# Chemical kinetics and half-life

## Chemical kinetics reaction

Chemical kinetics deals with chemical engineering for calculating the rate of a chemical reaction. In the experimental study, chemical kinetic provides the information for the formulation of mechanism or molecular pathways of a reaction.

But very fast or very slow reactions can not discuss under chemical kinetics only a moderate rate of reaction discus. All the redox reactions are fast and the formation of HCl in dark conditions is a slow reaction.

$H_{2}+Cl_{2}\xrightarrow{in\,&space;dark\,&space;condidion}2HCl$

Hydrolysis of ester, decomposition of N2O5, the reaction between PO4 and I, etc are the examples of moderate rate reaction.

### Importance of chemical kinetics

There is two main importance for studying the rate of chemical kinetics in chemical engineering.

1. The first is the practical importance for able to predict how quickly a reaction mixture moves to equilibrium. Thus it depends only on the external factor which used to proceeding the reaction.
2. The second is the theoretical importance of the formulation of the reaction mechanism. Thus we can analyze the chemical reaction to many elementary steps.

#### Rate of reaction in chemical kinetics

Kinetics rate of reaction defined as the rate of change of concentration of reactant and products at any instant of time.

Let a simple kinetic reaction
A → B

$\therefore&space;Rate\,&space;of\,&space;reaction\,&space;terms\,&space;of\,&space;reactant&space;=-\frac{dC_{A}}{dt}$

$Rate\,&space;of\,&space;reaction\,&space;terms\,&space;of\,&space;product&space;=\frac{dC_{B}}{dt}$

Negative signs used to show the decrease of concentration of reactants with time. But the rate is always a positive quantity and it decreases with time.

A complication arises if we consider the reaction

A + 2B → 3C + D

The above reaction rate of decrease of B is twice the rate of decrease of A. To resolve this complication we mention the rate of reaction in terms of a particular component. Thus the best way to write the rate of kinetic reaction

$Rate=\pm&space;\frac{1}{\gamma&space;_{i}}\left&space;(&space;\frac{dC_{i}}{dt}&space;\right&space;)$

where γi = stoichiometric coefficient of the balanced equation and positive sign used for product and negative sign for reactants. Thus the rate equation for the above reaction

$Rate=-\frac{dC_{A}}{dt}=-\frac{1}{2}\left&space;(&space;\frac{dC_{B}}{dt}&space;\right&space;)=\frac{1}{3}\frac{dC_{C}}{dt}=\frac{dC_{D}}{dt}$

Question

In a chemical reaction, N2 + 3H2 → 2NH3  the rate of (dNH3/dt) = 2×10-4. Calculate the value of (-dH2/dt).

The rate equation for the above reaction

$Rate&space;=-\frac{d_{N_{2}}}{dt}=-\frac{1}{3}\left&space;(&space;\frac{d_{H_{2}}}{dt}&space;\right&space;)=\frac{1}{2}\left&space;(&space;\frac{d_{NH_{3}}}{dt}&space;\right&space;)$

$\therefore&space;-\frac{1}{3}\left&space;(&space;\frac{d_{H_{2}}}{dt}&space;\right&space;)=\frac{1}{2}\left&space;(&space;\frac{d_{NH_{3}}}{dt}&space;\right&space;)$

$or,&space;-\left&space;(&space;\frac{d_{H_{2}}}{dt}&space;\right&space;)=\frac{3}{2}\left&space;(&space;\frac{d_{NH_{3}}}{dt}&space;\right&space;)$

$\therefore&space;\frac{d_{H_{2}}}{dt}=\frac{3}{2}\times&space;2\times&space;10^{-4}&space;\,&space;mol\,&space;lit^{-1}\,&space;sec^{-1}$

= 3 × 10-4 mol lit-1 sec-1

#### Units of reaction rate

In kinetics rate of reaction defined as the rate of change of concentration of reactant and products at any instant of time.

$\therefore&space;unit\,&space;of\,&space;reaction\,&space;rate=\frac{unit\,&space;of\,&space;conc.}{unit\,&space;of\,&space;time}$

Thus the normal and SI units of the rate of the reaction
mol lit-1sec-1
and mol m-3 sec-1

#### Factor on which reaction rate depends

There are several factors that influence the rate of a chemical reaction. Out of this some important factor are

1. Active mass or concentration of the reactants and products derived in mass action law.
2. The temperature of the equation derived in the Arrhenius equation.
3. Presence of catalyst in a chemical reaction.
4. Degree of fineness of reactants.
5. Absorption of radiation of suitable frequency derived in the photochemical reaction.

#### Rate law in chemical kinetics

Dependence of reaction rate with the concentration of reactants we express rate equation or rate law. Mass action law guides the basis of the formulation of rate equations. But the rate equation can only be formulated on the basis of experimental data.

Thus decomposition of H2O2 represented as

2H2O2 → 2H2O + O2

From the mass action law kinetics rate equation

$-\frac{d_{H_{2}O_{2}}}{dt}&space;=k\times&space;C_{H_{2}O_{2}}^{2}$

But the experiment shows the rate equation is

$-\frac{d_{H_{2}O_{2}}}{dt}&space;=k\times&space;C_{H_{2}O_{2}}$

The single power of concentration controlled the decomposition of H2O2. Thus this reaction is 1st order because of only one concentration term used in the rate equation. K = const, called rate constant of a specific reaction.

Thus for a general reaction,

aA + bB → cC + dD

Rate of reaction = k × CAm × CBn

Where values of m and n depend on the experimental data. It does not depend on the values of a and b.

### Unit of the rate constant

If a kinetic reaction expressed as,
A → Product

The rate equation for this reaction

$-\frac{dC_{A}}{dt}&space;=&space;k\,&space;C_{A}^{n}$

where n= order of the reaction

$\therefore&space;unit\,&space;of\,&space;k=\frac{unit\,&space;of\,&space;conc.}{\left&space;(&space;unit\,&space;of\,&space;conc.&space;\right&space;)^{n}\times&space;unit\,&space;of\,&space;time}$

$or,&space;unit\,&space;of\,&space;k=\frac{\left&space;(unit\,&space;of\,&space;conc.&space;\right&space;)^{1-n}}{unit\,&space;of\,&space;time}$

 Order of reaction Unit of the rate constant Zero-order mol lit-1 sec-1 first-order sec-1 second-order lit mol-1 sec-1

Thus the unit of k for the first-order reaction = sec-1

#### Half-life in chemical kinetics

When the chemical reaction proceeds the concentration of reactant decreases and productivity increases. But after a certain period of time the value of the reactant one haft of initial concentration. This period of time is called half-life in chemical kinetics.

The integrated rate law for the zero-order kinetics

x = k0 × t
but when t=t½ , x = x/2
∴ t½ = x/2k
where x = initial conc. of reactant.

But the integrated rate law for the first-order reaction

$t&space;=\frac{1}{k_{1}}\left&space;(&space;ln\frac{a}{a-x}&space;\right&space;)$

Thus when t = t½, x=a/2
∴ t½ = ln2/k1

#### Purpose of rate law in Kinetics

The statement of rate law serves mainly three purpose

1. It permits to predicts the rate at a given concentration from the knowledge of rate constant.
2. Rate laws help to build up the possible mechanism of a chemical reaction.
3. From the rate law, we can clarify the kinetics of the chemical reaction as first, second or third-order kinetics.