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Arrhenius Equation is a mathematical expression that shows the relationship between the rate constant (of a chemical reaction), the absolute temperature, and the A factor (also known as the pre-exponential factor). The factor is perceived as the frequency of oriented collisions between reactant particles. It elucidates the relationship between reaction rates and absolute temperature.
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Key Terms: Arrhenius equation, Boltzmann constant, Arrhenius plot, catalyzed reactions, Arrhenius equation, pre-exponential factor, rate constant, Chemical reaction
Arrhenius Equation: Expression
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The Arrhenius equation expression is shown as:
k=Ae-Ea/RT
Where,
- k denotes the rate constant of the reaction
- A denotes the pre-exponential factor. In terms of the collision theory, it is the frequency of correctly oriented collisions between the reacting species
- e is the base of the natural logarithm (Euler’s number)
- Ea signifies the activation energy of the chemical reaction (energy per mole)
- R denotes the universal gas constant
- T denotes the absolute temperature associated with the reaction (in Kelvin)
If the activation energy is expressed in terms of energy per reactant molecule, the Boltzmann constant (kB) must be substituted for the universal gas constant in the Arrhenius equation.
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Graphical Representation of the Arrhenius Equation
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For the decomposition reaction undergone by nitrogen dioxide (given by 2NO2 → 2NO + O2), a graph plotted with the rate constant (k) on the Y-axis and the absolute temperature (T) on the X-axis is provided below. The rate of the reaction increases when the temperature increases.
Arrhenius Plot
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An Arrhenius plot shows the logarithm of a reaction rate constant, In(k), on the ordinate axis alongside the reciprocal of temperature, 1/T, on the abscissa. The impact of temperature on the rates of chemical reactions is increasingly investigated using Arrhenius plots.
An Arrhenius plot for a single rate-limited thermally activated process yields a straight line. From this line, the activation energy and pre-exponential factor may be calculated. When logarithms are taken on both sides of the equation, the Arrhenius equation can be written as follows:
In k=In(Ae-Ea/RT)
Solving the equation further:
In k=In(A)+In(e-Ea/RT)
In k=In(A) + (-Ea/RT) = In(A)- (Ea/R) (1/T)
Since ln(A) is a constant, the equation corresponds to that of a straight line (y = mx + c) whose slope (m) is -Ea/R. When the logarithm of the rate constant (ln K) is plotted on the Y-axis and the inverse of the absolute temperature (1/T) is plotted on the X-axis, the resulting graph is called an Arrhenius plot.
Arrhenius Equation and Catalysts
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A catalyst's role is to reduce the activation energy required by a reaction. This allows for the usage of reduced activation energy (as accounted for by the catalyst) to solve the Arrhenius equation. It also helps obtain the rate constant for the catalyzed reaction.
Arrhenius Equation and Catalysts
The Arrhenius equation's exponential component (-Ea/RT) provides for an exponential increase in the value of the rate constant for any drop in activation energy. Since the rate of a chemical reaction is proportional to its rate constant, a drop in activation energy results in an exponential rise in the reaction rate.
The temperature has a significant impact on the rates of uncatalyzed reactions as compared to the rates of catalyzed reactions. This happens because the activation energy is in the numerator of the exponential expression -Ea/RT. Whereas, the absolute temperature is in the denominator. Since the catalyzed reaction has low activation energy, the influence of temperature on the rate constant is more evident in the corresponding uncatalyzed reaction.
Arrhenius Equation and the Pre-Exponential Factor (A)
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The symbol ‘A’ in the Arrhenius equation indicates the pre-exponential factor (or the frequency factor). This factor is associated with molecule collisions. It may be considered as the frequency of correctly oriented collisions between molecules that contain enough energy to initiate a chemical reaction.
The pre-exponential factor is represented by the following equation:
A=ρZ
Here, Z is the frequency factor (frequency of collisions) is the steric factor (that deals with the orientation of molecules). The units of A depend on the order of the reaction. For example, the value of ‘A’ for a second-order rate constant is expressed in L.mol-1.s-1 (or M-1s-1, since M = mol.L-1) and that of a first-order rate constant is expressed in s-1.
Eliminating the A Factor from the Arrhenius Equation
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Taking a chemical reaction at two distinct temperatures T1 and T2, with corresponding rate constants k1 and k2, the logarithmic version of the Arrhenius equation is:
In k1= In(A) - Ea/RT1
In (k2) =In(A) - Ea/RT2
The second equation can be rearranged to get the value of ln(A):
In(A)=Ink2+Ea/RT2
Substituting the value of ln(A) in the equation for ln(k1), the following equations can be obtained:
Ink1 = Ink2+Ea/RT2 - Ea/RT1
Shifting Ink2 to the LHS, the value of Ink1-Ink2 becomes:
Ink1- Ink2 = Ea/RT2 - Ea/RT1
The LHS of the equation is of the form In(x)-In(y), which can be simplified to In(x/y). Also, the term Ea/R is a common factor to both the terms in the RHS. Therefore, the entire equation can be simplified as follows:
In(k1/k2) = Ea/R (1/T1-1/T2)
Things to Remember
- The Arrhenius Equation is the formula used to determine the energy of activation and justify the impact of temperature on the pace of reaction.
- The formula for Arrhenius Equation is K = A e-Ea/RT
- The collision theory underpins the Arrhenius equation. A reaction, according to this theory, is a collision between two molecules (of the same or different substances) that results in the formation of an intermediate.
- The Arrhenius equation allows one to account for factors that influence the rate of a reaction that cannot be estimated using the rate law.
- The Arrhenius equation may be used to determine the influence of the energy barrier, temperature, frequency, presence of a catalyst, and collision orientation.
- Any chemical reaction, according to Arrhenius, is only feasible when reacting molecules are triggered with a minimum amount of energy, known as threshold energy. The kinetic energy of the majority of molecules is lower than this minimum energy. Activation energy is the additional energy required to activate reactant molecules.
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Sample Questions
Ques. For a reaction consider the plot of ln k versus 1/T given in the figure. If the rate constant of this reaction at 400K is 10−5 s−1, then calculate the rate constant at 500K. (3 Marks)
Ans. We know,
InK2 / K1=Ea/R [1/T1-1/T2]
Given that,
K1 = 10-5; T1 = 400K ; T2 = 500K ; -Ea/R= – 4606K
Now,
2.303log K2/10-5 = 4606 [1/400-1/500]
K2 = 10-4s-1
Ques. The decomposition of hydrocarbon follows the equation
k = (4.5 x 1011S-1)e-28000K/T
Calculate Ea . (3 Marks)
Ans. The given equation is k = (4.5 x 1011S-1)e-28000K/T ….(i)
The Arrhenius equation is given by,
k=Ae-Ea/RT….(ii)
From equations (i) and (ii), we obtain
Ea / RT=28000 K / T
Ea=R x 28000 K
=8.314 J K-1mol-128000
232791 J mol-1
=232.791 kJ mol-1
Ques. The rate of a reaction quadruples when the temperature changes from 290 to 330K. Find the energy of activation of the reaction assuming that it does not change with temperature. (3 Marks)
Ans. Given:
T1 = 290K
T2 = 330K
K2 =4K1
From the Arrhenius equation, we obtain
by substituting all the values we get,
Ea = 1,103276.8/40
=27,581.9 J/mol
Ea = 27.5819 kJ/mol
Ques. The activation energy of a chemical reaction is 100 kJ/mol and its A factor is 10 M-1s-1. Find the rate constant of this equation at a temperature of 300 K. (4 Marks)
Ans. Given,
Ea=100 kJ.mol-1=100000 J.mol-1
A=10 M-1s-1, In(A)=2.3(approx.)
T=300K
The value of the rate constant can be obtained from the logarithmic form of the Arrhenius equation, which is:
In k=In(A)-(Ea/RT)
In k = 2.3-(100000 J.mol-1)/ (8.314 J.mol-1.K-1) 300K
In k=2.3-40.1
In k=37.8
k= 3.8341 x 10-17M-1s-1(from the units of the A factor, it can be understood that the reaction is a second-order reaction, for which the unit of k is M-1s-1).
Therefore, the value of the rate constant for the reaction at a temperature of 300K is approximately 3.8341 x 10-17M-1s-1.
Ques. At a temperature of 600 K, the rate constant of a chemical reaction is 2.75 x 10-8M-1s-1. When the temperature is increased to 800K, the rate constant for the same reaction is 1.95 x 10-7M-1s-1. What is the activation energy of this reaction? (5 Marks)
Ans. Given,
T1 = 600K
k1 = 2.75 x 10-8M-1s-1
T2 = 800K
k2=1.95 x 10-7M-1s-1
When the A factor is eliminated from the Arrhenius equation, the following equation is obtained:
In(k1/k2)=-Ea/R (1/T1-1/T2)
Substituting the given values in the equation, the value of Ea can be determined:
In(2.75 x 10-8/1.95 x 10-7)= (-Ea/8.314 J.K-1.mol-1) x (0.00041 K-1)
In(0.141) =Ea x (-0.0000493)J-1.mol
Ea = (-1.958)/ (-0.0000493) J.mol-1 = 39716 J.mol-1
The activation energy of the reaction is approximately 39716 J.mol-1.
Ques. For the reaction: 2A+B → A2B is k[A][B]2with k=2.0 x 10-6mol-2L2s-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L–1, [B] = 0.2 mol L–1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L–1. (5 Marks)
Ans. The initial rate of reaction is
Rate =k[A][B]2
=(2.0 x 10-6mol-2L2s-1) (0.1 mol L-1) (0.2 mol L-12)
=8.0 x 10-9mol-2L2s-1
When [A] is reduced from 0.1 mol L-1 to 0.06 mol L-1, the concentration of A reacted =(0.1-0.06) mol L-1=0.04 mol L-1
Therefore, concentration of B reacted = ½ x 0.04 mol L-1=0.02 mol L-1
Then, concentration of B available, B=(0.2-0.02) mol L-1=0.18 mol L-1
After [A] is reduced to 0.06 mol L-1, the rate of the reaction is given by,
Rate = k[A][B]2
=(2.0 x 10-6mol-2L2s-1) (0.06 mol L-1) (0.18 mol L-12)
=3.89 x 10-9molL-1s-1
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