The rate constant for the decompoistion of a certain reaction is described by the equation:
`log k(s^(-1)) = 14 - (1.25 xx 10^(4) K)/(T)`
Pre-exponential factor for this reaction is

Options

A. `14 s^(-1)`
B. `10^(14) s^(-1)`
C. `10^(-14) s^(-1)`
D. `1.25 xx 10^(4) s^(-1)`

Explanation

To find the pre-exponential factor (A) for the given reaction, we will use the Arrhenius equation in its logarithmic form. The equation provided is:

\[ \log k (s^{-1}) = 14 - \frac{1.25 \times 10^{4} K}{T} \]

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation can be expressed in logarithmic form as:

\[ \log k = \log A - \frac{E_a}{R} \cdot \frac{1}{T} \cdot \frac{1}{2.303} \]

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

### Step 2: Compare the Given Equation with the Arrhenius Equation
From the given equation, we can rewrite it as:

\[ \log k = 14 - \frac{1.25 \times 10^{4}}{T} \]

This can be compared to the Arrhenius equation:

\[ \log k = \log A - \frac{E_a}{R} \cdot \frac{1}{T} \cdot \frac{1}{2.303} \]

### Step 3: Identify the Pre-exponential Factor
From the comparison, we can see that:

\[ \log A = 14 \]

### Step 4: Solve for A
To find \( A \), we need to convert from logarithmic form to exponential form:

\[ A = 10^{14} \]

### Step 5: Express A in Appropriate Units
Since the rate constant \( k \) is in \( s^{-1} \), the pre-exponential factor \( A \) will also have units of \( s^{-1} \).

Thus, the pre-exponential factor for this reaction is:

\[ A = 10^{14} \, s^{-1} \]

### Final Answer
The pre-exponential factor for this reaction is \( 10^{14} \, s^{-1} \).

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