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Exception(Arrhenius Theory), Effective Activation Energy is considered one of the most asked concept.
139 Questions around this concept.
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 constant K increases with
for a reaction, consider the plot of Int versus
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Rate of a reaction can be expressed by Arrhenius equation as : . In this equation,
represents
For a reaction, consider the plot of In k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10-5 s-1 , then the rate constant at 500 K is :
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 in chemical kinetics, which one of the following statements is correct?
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The graph between the log K versus 1/T is a straight line. The slope of the line is.
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 are
and
respectively the value of the rate constant
is -
We know that on increasing the temperature, the rate of the reaction or rate constant increases.
The rate equation is given as follows:
Rate
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:
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.
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.
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).
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:
On subtracting equation (i) from (ii), we get:
Thus,
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.
On the basis of mechanism, we have two types of reactions:
Simple or Elementary reaction
Complex Reaction
Important Facts:
This example will illustrate how to determine the rate law when the intermediate is involved in the rate-determining step.
Mechanism
In this case, step 1 is fast and step 2 is slow.
The rate law is given as follows:
rate
We know from equilibrium theory that:
Thus, Order
rate constant
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:
We know that the rate equations are given as follows:
Thus, overall rate of reaction is :
Thus, rate
Effective Rate Constant
Effective order of reaction
Thus, effective half life is given as :
NOTE: Effective activation energy, [A], [B], [C] with time (t) variation and % of [B] and % of [C] will be discussed in later concepts.
We know that Arrhenius equation is given as:
Thus,
Differentiate this equation with respect to temperature ' T '
Thus, we have:
We know that the rate equations are given as follows:
The overall rate of the reaction is:
On integrating both sides of the above equation, we get:
Thus,
Now, the rate equation for B can be given as follows:
Similarly, [C] is given as follows:
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