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Zero Order Reaction - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Zero Order Kinetics - Zero Order Reaction, Integrated Rate Law - Zero Order Reaction, Half Life and Life Time of Reaction, Graphs for Zero-Order Reaction is considered one of the most asked concept.

  • 39 Questions around this concept.

Solve by difficulty

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

In a zero- order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become :

The unit of rate constant for a zero order reaction is :

 for a zero-order reactor, the plot of concentration versus time is:

 Select the corret option regarding zero-ordes Reaction


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 Select the correct half-life for a 3 order reaction.

 select the correct option from the following about the unit of rate constant for zero order reaction?


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Which of the following represent the zero order reaction?


Consider the reaction follows zero order reaction. What is the t completion or t 100%  of the reaction if \mathrm{\text { (A) }_{\circ}} is initial concentration of reaction k is rate constant


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The correct of plot for first order reaction is :

Concepts Covered - 5

Zero Order Kinetics - Zero Order Reaction

In such reactions rate of reaction is independent of concentration of the reactants.
        \frac{-\mathrm{dx}}{\mathrm{dt}} \propto[\text { concentration }]^{0}
        \text { that is, } \mathrm{dx} / \mathrm{dt}=\mathrm{K}
On integration we get:
  \mathrm{x=K t+C}
if c = 0, t = 0 then
\begin{array}{l}{\mathrm{x}=\mathrm{Kt}} \\ {\mathrm{K}=\mathrm{x} / \mathrm{t}}\end{array}
\text { Unit of } \mathrm{K} \text { is } \mathrm{mol} \: \mathrm{L}^{-1} \text { time }^{-1}
Example: Photo Chemical reaction.
\bullet \: \: t_{\frac{1}{2}}=\text { time required for half completion of reaction }=\frac{a}{2 k}

Integrated Rate Law - 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,

\mathrm{A\rightarrow P}

\text { Rate }=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k[A]^{0}}

\text { Rate }=-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{k\: x\: 1}

\mathrm{d[A]\: =\: -kdt}

At t = 0, A = Ao
At t = t, A = A
Thus, on integrating both sides, we get:


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

Further simplifying the above equation, we get the rate constant, k as:



Half Life and Life Time of Reaction
  • Half-life of reaction: The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as t1/2.
    For a zero order reaction, rate constant is given as follows:

    At t = t1/2, [A] = [A]o/2
    Thus, the rate constant at t1/2 becomes:

    \mathrm{k=\frac{[\mathrm{A}]_{0}-1 / 2[\mathrm{A}]_{0}}{t_{1 / 2}}}
    \mathrm{t_{1 / 2}=\frac{[\mathrm{R}]_{0}}{2 \mathrm{k}}}
    Thus, it is clear that t1/2 for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant.
  • Life time of Reaction: It is time in which 100% of the reaction completes. It is represented as tlf.
    Thus, at t = tlf, A = 0
    Now, from zero order equation, we know:

    \mathrm{A\: =\: A_{o}\: -\: kt}

    \mathrm{0\: =\: A_{o}\: -\: kt_{lf}}

    \mathrm{Thus, t_{lf}\: =\:\frac{A_{o}}{k}}
Graphs for Zero-Order Reaction

Special Zero Order Reaction

This concept can be understood by the following example:

The reaction occurs as follows:
\mathrm{3A\: \rightarrow \: P}
Initial concentration of A is given as and rate of the reaction(r) is given as k[A]o. Find the concentration of A after time 't' and also determine the half life of A.

For zero-order reaction, the rate equation is given as follows:

\mathrm{A\: =\: A_{o}-kt}

According to question, we have given:

\mathrm{r\: =\: k[A]^{0}\: =\: -\frac{1}{3}\frac{dA}{dt}}

\mathrm{-\frac{1}{3}\frac{dA}{dt}\: =\: k\quad\quad\quad\quad\quad(Since\: [A]^{0}\: =\: 1)}

Now, integrating both sides we get:

\mathrm{[A]^{A}_{A_{o}}\: =\: -3k(t)^{t}_{o}}

\mathrm{A-A_{o}\: =\: -3kt}

\mathrm{\mathbf{A\: =\: A_{o}-3kt}}
This is the concentration of A after time 't'.

Now, for zero-order reaction, the half-life of a reaction is given as below:

\mathrm{t_{1/2}\: =\: \frac{A_{o}}{2k}\: =\: \frac{A_{o}}{2(3k)}}

\mathrm{Thus,\: \mathbf{t_{1/2}\:=\: \frac{A_{o}}{6k}}}
This is the half-life of A for this reaction.

Study it with Videos

Zero Order Kinetics - Zero Order Reaction
Integrated Rate Law - Zero Order Reaction
Half Life and Life Time of Reaction

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