For a reaction, activation energy `(E_a )` =0 and rate constant, `k = 1.5 x 10^4 s^(-1)` at 300 K. What is the value of rate constant at 320 K'?

Options

A. `3.2 x 10^6 s^(-1)`
B. `3.2 x 10^4 s^(-1)`
C. `1.5 x 10^4 s^(-1)`
D. `6.4 x 10^8 s^(-1)`

Answer

To solve the problem, we will use the Arrhenius equation, which relates the rate constants at two different temperatures. The equation is given by:

\[ \ln \left( \frac{k_2}{k_1} \right) = -\frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

Where:
- \( k_1 \) is the rate constant at temperature \( T_1 \)
- \( k_2 \) is the rate constant at temperature \( T_2 \)
- \( E_a \) is the activation energy
- \( R \) is the universal gas constant (approximately \( 8.314 \, \text{J/mol·K} \))
- \( T_1 \) and \( T_2 \) are the absolute temperatures in Kelvin

### Step 1: Identify the given values
- \( E_a = 0 \, \text{kJ/mol} \) (activation energy)
- \( k_1 = 1.5 \times 10^4 \, \text{s}^{-1} \) at \( T_1 = 300 \, \text{K} \)
- \( T_2 = 320 \, \text{K} \)

### Step 2: Substitute the values into the Arrhenius equation
Since the activation energy \( E_a = 0 \), the equation simplifies to:

\[ \ln \left( \frac{k_2}{k_1} \right) = -\frac{0}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]

This simplifies to:

\[ \ln \left( \frac{k_2}{k_1} \right) = 0 \]

### Step 3: Solve for \( k_2 \)
Taking the exponential of both sides gives:

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

Thus, we have:

\[ k_2 = k_1 \]

### Step 4: Substitute \( k_1 \) to find \( k_2 \)
Since \( k_1 = 1.5 \times 10^4 \, \text{s}^{-1} \):

\[ k_2 = 1.5 \times 10^4 \, \text{s}^{-1} \]

### Final Answer:
The value of the rate constant at \( 320 \, \text{K} \) is:

\[ k_2 = 1.5 \times 10^4 \, \text{s}^{-1} \]