In a zero-order reaction, the rate of the reaction is independent of the concentration of the reactant. This can be seen in the differential rate law which shows how the rate of a reaction depends on the concentration of the reactant(s):

A → Products

Rate = *k*[A]^{0 }= *k*

* *

*where **k** is the r**ate constant,** and the exponent 0 is the **reaction order**, which in a general formula is given by n.*

When n = 0, the [A]^{0 }term equals 1, and therefore, the rate is equal to the rate constant of the reaction, and the concentration of reactant decreases linearly with time.

So, **how do we understand that the rate does not depend on concentration?** In other words, it means the amount of reactant that is actually available for reaction does not change with the overall quantity of reactant. Most zero-order reactions either require a catalyst or occur between gases in saturated containers.

For example, the reaction of H_{2} and Cl_{2} gases is catalyzed with light (**photochemical reaction**):

\[{{\rm{H}}_{\rm{2}}}\left( g \right){\rm{ }} + {\rm{ C}}{{\rm{l}}_{\rm{2}}}\left( g \right)\;\mathop \to \limits^{hv} \,{\rm{2HCl}}\left( g \right)\]

The decomposition of nitrous oxide over the surface of a hot platinum catalyst is another zero-order reaction:

\[{\rm{2}}{{\rm{N}}_{\rm{2}}}{\rm{O}}\left( g \right){\rm{ }}\mathop \to \limits^{Pt} {\rm{ 2}}{{\rm{N}}_{\rm{2}}}\left( g \right)\;\, + \;{{\rm{O}}_{\rm{2}}}\left( g \right)\]

As long as there is enough surface of the catalyst to capture all the reactant molecules, it does not matter how much of it there is – the reaction continues at the same rate.

**Another example** that is not a chemical reaction and generally follows the **zero-order** rules is **sublimation**. This is because only molecules at the surface of a substance can sublime, and their concentration does not change with the sublimation.

**The Units of Rate Constant, ****k** **for a Zero-Order Reaction**

**k**

Zero-order indicates that the rate does not depend on the concentration, and therefore, the rate is equal to the concentration.

rate = *k*[A]^{0}

[A]^{0 }= 1, therefore,

rate = *k*

The units for the rate are mol/L, so it is the same as the rate constant:

*k = mol/L s **or M/s or M x s^{-1}*

^{ }

The following table **summarizes** the **rate laws**, **half-lives**, and ** k units** for first-, second-, and zero-order reactions:

There is also a formula which you can use as a shortcut to determine the units of a rate constant:

*k*** units = M ^{1-n }· t^{-1}**

*where n is the reaction order*

If we needed to determine the units of k for a zero order reaction, we would use 0 for the n:

*k*** units = M ^{1-0 }· t^{-1 }= M · t^{-1}**

**The** **Integrated Rate Law of a Zero-Order Reaction**

Integrating the differential rate law, we obtain the integrated rate law for the first-order reactions. You can check this article for more details, but for now, we will write the final form of the integrated rate law for zero-order reactions.

For a simple hypothetical reaction where molecule A transforms into products, the (differential) rate law for a zero-order reaction can be written as:

A → Products

Rate = *k*[A]^{0} = *k*

Integrating the differential rate law, we obtain the integrated rate law for the zero-order reactions:

Note that the zero-order integrated rate law is also in the form of an equation for a straight line:

Therefore, a straight line with a slope of –*k* and an intercept of [A]_{0} is obtained in a graph of [A] versus time:

Remember, this was not the case for first- and second-order reactions where a straight line was obtained when the ln [A] and 1/[A] needed to be plotted vs time.

**The Half-Life of Zero-Order Reactions**

The integrated rate low for zero-order reactions A → Products is:

\[{\left[ {\rm{A}} \right]_t}\; = \; – \;kt\, + \,{\left[ {\rm{A}} \right]_0}\]

Half-life is when the initial concentration [A]_{0 }dropped by 50% which means:

\[{\left[ {\rm{A}} \right]_{{t_{1/2}}}} = {\rm{ }}{\textstyle{1 \over 2}}{\left[ {\rm{A}} \right]_0}\]

So, replacing [A]_{t }with 1/2[A]_{0, }we get:

\[{\textstyle{1 \over 2}}{\left[ {\rm{A}} \right]_0}\; = \; – \;k{t_{1/2}}\, + \,{\left[ {\rm{A}} \right]_0}\]

W can now rearrange this equation to obtain the expression for the half-life of the zero-order reactions:

\[k{t_{1/2}}\; = \;{\textstyle{1 \over 2}}{\left[ {\rm{A}} \right]_0}\]

The equation indicates that **the smaller the [A] _{0, }the shorter the half-life** or, in other words, the half-life of a zero-order reaction gets shorter as the concentration decreases.

To summarize, this is what we learned about the half-life of a reaction and its correlation with the concentration for a first-, second-, and zero-order reactions:

** **

Notice that in all cases, the half-life depends on the rate constant which appears in the denominator. And this indicates that the faster the reaction, the shorter its half-life.

**Check Also**

- Reaction Rate
- Rate Law and Reaction Order
- How to Determine the Reaction Order
- Integrated Rate Law
- The Half-Life of a Reaction
- Half-Life and Radioactivity Practice Problems
- First-Order Reactions
- Second-Order Reactions
- Determining the Reaction Order Using Graphs
- Units of Rate Constant k
- How Are Integrated Rate Laws Obtained
- Activation Energy
- The Arrhenius Equation
**Chemical Kinetics Practice Problems**