second order reaction exmples of second order reaction differential rate equation units of second order reaction

1. Second order reaction
Sara Tariq 11/20/17 BS 3

2. 1
Contents
Second order reaction.........................................................................................................................2
Reaction Rate.................................................................................................................................2
Examples of second order reaction..................................................................................................2
Types of second order reaction....................................................................................................2
Derivative and Integral Forms ......................................................................................................2
Case 1: A + A → P (Second Order with same reacting molecules)....................................................3
Case 2: A + B → P (Second Order Reaction with different reacting molecules).................................4

3. 2
Second order reaction
It isthe sumof the exponentsof concentrationtermof the reactantsinvolvedinthe rate expression.
Reaction Rate
Integrationof the second-orderrate law
−
𝒅[𝑨]
𝒅𝒕
= k[A]2
Examples of second order reaction
Following are the examples of second order reaction;
1. 2 HI → I2 + H2
Hydrogen Iodide decomposing into iodine gas and hydrogen gas.
2. O3 → O2 + O2
During combustion, oxygen atoms and ozone can form oxygen molecules.
3. O2 + C → O + CO
Another combustion reaction, oxygen molecules react with carbon to form
oxygen atoms and carbon monoxide.
4. O2 + CO → O + CO2
This reaction often follows the previous reaction. Oxygen molecules react with
carbon monoxide to form carbon dioxide and oxygen atoms.
5. 2 NOBr → 2 NO + Br2
In the gas phase, nitrosyl bromide decomposes into nitrogen oxide and bromine
gas.
6. NH4CNO → H2NCONH2
Ammonium cyanate in water isomerizes into urea.
7. CH3COOC2H5 + NaOH → CH3COONa + C2H5OH
An example of the hydrolysis of an ester in the presence of a base. In this case,
ethyl acetate in the presence of sodium hydroxide.
8. H+ + OH- → H2O
Hydrogen ions and hydroxy ions form water.
9. 2 NO2 → 2 NO + O2
Nitrogen dioxide decomposing into nitrogen monoxide and oxygen molecule.
Types of second order reaction
1. Whenthe concentrationof bothreacting,moleculesare same
2. Whenthe concentrationof bothreacting,moleculesare differentfromeachother
Derivative and Integral Forms
To describe how the rate of a second-order reaction changes with concentration of reactants or
products, the differential (derivative) rate equation is used as well as the integrated rate

4. 3
equation. The differential rate law shows that how the rate of the reaction changes in time,
while the integrated rate equation shows that how the concentration of species changes over
time. Both equations can be derived from the above expression for the reaction rate. Plotting
these equations can also help in determining either a certain reaction is second-order or not.
Case 1: A + A → P (Second Order with same reacting molecules)
Two of the same reactant (A) combine in a single elementary step.
A+A⟶P
2A⟶P
The reaction rate for this step can be written as
Rate=−
1
2
𝑑[𝐴]
𝑑𝑡
= +
𝑑[𝑃]
𝑑𝑡
and the rate of loss of reactant A
𝑑𝐴
𝑑𝑡
=−k[A][A] = −k[A]2
The rate at which A decreases can be expressed using the differential rate equation.
−
𝑑[𝐴]
𝑑𝑡
= k[A]2
The equation can then be rearranged
−
𝑑[ 𝐴]
[ 𝐴]2 = −kdt
Since we are interested in the change in concentration of an over a period, we integrate
between t=0 and t, the time of interest.
∫
𝑑[ 𝐴]
[ 𝐴]2
[𝐴] 𝑡
[𝐴]0
= −k∫ 𝑑𝑡
𝑡
0
To solve this, we use the following rule of integration (power rule)
∫
𝑑𝑥
𝑥2 = −
1
𝑥
+ constant
We obtain the following integrated rate equation
1
[𝐴] 𝑡
−
1
[𝐴]0
= kt

5. 4
rearranging the integrated rate equation, we obtain an equation of the line:
1
[𝐴] 𝑡
= kt +
1
[𝐴]0
The above equation directly relates to the graph which provides a linearrelationship.Inthis case,
and for all second order reactions, the linear plot of
1
[𝐴] 𝑡
versus time will yield the below graph.
This graph is useful in many ways. If we only know the concentrations at specific times for a
reaction, we can draw a graph like the one above. If the graph yields a straight line, then the
reaction must be second order. In addition, with this graph we can find the slope of the line and
this slope is k, the reaction constant. The concentration of reactants approaches zero more
slowly in a second-order, compared to that in a first order reaction.
Case 2: A + B → P (Second Order Reaction with different reacting molecules)
the rate at which A decreases can be expressed using the differential rate equation

6. 5
𝑑[𝐴]
𝑑𝑡
= −k[A][B]
in this case the initial concentration of the two reactants are not equal. Let x be
the concentration of each species reacted at time t.
Let [𝐴]0= a and [𝐵]0=b Then, [A]=a−x; [B]=b−x
The expression of rate law becomes
−
𝑑𝑥
𝑑𝑡
= −k([𝐴]0−x) ([𝐵]0−x)
which can be rearranged As
𝑑𝑥
[𝐴]0−x)( [𝐵]0−x)
= kdt
We integrate between t=0 (when x=0) and t, the time of interest.
∫
𝑑𝑥
[𝐴]0−𝑥)( [𝐵]0−𝑥)
𝑥
0
= 𝑘 ∫ 𝑑𝑡
𝑡
0
To solve this integral, we use the method of partial fractions
∫
1
(𝑎−𝑥)(𝑏−𝑥)
𝑥
0
dx =
1
𝑏−𝑎
(𝑙𝑛
1
𝑎−𝑥
− 𝑙𝑛
1
𝑏−𝑥
)
Evaluating the integral gives us:
∫
𝑑𝑥
[𝐴]0−𝑥)( [𝐵]0−𝑥)
𝑥
0
=
1
[ 𝐵]0−[ 𝐴]0
(𝑙𝑛
[𝐴]0
[𝐴]0−𝑥
− 𝑙𝑛
[𝐵]0
[𝐵]0−𝑥
)
Applying the rule of logarithm, the equation simplifies to:
∫
𝑑𝑥
[𝐴]0−𝑥)( [𝐵]0−𝑥)
𝑥
0
=
1
[ 𝐵]0−[ 𝐴]0
𝑙𝑛
[𝐵][𝐴]0
[𝐴][𝐵]0
We then obtain the integratedrateequation (underthe condition that [A] and [B]are not equal).
1
[ 𝐵]0 − [ 𝐴]0
𝑙𝑛
[𝐵][𝐴]0
[𝐴][𝐵]0
= 𝑘𝑡
Upon rearrangement of the integrated rate equation, we get
𝑙𝑛
[𝐵][𝐴]0
[𝐴][𝐵]0
= 𝑘([ 𝐵]0 − [ 𝐴]0 ) 𝑡

7. 6
Hence, from the last equation, we can see that a linear plot of 𝑙𝑛
[𝐴]0[𝐵]
[𝐴][𝐵]0
versus time is
characteristic of second-order reactions.
This graph can be used to find slope.
in form y = ax + b with a slope of a=k([B]0−[A]0) and a y-intercept of b = 𝑙𝑛
[𝐴]0
[𝐵]0
Unitsfor second orderreaction
Rate = k[A][B]
rate (𝑚𝑜𝑙 𝑑𝑚−3
𝑠−1
) [A][B] (𝑚𝑜𝑙 𝑑𝑚−3
)
𝑚𝑜𝑙 𝑑𝑚−3
𝑠−1
= k x 𝑚𝑜𝑙 𝑚−3
x 𝑚𝑜𝑙 𝑑𝑚−3
Rate = 𝒎𝒐𝒍−𝟏
𝒅𝒎 𝟑
𝒔−𝟏