Zero Order Reaction - Graph, Examples, Relationship, FAQs

In chemical kinetics, reactions are usually classified by how their rates are influenced by the concentration of reactants. In most instances, among these, reactions have a changing rate that is directly proportional to an increasing or decreasing concentration of reactants. However, there is a special class of reactions called zero-order reactions, whereby the rate of reaction shows independence in regard to a change in reactant concentration. That's an interesting property that sets zero-order reactions apart from first-order, second-order, and higher-order reactions in which the rate of reaction is directly proportional to the concentration of one or more reactants.

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Zero-order reactions are especially important in catalysis when the reaction rate is limited by the number of active sites on a catalyst, and not by the concentration of reactants. What limits such reactions is not the availability of reactants but that of catalytic sites. This is seen to follow a constant reaction rate until all active sites have been occupied, otherwise known as catalyst saturation.

An understanding of zero-order reactions is critical in the design of industrial processes, particularly those that which incorporate catalytic systems. Thirdly, such reactions are extremely useful in the knowledge of surface chemistry and the role of a catalyst in hastening chemical transformations. Hence, the study of zero-order kinetics allows chemists to make better predictions of, and to control, the behavior of a complex reaction in applications such as pharmaceuticals and environmental engineering

Zero order Reactions

In such reactions rate of reaction is independent of concentration of the reactants.

Rate $\propto[\text { concentration }]^0$
For example, suppose we have a reaction

$A \longrightarrow B$

then, the rate of reaction can be written as

Rate $=-\frac{\mathrm{dA}}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^0$

From the above equation it is evident that for a Zero order reaction,

(1) The rate of reaction is equal to the rate constant

(2) The rate of reaction is constant and independant of time

(3) The unit of rate constant is $\mathrm{molL}^{-1}$ time $^{-1}$

(4) The rate of reaction cannot be changed by changing the concentration of reactant.

Integrated Rate law for a Zero Order reaction

Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactants. Consider the reaction,

$\begin{aligned} & \mathrm{A} \rightarrow \mathrm{P} \\ & \text { Rate }=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^0 \\ & \Rightarrow \text { Rate }=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k} \\ & \Rightarrow \mathrm{d}[\mathrm{A}]=-\mathrm{kdt} \\ & \Rightarrow \int_{\left[\mathrm{A}_0\right]}^{[\mathrm{A}]} \mathrm{d}[\mathrm{A}]=-\mathrm{k} \int_0^{\mathrm{t}} \mathrm{dt}\end{aligned}$

Thus, on integrating both sides, we get:

$\left[\mathrm{A}_{\mathrm{t}}\right]=[\mathrm{A}]_0-\mathrm{kt}$

Comparing the above equation with the equation of a straight line, y = mx + c, if we plot [A] against t, we get a straight line as shown in the above figure with slope = –k and intercept equal to [A]_o.

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Half-life of reaction:

The half-life of a reaction is the time in which the concentration of a reactant is reduced to half of its initial concentration. It is represented as t_1/2.
For a zero order reaction, rate constant is given as follows:
$\begin{aligned} & \mathrm{A}_{\mathrm{t}}=\mathrm{A}_0-\mathrm{kt} \\ & \text { When } \mathrm{t}=\mathrm{t}_{\frac{1}{2}},[\mathrm{~A}]_{\mathrm{t}}=\frac{[\mathrm{A}]_0}{2}\end{aligned}$

Putting these values in the integrated rate expression,

$\frac{[\mathrm{A}]_0}{2}=[\mathrm{A}]_0-\mathrm{kt}_{\frac{1}{2}}$

Upon solving the above expression we have,

$\mathrm{t}_{\frac{1}{2}}=\frac{[\mathrm{A}]_0}{2 \mathrm{k}}$
Thus, it is clear that half life for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant.

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Life time of Reaction: It is time in which 100% of the reaction completes. It is represented as t_LF.
Thus, at t = t_LF, A = 0
Now, from integrated rate equation for zero order, we know:$\begin{aligned} & \mathrm{A}=\mathrm{A}_{\mathrm{o}}-\mathrm{kt} \\ & 0=\mathrm{A}_{\mathrm{o}}-\mathrm{kt}_{\mathrm{LF}} \\ & \text { Thus, } \mathrm{t}_{\mathrm{LF}}=\frac{\mathrm{A}_{\mathrm{o}}}{\mathrm{k}}\end{aligned}$

Graphs for Zero-Order Reaction

Some Solved Examples

Example 1:

Question:For a reaction:

nA $\longrightarrow$ Product
If the rate constant and the rate of reactions are equal, what is the order of the reaction?

1) (correct)0

2)2

3)1

4)3

Solution

In such reactions rate of reaction is independent of the concentration of the reactants.
$\begin{aligned} & \frac{-\mathrm{dx}}{\mathrm{dt}} \propto[\text { concentration }]^0 \\ & \text { that is, } \mathrm{dx} / \mathrm{dt}=\mathrm{K}\end{aligned}$

If the rate constant and the rate of reactions are equal the rate of reaction does not depend on reactant concentration. This is the definition of Zeroth's Order reaction.

Hence, the answer is the option (1).

Example 2:

Question:

The formation of gas at the surface of tungsten due to adsorption is the reaction of order

1) (correct)0

2)1

3)2

4)3

Solution:

Adsorption on the metal surface does not depend on the concentration of gas. So, it will be a zeroth-order reaction.

Hence, the answer is the option (1).

Example 3:

Question

Units of the rate constant of first and zero-order reactions in terms of molarity M unit are respectively.

1) (correct)$\sec ^{-1} \cdot M \sec ^{-1}$

2)$\sec ^{-1}, M$

3)$M \sec ^{-1}, \sec ^{-1}$

4)$M, \sec ^{-1}$

Solution

For zero-order reaction

$-\frac{\Delta[R]}{\Delta t}=k[R]^0$

unit of k is $M \sec ^{-1}$

For first-order reaction

$-\frac{\Delta[R]}{\Delta t}=k[R]^1$

unit of k is $\sec ^{-1}$

Hence, the answer is the option (1).

Example 4:

Question:

Which graph represents zero order reaction.

Solution

For a zero-order reaction:

$[\mathrm{A}]=[\mathrm{A}]_0-\mathrm{kt}$

Now, $\mathrm{t}_{\frac{3}{4}}$ represents the time taken for 75% completion of the reaction,

i.e. $[\mathrm{A}]=\frac{[\mathrm{A}]_0}{4}$

Putting these values in the integrated rate equation

$\frac{[\mathrm{A}]_0}{4}=[\mathrm{A}]_0-\mathrm{kt}_{\frac{3}{4}}$

$\Rightarrow \mathrm{t}_{\frac{3}{4}}=\frac{3[\mathrm{~A}]_0}{4}$

which represents a straight line passing through the origin and having a positive slope

Therefore, option(4) is correct.

Summary

Zero-order reactions are those chemical kinetics where the reaction progresses at a constant rate, which is independent of the concentration of reactants. In contrast to first- or higher-order reactions, a zero-order reaction basically depends on some other factor, like catalyst availability, and not on the concentration of reactants. It makes zero-order kinetics very important in industrial catalysis and surface chemistry applications, thus giving influence to processes related to pharmaceuticals and environmental engineering. These are based on knowledge about their rate laws, integrated rate laws, and associated concepts like half-life that depend only on the initial reactant concentrations and rate constants. Graphically, they explain that there is a linear relationship between reactant concentration with time. Knowing how to work out the best reaction conditions and output in complex systems of chemical reactions will result in many zero-order reactions.

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