Question

In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct option(s) following among is/are:
This question has multiple correct options.
(A) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation.
(B) The value of frequency factor is higher than that higher than that determined experimentally.
(C) The activation energy of the reaction is unaffected by the value of steric factor
(D) Since P=4.5, the reaction will not proceed unless an effective catalyst is used

Answer 1

Hint: The bimolecular reaction is a chemical reaction in which two molecular species participate and it can be either reversible or irreversible. It is a result of a collision between two molecular reactants.
Steric factor is referred to as the probability of collision between reacting molecular species which have the right orientation and positioning in order to produce products with required geometry and stereospecificity. It is determined experimentally, not theoretically. It is denoted by letter P.

Complete step by step solution:
During a bimolecular reaction, two reactants collide, and the rate of their collision is termed as frequency factor. It is calculated experimentally. It is denoted by A.
The relation between the rate of a chemical reaction, the absolute temperature and the preexponential factor is given by expression called the Arrhenius equation.
$k=A{e}^{-{\displaystyle \frac{E_a}{RT}}}$
k is a kinetic reaction rate constant, A represents pre-exponential factor or frequency factor, Ea is an activation energy, R is the gas constant and T is an absolute temperature.
For a chemical reaction, the Arrhenius equation can be written as:
$k_{collision}=A{e}^{-{\displaystyle \frac{E_a}{RT}}}$
We know, $P=4.5$
$\therefore P>1$
As we know the ratio of kexp to collision is equal to the steric factor, P. This can be represented as:
$${\displaystyle \frac{k_{\exp }}{k_{collision}}}=P$$
If we need to write Arrhenius equation for experimental k value, it can be expressed as:
$k_{\exp }=P{A}_{\exp }{e}^{-{\displaystyle \frac{E_a}{RT}}}$
On equating from both sides, we get
$A_{\exp }=AP$
Therefore, the value of frequency factor determined by an experiment must be higher than that determined theoretically, by Arrhenius equation.
$A_{\exp }>A$

Therefore, option (A) is correct.

The pre-exponential factor or frequency factor, A can be determined by using the following expression:
$A=Z\times P$
In the above expression, Z is the frequency factor and P is the steric factor.
In order to convert a reactant molecule into a product, the reactant molecule require a minimum amount of energy to excite molecules so that they can participate in the chemical reaction. This minimum amount of extra energy required is known as activation energy.
As given above, $A=Z\times P$
Therefore, the pre-exponential factor or steric factor for a chemical reaction does not depend upon the activation energy for that chemical reaction.
Hence, the steric factor will not affect the value of activation energy of that reaction.

Therefore, option (C) is the correct option.

Note :
The Arrhenius equation also has universal gas constant R in it.
$k=A{e}^{-{\displaystyle \frac{E_a}{RT}}}$
Generally, activation energy is denoted as energy per mole. When activation energy is expressed as reactant per mole, then the universal gas constant cannot be used in Arrhenius equation. Instead, the Boltzmann constant (kB) should be used.
$k=A{e}^{-{\displaystyle \frac{E_a}{k_{B}T}}}$