Arrhenius studies the effect of temperature on the rate of a reaction and postulted that rate constant varies with temperature exponentially as `k=Ae^(E_(a)//RT)` . Thuis method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied.
If x is the fraction of molecules having energy greater than `E_(a)` it will be given by :

Options

A. `x=-(E_(a))/(RT)`
B. In `x=-(E_(a))/(RT)`
C. `x=e^(E_(a)//RT)`
D. Any of these

Answer

To solve the problem regarding the fraction of molecules having energy greater than the activation energy \( E_a \), we can follow these steps:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 2: Relate the Fraction of Molecules to Activation Energy
The fraction of molecules \( x \) that have energy greater than \( E_a \) can be expressed as:
\[ x = e^{-\frac{E_a}{RT}} \]
This equation indicates that \( x \) decreases exponentially with increasing activation energy \( E_a \) or decreasing temperature \( T \).

### Step 3: Take the Natural Logarithm of Both Sides
To express the relationship in logarithmic form, we can take the natural logarithm (ln) of both sides:
\[ \ln x = -\frac{E_a}{RT} \]
This step allows us to isolate the activation energy in terms of the fraction of molecules.

### Step 4: Rearranging the Equation
From the equation \( \ln x = -\frac{E_a}{RT} \), we can rearrange it to express \( E_a \):
\[ E_a = -RT \ln x \]
This shows how the activation energy relates to the fraction of molecules with energy greater than \( E_a \).

### Conclusion
Thus, the expression for the fraction of molecules having energy greater than \( E_a \) is:
\[ x = e^{-\frac{E_a}{RT}} \]