The classic example of a firstorder reaction is the process of radioactive decay. The concentration of radioactive substance A at any time t an be expressed mathematically as:
[A_{t}] = [A_{0}]e^{kt}
where [A_{0}] = initial concentration of A
[A_{t}] = concentration of A at time t
k = rate constant
t = elapsed time
The halflife of a first order reaction can be calculated in a similar fashion as with the halflife of the zero order reaction and one would obtain the following:
where k is the first order rate constant. Notice that the halflife associated with the firstorder reaction is the only case where halflife is independent of concentration of a reactant or product. In other words, [A] does not appear in the halflife formula above.
A secondorder reaction has a rate proportional to the product of the concentration of two reactants, or to the square of the concentration of a single reactant. For example:
rate = k[A]^{2}
rate = k[B]^{2}
rate = k[A][B]
are all secondorder reactions. Therefore, a secondorder reaction has rate law in which the sum of the exponents are equal to 2.
The integrated rate law of a secondorder reaction is as follows:
(See page on secondorder reactions to see how this is derived)
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The halflife of a secondorder reaction is:
In the laboratory, one may collect a sample of data consisting of measured concentrations of a certain reactant A at different times. This sample data may look like the following (Sample data obtain from ChemElements PostLaboratory Exercises):
Time (min)  Concentration (M) 
0  0.906 
0.9184  0.8739 
9.0875  0.5622 
11.2485  0.5156 
17.5255  0.3718 
23.9993  0.2702 
27.7949  0.2238 
31.9783  0.1761 
35.2118  0.1495 
42.973  0.1029 
46.6555  0.086 
50.3922  0.0697 
55.4747  0.0546 
61.827  0.0393 
65.6603  0.0324 
70.0939  0.026 
One can then plot [A] versus time, ln[A] versus time, and 1/[A] versus time to see which plot yields a straight line. The reaction order will then be the order associated with the plot that gives a straight line. While it may seem that doing this seems tedious and difficult, the process becomes quite simple with the use of Excel, or any other similar program.
By utilizing the formula capabilities of Excel, we can obtain two more data tables of ln[A] vs. time and 1/[A] vs. time very easily.
Time (min)
 ln([A])

0


0.9184
 0.085122253

9.0875
 2.206899843

11.2485
 2.420234786

17.5255
 2.863656963

23.9993
 3.178024663

27.7949
 3.324852551

31.9783
 3.465057548

35.2118
 3.561381254

42.973
 3.760572012

46.6555
 3.84279082

50.3922
 3.919836401

55.4747
 4.015927061

61.827
 4.124340162

65.6603
 4.184494481

70.0939
 4.249835772

And
Time (min)
 1/[A]

0


0.9184
 1.088850174

9.0875
 0.110041265

11.2485
 0.088900742

17.5255
 0.057059713

23.9993
 0.041667882

27.7949
 0.035977823

31.9783
 0.031271206

35.2118
 0.028399571

42.973
 0.023270426

46.6555
 0.0214337

50.3922
 0.019844341

55.4747
 0.018026235

61.827
 0.016174163

65.6603
 0.015229903

70.0939
 0.014266577

We now plot the three data sets to get
We can see clearly that the graph of ln[A] vs time is a straight line. Therefore the reaction associated with the given data is a first order reaction.
Additional Problems:
1. In a thirdorder reaction involving two reactants and two products, doubling the concentration of the first reaction causes the rate to increase by a factor of 2. If the concentration of the second reactant is cut in half, the rate of this reaction will be?
Solution: The rate is directly proportional to the concentration of the first reactant. When the concentration of the reactant doubles, the rate also doubles. Because the reaction is thirdorder, the sum of the exponents in the rate law must be equal to 3. Therefore, the rate law is defined as follows: rate  k[A][B]^{2. }Reactant A has no exponent because its concentration is directly proportional to the rate. For this reason, the concentration of reactant B must be squared in order to write a law that represents a thirdorder reaction. when the concentration of reactant B is multiplied by 1/2, the rate will be multiplied by 1/4. Therefore, the rate of reaction will decrease by a factor of 4.
2. A certain chemical reaction follows the rate law, rate = k[NO][Cl_{2}]. Which of the following statements describe the kinetics of this reaction:
3. The data in the following table is collected for the combustion of the theoretical compound XH_{4}:
XH_{4} + 2O_{2} → XO_{2} + 2H_{2}O
What is the rate law for the reaction described?
Trial  XH_{4}_{ (initial)}  O_{2 (initial)}  Rate 
1  .6  .6  12.4 
2  .6  2.4  49.9 
3  1.2  2.4  198.3 
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