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 solve the problem of finding the rate constant at 310 K given that the activation energy \( E_a = 0 \) and the rate constant at 300 K is \( k_1 = 3.2 \times 10^6 \, \text{s}^{-1} \), we can use the Arrhenius equation. Here’s a step-by-step solution:

### Step 1: Write the Arrhenius Equation
The Arrhenius equation is given by:

\[ k = A e^{-\frac{E_a}{RT}} \]

Where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor
- \( E_a \) = activation energy
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( T \) = temperature in Kelvin

### Step 2: Set Up the Equations for Two Temperatures
For the two temperatures (300 K and 310 K), we can write:

1. For \( T_1 = 300 \, \text{K} \):
\[ k_1 = A e^{-\frac{E_a}{RT_1}} = 3.2 \times 10^6 \, \text{s}^{-1} \]

2. For \( T_2 = 310 \, \text{K} \):
\[ k_2 = A e^{-\frac{E_a}{RT_2}} \]

### Step 3: Divide the Two Equations
Dividing the equation for \( k_1 \) by the equation for \( k_2 \):

\[ \frac{k_1}{k_2} = \frac{A e^{-\frac{E_a}{RT_1}}}{A e^{-\frac{E_a}{RT_2}}} \]

This simplifies to:

\[ \frac{k_1}{k_2} = e^{-\frac{E_a}{RT_1} + \frac{E_a}{RT_2}} \]

### Step 4: Substitute \( E_a = 0 \)
Since \( E_a = 0 \):

\[ \frac{k_1}{k_2} = e^{0} = 1 \]

This implies:

\[ k_1 = k_2 \]

### Step 5: Conclusion
Since \( k_1 = k_2 \), the rate constant at 310 K is the same as the rate constant at 300 K:

\[ k_2 = 3.2 \times 10^6 \, \text{s}^{-1} \]

### Final Answer
The value of the rate constant at 310 K is:

\[ k_2 = 3.2 \times 10^6 \, \text{s}^{-1} \]

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