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# Chemical Kinetics

Chemical kinetics is the branch of physical chemistry which deals with the study of the rate of chemical reactions. Such studies enable us to understand the mechanism by which the reaction occurs.

In chemical equilibrium, only the initial and final states of chemical reactions are considered. Furthermore, the energy relations between the reactants and the products are governed by the laws of thermodynamics, but no attempt is made to indicate the intermediate stages of the reaction.

There are many reactions for which kinetic studies are not possible. Many ionic and explosive reactions occur instantaneously and in such cases, it is not easy to measure their rates. Special techniques have been devised to study their kinetics.

On the other hand, there are reactions which are so slow that it is difficult to observe appreciable changes in concentrations at room temperature even after months or years. Between these two extremes are reactions for which it is easy to determine the rates. The gaseous reactions like decomposition of HI or N2O5, the reactions in liquid phase like the hydrolysis of ester etc. come under this category.

Experimentally, it is found that the rate of a chemical reaction is dependent on factors like temperature, pressure and the concentration of the reacting species. The presence of a catalyst can increase or decrease the rate of the reaction by many powers of ten.

From the kinetic stand point, the reactions are classified into two groups:
• Homogeneous reactions which occur in one phase only. It may be a gaseous phase or a liquid phase.
• Heterogeneous reactions which take place in two or more phases, e.g., gaseous reactions taking place on the surface of a solid catalyst or on the walls of the container.
Rate, speed or velocity of a reaction have the same meaning in chemical kinetics. The rate of the reaction at any instant during the course of the reaction is defined as the rate at which the concentrations of the reacting species change with time, and it is represented by dC/dt where C is the concentration of the reacting substance at any time t.

The average rate of the reaction during a finite time interval, $\Delta$t is given by $\Delta$C/$\Delta$t. In the case of the reactants, concentrations decrease with time and, therefore, a negative sign is put before dC/dt, i.e., -dC/dt. On the other hand, the concentrations of the products increase with time and consequently the rate is written as dC/dt.

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## Chemical Kinetics Definition

Chemical reactions are spontaneous reactions. A spontaneous reaction may be slow or fast, depending upon the reactant involved, conditions available, etc. The branch of physical chemistry which deals with the rate of chemical reactions is called as chemical kinetics.
The study of chemical kinetics includes,
1. The rate of a reaction and rate laws
2. The factors that affect the rate of a reaction like temperature, pressure, concentration, catalyst, etc.
3. The mechanism of a particular reaction.
The rate of a particular reaction is very helpful to understand a chemical reaction. It helps in deciding favorable conditions for a reaction to be carried out. The rate of a reaction also helps us in determining a suitable catalyst for a particular reaction.

## Reaction Rate

The rate of a reaction tells us what speed a reaction is proceeding with. Let us consider the reaction,
A B

The concentration of the reactant decreases as time passes. This can be written as,
$\frac{Change\ in\ concentration\ of\ A}{Time\ taken\ for\ the\ change} = \frac{\left \lfloor A \right \rfloor _{2} - \left \lfloor A \right \rfloor _{1}}{t_{2}-t_{1}}$

So, if ΔA represents the change in concentration and ΔT represents change in time, then the rate of the reaction is given by,
Rate of the reaction =$\frac{\Delta \left \lfloor A \right \rfloor}{\Delta t}$

Since the concentration of a reactant decreases, rate of a reaction calculated with respect to reactants is negative. So,
Rate of the reaction =$-\frac{\Delta \left \lfloor A \right \rfloor}{\Delta t}$

### Unit of Rate:

Since rate indicates concentration vs. time, rate is given by moles/liter/s or mol.lit-1.sec-1. So, with respect to various reactions, it can be summarized that the rate of a reaction is directly proportional to the reactant concentration, each concentration being raised to some power.

This can be written as,
Rate α [A]n
Rate = k [A]n

where ‘k’ is called the rate constant.

### Rate Law:

An expression which shows how the reaction rate is related to concentration is called the rate law or rate equation.

Example:
For the equation: N2O5 → 4NO2 + O2
Rate law can be written as
Rate = k[N2O5]
Rate law for the chemical equation
H2 + I2 → 2HI is
Rate = k [H]2[I]2

## Order of a Reaction

The order of a reaction is defined as the sum of the powers of the concentrations in the rate law.

For Example,
If the rate law of an equation says: Rate = [A]2 [B]
the order of this reaction would be,

Order = 2+ 1 = 3
Order of a chemical reaction can be,
1. First order - Order is 1 - Rate = k[N2O5]
2. Second order - Order is 2 - Rate = k[H2][I2]
3. Third order - Rate = k[H2] [NO]2
4. Zero order - Rate = k.
Order of certain reactions is sometimes fractional too. These reactions are called fractional order reactions.

### Zero Order Reaction:

If the order of a reaction is independent of the concentrations of all the reactants present, that reaction is said to be a zero order reaction.
The rate law for the reaction,

NO2 + CO NO + CO2
Is Rate = k [NO2]2. The reactant, CO does not appear in the rate law, because rate is independent of concentration of carbon monoxide. This reaction, therefore, is of zeroth order with respect to CO.

## Specific Rate Law

Depending upon the order of a chemical reaction, there is a specific rate law expression for each order.

### First order reactionA $\to$ products:

K = $\frac{1}{t}log\tfrac{V_{\infty }}{V_{\infty }-V_{t}}\ time^{-1}$

where,

$V_{\infty }$ = Final volume.
$V_{t}$
= Volume at a time ‘t’.

Second Order 2A $\to$ Products :

K = $\frac{1}{t} \frac{x}{a(a-x)}$$mol^{-1}L\ time^{-1} where, a = Initial concentration. (a-x) = Concentration at time ‘t’. Third Order 3A \to Products: K = \frac{1}{t} \frac{x(2a-x)}{2a^{2}(a-x)^{2}}$$mol^{-2}L^{-2}\ time^{-1}$

### Half-life of a reaction:

The time required for the concentration of a reactant to decrease to half of its initial value. It is represented by $t_{\frac{1}{2}}$.

For a first order reaction, half-life of a reaction is calculated using the formula,

$t_{\frac{1}{2}}$ = 0.693 / K

For a first order reaction, therefore, the half-life is independent of the initial concentration.

## Chemical Kinetics Practice Problems

The problems on chemical kinetics are given below :

### Solved Examples

Question 1: Nitrogen monoxide and oxygen were combined in a flask at 25 degree Celsius and allowed to react as follows:
2NO(g) + O2(g) → 2NO2(g)
Write the rate law for the reaction with following information given,
 Experiment Initial concentration of O2 Initial concentration of NO Initial rate (mol/L s) A 0.020M 0.050M 0.038 B 0.020M 0.100M 0.152 C 0.080M 0.100M 0.608

Solution:
Rate = k [O2]x [NO]y
To find ‘x’ and ‘y’, we need to study the experiments individually.
In experiments A and B, concentration of oxygen is same, but concentration of NO is doubled.
Looking at the rate of experiment A and B, we see that the rate of ‘B’ is four times rate of ‘A’.
So, on increase of concentration of B by two times, rate has increased four times. This makes ‘y’, order of NO as 2. (y = 2).
Rate = k [O2]x [NO]2
Now let us consider experiments B and C, where concentration of A is increased four times and the rate has increased 4 times. So, the order of Oxygen is one. (x = 1).

Rate = k [O2]1 [NO]2
Rate equation would be,

Rate = k [O2] [NO]2

Question 2: A sample of the unstable isotope 20 Na is generated in the laboratory. Initially there are 2.40 milli moles of the isotope detected. After 0.30 seconds, amount of isotope detected was 0.30 mmol. Calculate the half-life of this reaction. Half-life is given by the expression: Half-life = 0.693 / k.
Solution:
Since initial taken was 2.40 and it has decreased 2.40 / 0.30 = 8 times, and since it is independent of concentration, let us tabulate the time and amount reduced to find the half-life of the reaction.

 Time Half-life Amount (m mole) 0 0 2.40 (initial) 0.10sec 1 1.20 0.20sec 2 0.60 0.30sec 3 0.30

So, the time taken to reduce the initial concentration to one half is = 0/10 seconds.
Thus, half-life of the reaction is 0.10 seconds

Question 3: The half-life of a first order reaction is 15 minutes. Calculate the rate constant of the reaction.
Solution:
For a first order reaction, Half-life = 0.693 / k.
So, since t1/2= 15 minutes,
15 minutes = 0.693 / k
k = 0.693 / 15 min = 0.0462 min-1

 More topics in Chemical Kinetics Reaction Rates Equilibrium Constants The Kinetic Theory of Gases
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