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)`
At what temperature, rate constant is equal to pre-exponential factor?

Options

A. `T = (1.25 xx 10^(4) K)/(14)`
B. `T = oo`
C. `T = 0`
D. `(14-1.25 xx 10^(4))K`

Explanation

To solve the problem, we need to determine the temperature at which the rate constant \( k \) is equal to the pre-exponential factor \( A \) in the Arrhenius equation. The given equation for the rate constant is:

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

### Step 1: Understand the relationship between \( k \) and \( A \)

In the Arrhenius equation, the rate constant \( k \) can be expressed as:

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

Taking the logarithm of both sides gives:

\[ \log k = \log A - \frac{E_a}{2.303RT} \]

### Step 2: Set \( k \) equal to \( A \)

We need to find the temperature \( T \) at which \( k = A \). When \( k = A \), we have:

\[ \log k = \log A \]

This implies that the term involving \( T \) in the logarithmic equation must be zero:

\[ \log k = \log A \implies -\frac{E_a}{2.303RT} = 0 \]

### Step 3: Analyze the condition for zero

For the term \(-\frac{E_a}{2.303RT}\) to be zero, the only way this can happen is if \( T \) approaches infinity. Therefore, we conclude:

\[ T \to \infty \]

### Conclusion

The temperature at which the rate constant \( k \) is equal to the pre-exponential factor \( A \) is when \( T \) approaches infinity.

### Final Answer

The temperature at which the rate constant is equal to the pre-exponential factor is:

\[ T \to \infty \]

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