# Zero order kinetics reactions

### Study kinetics of the zero-order reaction

The zero-order kinetics reactions deal with the speed of a chemical reaction in terms of reactants and products. Thus the main reasons for studying the rate of a chemical reaction

1. The first is the practical importance to predict how quickly a reaction mixture moves to the equilibrium state.
2. The second reason is the theoretical importance of formulating the mechanism of a chemical reaction.

In chemical kinetics, the rate of a chemical reaction generally expressed as the decrease of the concentration of the reactant or increase of concentration of the product per unit time.

A → Product

$-\left&space;(&space;\frac{dC_{A}}{dt}&space;\right&space;)=kC_{A}^{n}$

Some surface reactions the rate has been found to be independent of concentration. These reactions are known as the zero-order kinetics reaction.

#### Definition of the zero-order reaction

The differential rate shows the dependence of the rate with the concentration of the reacting species. But the integrated form of rate laws provides the concentration of these species at any time from the start of the reaction.

Rte of the zero-order reactions does not depend on the concentration of the reactants. Let us take a chemical reaction represented as

 A → Product Initial concentration a 0 After time t a – x x

Thus after time t concentration of reactant decreases in a zero-order reaction = x.

Question
If the rate of the reaction is equal to the rate constant. What is the order of the reaction?

Zero-order reaction.

#### Zero-order reaction derivation

The rate equation in terms of product,

$\frac{dx}{dt}=k_{0}$

where k₀ = rate constant of the zero-order reaction.
or, dx = k0dt

Integrating the above equation within limits,

$\int_{0}^{x}dx=k_{0}\int_{0}^{t}dt$

∴ x = k0t + c
where c = integration constant of the reaction.

But when t = o, that is the initial of the reaction, x is also zero. So the integration constant c is also zero.

Thus from the above equation

x = k0t

This is the relationship between decreases of concentration of the reactant within time.

#### Rate of reaction in terms of reactant

Rate equation in terms of reactant,

$-\frac{dC_{A}}{dt}&space;=&space;k_{0}\times&space;C_{A}^{0}=k_{0}$

where CA = concentration of the reactant at the time t.

or, – dCA = k0dt

Integrating the above equation

$-\int&space;\left&space;d\left&space;(&space;C_{A}&space;\right&space;)=k_{0}\int&space;dt$

∴ – CA = k0t + c
where c = integration constant of the reaction.

If the initial concentration of the reactant = C0 then from the above equation,

–  C0 = 0 + c
∴ c = – C0

Thus the rate equation for the zero-order reaction

–  CA = k0t – C0
or, C0 – CA = k0t

This is another form of the rate equation of the zero-order kinetics reaction.

#### Unit of the rate constant for the zero-order reaction

The rate equation in terms of product for the nth-order reaction

$-\frac{dC_{A}}{dt}&space;=&space;k_{0}\times&space;C_{A}^{n}$

$\therefore&space;unit\,&space;of\,&space;k_{0}=\frac{unit\,&space;of&space;conc.}{\left&space;(unit\,&space;of&space;conc.&space;\right&space;)^{n}\times&space;unit\,&space;of&space;time}$

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

For a chemical reaction unit molar concentration = mol lit-1 and unit of time = sec.

Thus the rate constant for the zero-order reaction

$=\frac{\left&space;(&space;mol\,&space;lit^{-1}&space;\right&space;)^{1-0}}{sec}$

= mol lit sec-1

#### Zero-order reaction half-life equation

Half-life means 50% of reactants disappear in that time interval.

The rate equation for zero-order kinetics reactions

C0 – CA = k0t

But when t = t½, that is the half-life of the reaction completed, the concentration of the reactant

CA = CA/2

$\therefore&space;C_{A}-\frac{C_{A}}{2}=k_{0}t_{\frac{1}{2}}$

$or,&space;t_{\frac{1}{2}}=\frac{C_{A}}{2k}$

Thus the half-life of the zero-order reactions depends on the initial concentration of the reactant.

Question
The half-life of a zero-order reaction = x. If the reaction completed on t1 time, what is the relation between x and t1?

t½ = x = a/2k and completion time t1 = a/k

$\therefore&space;\frac{x}{t_{1}}=\frac{a}{2k}\times&space;\frac{k}{a}$

or, t1 = 2x

#### Zero-order kinetics examples

Although the reactions which have an overall order of zero are rare. The enzyme-catalyzed reaction is the example of zero-order kinetics with respect to the substrate.

$substrate\,&space;\overset{enzyme}{\rightarrow}\,&space;product$

Heterogeneous catalyzed reactions are also the zero-order reaction.

#### Rate of formation of NO2

The chemical reaction for the formation of NO2

$N_{2}O_{5}\rightarrow&space;2NO_{2}+\frac{1}{2}O_{2}$

The rate equation for this reaction

$-\frac{d\left&space;(&space;N_{2}O_{5}&space;\right&space;)}{dt}=\frac{1}{2}\,&space;\frac{d\left&space;(&space;NO_{2}&space;\right&space;)}{dt}=2\,&space;\frac{d\left&space;(&space;O_{2}&space;\right&space;)}{dt}$

If the rate of disappearance of N2O5 = x, then the rate of formation of NO2 = 2x.

#### Characteristics of zero-order kinetics

1. The rate of the reaction is independent of concentration.
2. Half-life is proportional to the initial concentration of the reactant.
3. The rate of the reaction is always equal to the rate constant of the reaction at all concentrations.