## What is zero order reaction?

**Zero order reaction kinetics** in chemistry define the rate of chemical reaction in terms of reactant and product per unit time. It is independent of the concentration of reacting species.

Chemical kinetics deals with the speed and mechanism of reaction in terms of reactant and product molecules.

In chemical equilibrium, only the initial and final states are considered. The relation between reactant and products given from Law of mass action and energy relation governed from thermodynamics Laws. It does not consider the time and intermediate states of a chemical reaction.

## Example of zero order reaction

The zero order reaction kinetics are rare in reality. Generally, the chemical reaction carried out by a chemical catalyst is zero order. The enzyme catalysis reaction is an example of zero order reaction with respect to the substrate. In catalyzed reactions, the transformation takes place on the surface of the catalyst or the walls of the container.

The most common examples of zero-order sections are,

- The photochemical reaction between hydrogen with chlorine.
- Decomposition of nitrous oxide over a hot platinum surface.
- Decomposition of ammonia in presence of molybdenum or tungsten surface.
- Dissociation of hydrogen iodide in the gold surface with high pressure. It is a bimolecular reaction.

## Rate Law expression

In kinetics, the rate of a chemical reaction is generally expressed in terms of reactant and product. Therefore, the concentration of reactant decreases, and the product increase. When these decrease or increase concentration expressed per unit time, we find the rate law in chemical kinetics.

A â†’ Product

Rate of reaction = âˆ’ d[A]/dt = k [A]^{n}

In some surface reactions, the rate has been found to be independent of concentration. These reactions are known as zero order kinetics.

## Zero order reaction formula

Rate laws in chemical kinetics define the dependence of the concentration of the reacting species. Whereas the integrated law provides the increase or decrease of the concentration of reacting species at any time from the start of the reactions.

### Integrated rate law for zero order reaction

Let us take a chemical reaction with the initial concentration of reactant = a and product = 0. After time t concentration of reactant = (a âˆ’ x) and product = x. Therefore, concentration decrease after t time = x.

The rate in terms of product,

dx/dt = k

where k = rate constant

or, dx = kdt

When we integrating the above equation within limits, produce the rate equation for the reactions,

x = kt + c

where c = integration constant

But when t = 0, that is the initial state of the reaction, x = 0. The integration constant (c) of the above equation equal to zero.

âˆ´ x = k Ã— t

This is the formula that expresses the decreases or increases of concentration of the reactant or product within time.

### Differential form of zero order reaction

Let the initial concentration of reactant = [A]_{0} and after t time = [A], thus the rate equation in terms of reactant,

âˆ’ d[A]/dt = k Ã— [A]^{0} = k

Negative sign because the concentration decreases with time and initial concentration reactant [A]_{0} = 0.

Again, âˆ’ d[A] = kdt

Integrating the above equation

âˆ«d[A] = kÂ âˆ«dt

âˆ´ âˆ’ [A] = kt + c

where c = integration constant

If the initial concentration of the reactant = [A]_{0} when time = 0, that is the initial time,

âˆ’ [A]_{0 }= 0 + c

or, c = âˆ’ [A]_{0}

Therefore the integrated rate equation for the zero order kinetics,

âˆ’ [A] = kt âˆ’ [A]_{0}

or, [A]_{0} âˆ’ [A] = kt

It is another form of the rate equation for zero order kinetics.

## Zero order reaction graph

The rate law formula for zero order reaction,

[A]_{0} âˆ’ [A] = kt

When the rate of reaction is plotted against concentration and time, the following graph obtained below the picture,

## Unit of rate constant

The rate equation for nth-order reaction = k_{n} Ã— [A]^{n}

Therefore, the unit of rate constant,

k_{n}Â = unit of concentration/(unit of concentration)^{n} Ã— unit of time = (unit of concentration)^{1âˆ’n}/unit of time

### Unit of rate constant for zero order reaction

For zero order kinetics, n = 0 and molar concentration = mol lit^{âˆ’1.}

Therefore, the unit rate constant for the zero-order reaction,

= (mol lit^{âˆ’1})^{1âˆ’0}/s

= mol lit^{âˆ’1} s^{âˆ’1}

## Half life in zero order reaction

**Half life** means 50 percent of reactants disappear in that time interval. When t = t_{Â½}, that is the half-life of the reaction completed, the concentration of the reactant, [A] = [A]/2.

Therefore, [A]/2 = kÂ Ã— t_{Â½}

or, t_{Â½} = [A]/2k

From the above formula, the half-life of the zero order kinetics depends on the initial concentration of the reactant.

## Enzyme catalyzed reaction

Enzymes are complex protein produces by the living organisms of our environment. These are responsible for catalyzing infinite types of chemical changes occurring in our living system.

- For example, catalase obtained from living plant cells catalyzes the decomposition of hydrogen peroxide.
- Yeast contains various enzymes, among which maltase converts maltose into glucose, and zymase further converted glucose into ethyl alcohol.

All the reactions will be in zero order for a given amount of enzyme with respect to the substrate.

## Factors affecting rate of reaction

- The rate of the reactions is independent of concentration.
- Half-life is proportional to the initial concentration of the reactant.
- The rate of zero order kinetics in chemistry or chemical science is always equal to the rate constant at all concentrations.

## Zero order reaction problems

**Problem:** When is the rate of the reaction equal to the rate constant?

**Solution:** For zero order reaction, the rate of the reaction is equal to the rate constant or rate of reaction = k.

**Problem:** The half-life of zero order reaction = x. If the reaction is completed on t_{1} time, what is the relation between x and t_{1}?

**Solution:** t_{Â½} = x = a/2k and completion time t_{1} = a/k. Therefore, x/t_{1} = (a/2k) Ã— (k/a)

or, t_{1} = 2x