For a reaction the activation energy `E_a=0` and the rate constant=`3.2 times 10^6 sec^-1` at 300K. What is the value of rate constant at 310K?

Options

A. `6.4 times 10^6 sec^-1`
B. `6.4 times 10^12 sec^-1`
C. `3.2 times 10^6 sec^-1`
D. 0

Answer

To find the value of the rate constant at 310 K given that the activation energy \(E_a = 0\) and the rate constant \(k_1 = 3.2 \times 10^6 \, \text{s}^{-1}\) at 300 K, we can use the Arrhenius equation.

### Step-by-Step Solution:

1. **Understanding 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 (approximately \(8.314 \, \text{J mol}^{-1} \text{K}^{-1}\)),
- \(T\) is the temperature in Kelvin.

2. **Using the Logarithmic Form**:
We can also express the change in rate constants with temperature using the logarithmic form:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

3. **Substituting Known Values**:
Given:
- \(E_a = 0\)
- \(k_1 = 3.2 \times 10^6 \, \text{s}^{-1}\) at \(T_1 = 300 \, \text{K}\)
- \(T_2 = 310 \, \text{K}\)

Substituting these values into the equation:
\[ \log \frac{k_2}{3.2 \times 10^6} = \frac{0}{2.303R} \left( \frac{1}{300} - \frac{1}{310} \right) \]

4. **Simplifying the Equation**:
Since \(E_a = 0\), the right-hand side becomes zero:
\[ \log \frac{k_2}{3.2 \times 10^6} = 0 \]

5. **Exponentiating Both Sides**:
This implies:
\[ \frac{k_2}{3.2 \times 10^6} = 10^0 = 1 \]
Therefore:
\[ k_2 = 3.2 \times 10^6 \, \text{s}^{-1} \]

6. **Final Answer**:
Thus, the value of the rate constant at 310 K is:
\[ k_2 = 3.2 \times 10^6 \, \text{s}^{-1} \]

### Conclusion:
The correct answer is \(3.2 \times 10^6 \, \text{s}^{-1}\), which corresponds to option three.