Zero Order Reaction - Graph, Examples, Relationship, FAQs

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.

Also read :

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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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NCERT Chemistry Notes:

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