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Arrhenius Equation - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

  • Exception(Arrhenius Theory), Effective Activation Energy is considered one of the most asked concept.

  • 121 Questions around this concept.

Solve by difficulty

Two reactions R1 and R2 have identical pre-exponential factors.  The activation energy of R1 exceeds that of R2 by 10 kJ mol−1.  If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to :

(R=8.314 J mol−1K−1)


Rate of a reaction can be expressed by Arrhenius equation as : k=Ae^{-E/RT}  . In this equation, E represents

The rate of a reaction increases by 3 times when the temperature is raised from 300 K to 310 K. If K is the rate constant at 300 K, then the rate constant at 310 K will be equal to

In the respect of the equation k=Ae^{-E_{a}/RT}  in chemical kinetics, which one of the following statements is correct?

The graph between the log K versus 1/T is a straight line. The slope of the line is.

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The graph between the log K versus 1/T is a straight line. The intercept of the line on y axis is.

The rate constant, the activation energy, and the Arrhenius parameter (A) of a chemical reaction at  25^{\circ} \mathrm{C}  are  3.0 \times 10^{-4} \mathrm{sec}^{-1}, 104.4 \mathrm{~kJ} \mathrm{~mol}^{-1}  and  6.0 \times 10^{-4} \mathrm{sec}^{-1}  respectively the value of the rate constant  \mathrm{T \rightarrow \infty}  is - 

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The mean life of a radio nuclide which decays by given parallel path is:

X \stackrel{\lambda_{1}}{\longrightarrow} Y \lambda_{1}=3.6 \times 10^{-3} \mathrm{~s}^{-1}
2 \times \stackrel{\lambda_{2}}{\longrightarrow} Y \lambda_{2}=10^{-3} \mathrm{~s}^{-1}

Which of the following reactions will have the fastest rate at room temperature (25 { }^{\circ} \mathrm{C} )?

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What is the activation energy \mathrm{( E_{a} )} in a chemical reaction?

Concepts Covered - 9

Effect of Temperature on Rate of Reaction: Temperature Coefficient

We know that on increasing the temperature, the rate of the reaction or rate constant increases.
The rate equation is given as follows:

\mathrm{Rate\: =\: k[conc]^{n}}

Here k is the rate constant

Now, we will see the relation between k and T or also known as 'approximate dependency of k on T'.

Generally on 10oC rise in temperature, rate constant nearly doubles.

Temperature Coefficient: It is the ratio of two rate constants. Thus, mathematically it is given as:

\mathrm{T_{coeff}\: =\: \frac{k_{(t\: +\: 10)^{o}C}}{k_{t^{o}C}}}

Thus, the temperature coefficient is showing the dependency of the rate constant(k) on temperature(T).

NOTE: The standard value of the temperature coefficient is given at t = 25oC and (t+10) = 35oC.

Effect of Temperature on Rate of Reaction: Accurate Dependency of K on T

The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation. It was first proposed by Dutch chemist, J.H. van’t Hoff but Swedish chemist, Arrhenius provided its physical justification and interpretation.

        \mathrm{k=A e^{-E a / R T}}

where A is the Arrhenius factor or the frequency factor. It is also called pre-exponential factor. It is a constant specific to a particular reaction. R is gas constant and Ea is activation energy measured in joules/mole(J mol–1).

Ratio of Two Rate Constants at Two Different Temperatures

We have the rate constant K1 at temperature T1 and rate constant K2 at temperature T2

We know that the Arrhenius equation is given as follows:

\mathrm{log_{10}K_{1}\: =\: log_{10}A\: -\: \frac{E_{a}}{2.303RT_{1}}\quad\quad\quad\quad\quad............(i)}

\mathrm{log_{10}K_{2}\: =\: log_{10}A\: -\: \frac{E_{a}}{2.303RT_{2}}\quad\quad\quad\quad\quad............(ii)}

On subtracting equation (i) from (ii), we get:

\mathrm{log_{10}K_{2}\: -\: log_{10}K_{1}\: =\: \frac{E_{a}}{2.303RT_{1}}\: -\: \frac{E_{a}}{2.303RT_{2}}}

\mathrm{\mathbf{Thus,}\: log\frac{K_{2}}{K_{1}}\: =\: \frac{Ea}{2.303R}\left [ \frac{1}{T_{1}}\: -\: \frac{1}{T_{2}} \right ]}

Exception(Arrhenius Theory)

Although the Arrhenius equation explains the exact relationship between the rate of reaction and the temperature but there are still some exceptions in this theory. Actually, on increasing the temperature rate may decrease sometimes and may not follow Arrhenius equation. Following examples will illustrate these exceptions.

  • Bacterial decomposition

    From the graph, it is clear that first on increasing the temperature the rate also increases but at Ti (also known as inversion temperature) rate starts to decrease on further increasing the temperature.
Complex Reaction - Mechanism of Reaction

On the basis of mechanism, we have two types of reactions:

  • Simple or elementary reaction
  • Complex or multi-step reaction

Simple or Elementary reaction

  • The reactions, which occur in single step, are called simple or elementary reactions. For example:

    \mathrm{NO(g)\: +\: O_{3}(g)\: \rightarrow \: NO_{2}(g)\: +\: O_{2}(g)}
  • An elementary reaction is an individual molecular event that involves breaking or making of chemical bonds. The overall reaction describes the stoichiometry of the overall process but provides no information how the reaction occurs.

Complex Reaction

  • A complex reaction takes place in a sequence of a number of elementary steps. 
  • Molecularity of complex reaction is not defined. Molecularity of each step can be defined but not for overall.
  • Overall rate of reaction is given by slowest step of the complex reaction.
    For example, combination of NO2 and CO occurs in a sequence of elementary steps.

    \\\mathrm{NO_{2}(g)\: +\: NO_{2}(g)\: \rightarrow \: NO_{3}(g)\: +\: NO(g)\quad\quad\quad\quad(elementary\: step\: I)}\\\\\mathrm{NO_{3}(g)\: +\: CO(g)\: \rightarrow \: NO_{2}(g)\: +\: CO_{2}(g)\quad\quad\quad\quad(elementary\: step\: II)}

Important Facts:

  • The numbeer of reactant molecules taking part in an elementary step or in an elementary reaction is expressed as molecularity of that step of molecularity of that reaction respectively.
  • For elementary reactions usually order of reaction and molecularity are same. Thus, it can be said that if order of a reaction for a change is fractional it cannot be an elementary reaction.
Complex Reaction (When Intermediate is Incorporated)

This example will illustrate how to determine the rate law when the intermediate is involved in rate-determining step.

\mathrm{2O_{3}\: \rightarrow \: 3O_{2}}


\\\mathrm{O_{3}\: \rightleftharpoons \: O_{2}\: +\: O\quad\quad\quad\quad\quad K_{1}\: and \: K_{2}\: are\: the \:forward\: and\: backward\: reaction\: constants\: respectively}\\\\\mathrm{O\: +\: O_{3}\: \rightarrow \: 2O_{2}\quad\quad\quad\quad\quad K_{3}\: is\: the\: rate\: constant}

In this case, step 1 is fast and step 2 is slow.

The rate law is given as follows:

\mathrm{rate\: =\: K_{3}[O]^{1}[O_{3}]^{1}}

We know from equilibrium theory that:

\\\mathrm{K_{eq}\: =\: \frac{K_{1}}{K_{2}}\: =\: \frac{[O_{2}][O]}{[O_{3}]}}\\\\\mathrm{[O]\: =\: \frac{K_{1}[O_{3}]}{K_{2}[O_{2}]}}\\\\\mathrm{Thus, rate\: =\: \frac{K_{3}.K_{1}}{K_{2}}\frac{[O_{3}][O_{3}]}{[O_{2}]}\: =\: \frac{K_{3}.K_{1}}{K_{2}}\frac{[O_{3}]^{2}}{[O_{2}]}}

\\\mathrm{Thus,\: Order\: =\: 2-1=1}\\\\\mathrm{rate\: constant\: =\: \frac{K_{3}.K_{1}}{K_{2}}}

Parallel First Order Kinetics

In this situation, B and C both are forming. These types of reactions are known as parallel reactions. Both these reactions are first order reactions with rate constants K1 and K2 respectively and half-lives as t(1/2)1 and t(1/2)2.

For these parallel reactions, we need to find:

  • Effective order
  • Effective rate constant
  • Effective t1/2
  • Effective Activation energy
  • [A], [B], [C] with time (t) variation
  • % of [B] and % of [C]

We know that the rate equations are given as follows:

\\\mathrm{r_{1}\:=\: \frac{-dA}{dt}\:=\: K_{1}[A] }\\\\\mathrm{r_{2}\: =\: \frac{-dA}{dt}\: =\: K_{2}[A]}

\\\mathrm{Thus,\: overall\: rate\: of\: reaction\: is:}\\\\\mathrm{\frac{-dA}{dt}\: =\: K_{1}[A]\: +\: K_{2}[A]\: =\: (K_{1}\: +\: K_{2})[A]}

\mathrm{Thus,\: rate\: =\: (K_{1}\: +\: K_{2})[A]^{1}}

\mathrm{\mathbf{Effective\: Rate\: Constant(K_{eff})}\: =\: (K_{1}\: +\: K_{2})}

\mathrm{\mathbf{Effective\: order\: of\: reaction}\: =\: 1}

\mathrm{Now, effective\: half-life(t_{1/2})\:=\: \frac{0.693}{K_{eff}}\: =\: \frac{0.693}{K_{1}\: +\: K_{2}}}

                                                           \mathrm{\Rightarrow\: \frac{0.693}{\frac{0.693}{(t_{1/2)_{1}}}+\frac{0.693}{(t_{1/2})_{2}}}}

\\\mathrm{Thus,\: effective\: half\: life\: is\: given\: as:}\\\\\mathrm{\frac{1}{(t_{1/2})_{eff}}\: =\: \frac{1}{(t_{1/2})_{1}}\: +\:\frac{1}{(t_{1/2})_{2}} }

NOTE: Effective activation energy, [A], [B], [C] with time (t) variation and % of [B] and % of [C] will be discussed in later concepts.

Effective Activation Energy

We know that Arrhenius equation is given as:

\mathrm{K\: =\: A.e^{-Ea/RT}}

\mathrm{K_{eff}\: =\: K_{1}\: +\: K_{2}}

\mathrm{Thus, A_{eff}.e^{-Ea_{eff}/RT}\: =\: A_{1}.e^{-Ea_{1}/RT}\: +\: A_{2}.e^{-Ea_{2}/RT}}

\mathrm{Differentiate\: this\: equation\: with\: respect\: to\: temperature\: 'T'}

\mathrm{Thus,\: we\: have:}

\mathrm{A_{eff}.e^{-Ea_{eff}/RT}\left ( \frac{+Ea_{eff}}{RT^{2}} \right )\: =\:A_{1}.e^{-Ea_{1}/RT}\left ( \frac{+Ea_{1}}{RT^{2}} \right ) \: +\: A_{2}.e^{-Ea_{2}/RT}\left ( \frac{+Ea_{2}}{RT^{2}} \right )}

\mathrm{K_{eff}\, E_{eff}\: =\: K_{1}Ea_{1}\: +\: K_{2}Ea_{2}}

\mathrm{E_{eff}\: =\: (K_{1}Ea_{1}\: +\: K_{2}Ea_{2})/K_{eff}}

[A], [B], [C] Vs time(t)

We know that the rate equations are given as follows:

\\\mathrm{r_{1}\:=\: \frac{-dA}{dt}\:=\: K_{1}[A] }\\\\\mathrm{r_{2}\: =\: \frac{-dA}{dt}\: =\: K_{2}[A]}

The overall rate of the reaction is:

\mathrm{\frac{-dA}{dt}\: =\: K_{1}[A]\: +\: K_{2}[A]\: =\: (K_{1}\: +\: K_{2})[A]}

\mathrm{\Rightarrow\: \frac{dA}{dt}\: =\: (K_{1}\: +\: K_{2})[A]^{1}}

\mathrm{\Rightarrow\: \frac{dA}{A}\: =\: -(K_{1}\: +\: K_{2})dt}

On integrating both sides of the above equation, we get:

\mathrm{\int_{A_{o}}^{A}\frac{dA}{A}\: =\: -\int_{0}^{t}(K_{1}\: +\: K_{2})dt}

\Rightarrow \mathrm{\left [ logA \right ]^{A}_{A_{o}}\: =\, -(K_{1}\: +\: K_{2})\left [ t \right ]^{t}_{0}}

\mathrm{\Rightarrow log\, A\: -\: log\, A_{o}\: =\: -k_{eff}t}

\mathrm{\Rightarrow log\, \frac{A}{A_{o}}\: =\: -k_{eff}t}

\mathrm{\Rightarrow \frac{A}{A_{o}}\: =\: e^{-k_{eff}t}}

\mathrm{Thus,[A]\: =\: [A_{o}] e^{-k_{eff}t}\quad\quad\quad\quad\quad\quad...........(i)}

Now, the rate equation for B can be given as follows:

\mathrm{\frac{dB}{dt}\: =\: k_{1}[A]\: =\: k_{1}[A_{o}]e^{-(k_{1}\: +\: k_{2})t}\quad\quad\quad\quad\quad(From\: equation\: (i))}

\mathrm{\Rightarrow\int_{0}^{B}{dB}\: =\: \int_{0}^{t}k_{1}[A_{o}]e^{-(k_{1}\: +\: k_{2})t}dt}

\mathrm{Thus, [B]\: =\: \frac{k_{1}A_{o}}{k_{1}+k_{2}}\left [ 1\: -\: e^{-(k_{1}\: +\: k_{2})t} \right ]}

Similarly, [C] is given as follows:

\mathrm{[C]\: =\: \frac{k_{2}A_{o}}{k_{1}+k_{2}}\left [ 1\: -\: e^{-(k_{1}\: +\: k_{2})t} \right ]}

Study it with Videos

Effect of Temperature on Rate of Reaction: Temperature Coefficient
Effect of Temperature on Rate of Reaction: Accurate Dependency of K on T
Ratio of Two Rate Constants at Two Different Temperatures

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