K is rate constant at temp T then value of `underset(Ttooo)lim` log K is equal to :-

Options

A. `A//2.303`
B. A
C. 2.303 A
D. log A

Explanation

To solve the problem, we need to analyze the relationship between the rate constant \( K \) and the temperature \( T \) using the Arrhenius equation. The Arrhenius equation is given by:

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

Taking the logarithm of both sides, we have:

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

Now, we need to evaluate the limit as \( T \) approaches infinity:

\[ \lim_{T \to \infty} \log K \]

Substituting the expression for \( \log K \):

\[ \lim_{T \to \infty} \log K = \lim_{T \to \infty} \left( \log A - \frac{E_a}{2.303RT} \right) \]

Now, we can separate the limit:

\[ \lim_{T \to \infty} \log K = \lim_{T \to \infty} \log A - \lim_{T \to \infty} \frac{E_a}{2.303RT} \]

Since \( \log A \) is a constant, its limit as \( T \) approaches infinity remains \( \log A \):

\[ \lim_{T \to \infty} \log A = \log A \]

Next, we analyze the second term:

\[ \lim_{T \to \infty} \frac{E_a}{2.303RT} \]

As \( T \) approaches infinity, \( \frac{1}{T} \) approaches \( 0 \). Therefore:

\[ \lim_{T \to \infty} \frac{E_a}{2.303RT} = 0 \]

Putting it all together, we have:

\[ \lim_{T \to \infty} \log K = \log A - 0 = \log A \]

Thus, the final result is:

\[ \lim_{T \to \infty} \log K = \log A \]

### Final Answer:
\[ \lim_{T \to \infty} \log K = \log A \]