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Question: The rate constant of a reaction:\(A\to B+C\) at \({{27}^{0}}C\) is \(3.0\times {{10}^{-5}}{{s}^{-1}}...

The rate constant of a reaction:A→B+CA\to B+C at 270C{{27}^{0}}C is 3.0×10−5s−13.0\times {{10}^{-5}}{{s}^{-1}} and at this temperature1.5×10−41.5\times {{10}^{-4}} percent of the reactant molecules are able to cross-over the P.E. barrier. The maximum rate constant of the reaction is
A) 4.5×10−9s−14.5\times {{10}^{-9}}{{s}^{-1}}
B) 4.5×10−11s−14.5\times {{10}^{-11}}{{s}^{-1}}
C) 0.2s−10.2{{s}^{-1}}
D) 20s−120{{s}^{-1}}

Explanation

Solution

The correct answer to this question is based on the Arrhenius equation formula which is given by k=Ae−EaRTk=A{{e}^{-\dfrac{{{E}_{a}}}{RT}}} where k is the rate constant and the answer is obtained by substituting the values in this equation.

Complete step by step answer:
In the concepts of physical chemistry in our previous classes, we have come across the equations which include the rate of the reaction that depends on the temperature.
- In the thermodynamics part we have studied about the Arrhenius equation given for the temperature dependent reaction rates.

Let us understand the base of Arrhenius' equation. This equation was given by Arrhenius for both the forward and backward reactions where the rate of the reaction having equilibrium constant is dependent on the temperature.
- This Arrhenius equation has a wide range of applications for calculating the energy of activation and also the rate of the reaction.
- The Arrhenius equation is given by the formula,
k=Ae−EaRTk=A{{e}^{-\dfrac{{{E}_{a}}}{RT}}}
where, k is the rate constant of a reaction
A is the Arrhenius constant called the pre-exponential factor which is a constant
T is the absolute temperature
Ea{{E}_{a}}is the activation energy and
R is the real gas constant
Now, the whole quantity e−EaRT{{e}^{-\dfrac{{{E}_{a}}}{RT}}} is nothing but the fraction of molecules crossing over the potential energy (P.E.) barrier.

Thus, e−EaRT{{e}^{-\dfrac{{{E}_{a}}}{RT}}}=1.5×10−41.5\times {{10}^{-4}} % as per the given data.
⇒e−EaRT=1.5×10−4100=1.5×10−6s−1\Rightarrow {{e}^{-\dfrac{{{E}_{a}}}{RT}}}=\dfrac{1.5\times {{10}^{-4}}}{100}=1.5\times {{10}^{-6}}{{s}^{-1}}
Also, k = 3.0×10−5s−13.0\times {{10}^{-5}}{{s}^{-1}}

Therefore, substituting these values, we get
3.0×10−5s−13.0\times {{10}^{-5}}{{s}^{-1}}=A×A\times 1.5×10−61.5\times {{10}^{-6}}
⇒A=3.0×10−5s−11.5×10−6\Rightarrow A=\dfrac{3.0\times {{10}^{-5}}{{s}^{-1}}}{1.5\times {{10}^{-6}}}
Thus, A=2×101=20s−1A=2\times {{10}^{1}}=20{{s}^{-1}}
So, the correct answer is “Option D”.

Note: The important point to be noted here is that the conversion from percentage to the numerical form and for this question is to be read carefully and if it is not converted then the answer would be option C). Therefore, be careful while reading questions and do not mark wrong answers.