The rate of reaction increases isgnificantly with increase in temperature. Generally, rate of reactions are doubled for every `10^(@)C` rise in temperature. Temperature coefficient gives us an idea about the change in the rate of a reaction for every `10^(@)C` change in temperature.
`"Temperature coefficient" (mu) = ("Rate constant of" (T + 10)^(@)C)/("Rate constant at" T^(@)C)`
Arrhenius gave an equation which describes aret constant `k` as a function of temperature
`k = Ae^(-E_(a)//RT)`
where `k` is the rate constant, `A` is the frequency factor or pre-exponential factor, `E_(a)` is the activation energy, `T` is the temperature in kelvin, `R` is the universal gas constant.
Equation when expressed in logarithmic form becomes
`log k = log A - (E_(a))/(2.303 RT)`
For a reaction `E_(a) = 0` and `k = 3.2 xx 10^(8)s^(-1)` at `325 K`. The value of `k` at `335 K` would be

Options

A. `3.2 xx 10^(8) s^(-1)`
B. `6.4 xx 10^(8) s^(-1)`
C. `12.8 xx 10^(8) s^(-1)`
D. `25.6 xx 10^(8) s^(-1)`

Explanation

To solve the problem, we need to find the value of the rate constant \( k \) at \( 335 \, K \) given that the activation energy \( E_a = 0 \) and \( k = 3.2 \times 10^{8} \, s^{-1} \) at \( 325 \, K \).

### Step-by-Step Solution:

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 frequency factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

2. **Substitute the Given Values**:
We know that \( E_a = 0 \). Therefore, the equation simplifies to:
\[ k = A e^{0} = A \]
This means that the rate constant \( k \) is equal to the frequency factor \( A \) at any temperature.

3. **Calculate \( k \) at \( 325 \, K \)**:
Given \( k_1 = 3.2 \times 10^{8} \, s^{-1} \) at \( 325 \, K \), we can conclude that:
\[ A = k_1 = 3.2 \times 10^{8} \, s^{-1} \]

4. **Calculate \( k \) at \( 335 \, K \)**:
Since \( E_a = 0 \), the value of \( k \) does not change with temperature. Therefore, at \( 335 \, K \):
\[ k_2 = A = 3.2 \times 10^{8} \, s^{-1} \]

5. **Final Answer**:
The value of \( k \) at \( 335 \, K \) is:
\[ k = 3.2 \times 10^{8} \, s^{-1} \]