A reaction is of zero order when the rate of reaction is independent of the concentration of materials. The rate of reaction is a constant. When the limiting reactant is completely consumed, the reaction stops abruptly.
The zero order rate law for the general reaction
is written as the equation
which on integration of both sides gives
When t = 0 the concentration of A is [A]0. The constant of integration must be [A]0.
Now the integrated form of zero-order kinetics can be written as follows
Plotting [A] versus t will give a straight line with slope -k.
A general unimolecular reaction
where A is a reactant and P is a product is called a first-order reaction.
The rate is proportional to the concentration of a single reactant raised to the first power.
The decrease in the concentration of A over time can be written as:
Equation (2) represents the differential form of the rate law. Integration of this equation and determination of the integration constant C produces the corresponding integrated law.
Integrating equation (2) yields:
The constant of integration C can be evaluated by using boundary conditions. When t = 0, [A] = [A]0. [A]0 is the initial concentration of A.
Substituting into equation (3) gives:
Therefore the value of the constant of integration is:
Substituting (5) into (4) leads to:
Plotting ln[A] or ln[A] / [A]0 against time creates a straight line with slope -k. The plot should be linear up to a conversion of about 90%.
Equation (6) can also be written as:
This means that the concentration of A decreases exponentially as a function of time.
The rate constant k can also be determined from the half-life t1/2. Half-life is the time it takes for the concentration to fall from [A]0 to [A]0 / 2 ==> [A] = 1/2 [A]0 .
According to equation (6) is obtained:
A and B react to produce P:
If the initial concentration of the reactant A is much larger than the concentration of B, the concentration of A will not change appreciably during the course of the reaction The concentration of the reactant in excess will remain almost constant. Thus the rate's dependence on B can be isolated and the rate law can be written
Equation (1) represents the differential form of the rate law. Integration of this equation and evaluation of the integration constant C produces the corresponding integrated law.
Substituting [B] = c into equation (1) yields:
Integrating equation (2) gives:
The constant of integration C can be evaluated by using boundary conditions. At t = 0 the concentration of B is c0.
Accordingly is obtained:
If the decrease in concentration of B is followed by photometric measurement the Beer' Law must be taken into account.
Combining equation (4) and Beer' Law
c = concentration of the compound in solution, expressed in mol · L -1
P0= radiant power for radiation entering; P= radiant power for radiation leaving
gives the relationship between k' and lnA:
According to equation (7), a plot of lnA versus time should lead to a straight line whose slope is the pseudo-first order rate constant k'. The value of k' can then be divided by the known, constant concentration of the excess compound to obtain the true constant second order k:
The pseudo-first order rate constant k' can be also determined from the half-life t1/2.
The rate of a second order reaction is proportional to either the concentration of a reactant squared, or the product of concentrations of two reactants.
For the general case of a reaction between A and B, such that
the rate of reaction will be given by
1. Initial concentrations of the two reactants are equal:
Equation (1) can be written as:
Separating the variables and integrating gives:
Provided that [A] = [A]0 at t = 0 the constant of integration C becomes equal to 1 / [A]0.
Thus the second order integrated rate equation is
A plot of 1 / [A] vs t produces a straight line with slope k and intercept 1 / [A]0 . The plot should be linear up to a conversion of about 50%.
2. Starting concentrations of the two reactants are different:
If [A]0 and [B]0 are different the variable x is used.
Equation (1) becomes
where [A]0 - x = [A], [B]0 - x = [B] and x is the decrease in the concentration of A and B.
Equation (5) can be integrated after separation of the variables and partial fraction expansion. The result is:
where C is the constant of integration.
Using the condition that x = 0, when t = 0, the value of C can be found
and equation (6) becomes
If [A]0 > [B]0, then a plot of
The Arrhenius equation is often written in the logarithmic form:
If the experimental method yields reactant concentrations rather than x, the equivalent form of equation (8) is
against t will have a positive slope, equal ([A]0 - [B]0) k.
|It is a well-known fact that raising the temperature increases the reaction rate. Quantitatively this relationship between the rate a reaction proceeds and its temperature is determined by the Arrhenius Equation: |
R = 8.314 J/mol·K
T = absolute temperature in Kelvins
A = pre-exponential or frequency factor
A = p · Z, where Z is the collision rate and p is a steric factor.
Z turns out to be only weakly dependant on temperature. Thus the frequency factor is a constant,
specific for each reaction.
Because equivalent amounts of A and B are reacting, [A] can be expressed in terms of [B].
If [B] = x , [A] = [A]0 - (x0 - x)
Provided that the initial concentration of A is twice the initial concentration of B (see Kinetic equations - Reaction Second Order - Download PDF file) equation (10) becomes
|The Arrhenius equation is based on the collision theory which supposes that particles must collide with both the correct orientation and with sufficient kinetic energy if the reactants are to be converted into products. |
The two animations are taken from Effects of temperature, concentration, catalysts, inhibitors on reaction rates
Bestimmung von Ea
"Two-Point Form" of the Arrhenius Equation
The activation energy can also be found algebraically by substituting two rate constants (k1, k2) and the two corresponding reaction temperatures (T1, T2) into the Arrhenius Equation (2).
Substracting equation (4) from equation (3) results in
Rerrangement of equation (5) and solving for E a yields
A plot of lnk versus 1/T produces a straight line with the familiar form y = -mx + b, where
x = 1/T
y = lnk
m = - Ea/ R
b = lnA
The activation energy Ea can be determined from the slope m of this line: Ea = -m · R
The value of the activation energy Ea is rounded to one decimal place. The value of lnA shall be expressed with an accuracy of two decimal places.
An accurate determination of the activation energy requires at least three runs completed at different reaction temperatures. The temperature intervals should be at least 5°C.