For a reaction having rate constant k at a temperature T, when a graph between ln k and 1/T is drawn a straight line is obtained. The point at which line cuts y -axis and x -axis respectively correspond to the temperature:

Options

A. `0, (Ea)/(2.303R)log A`
B. `oo , (E_(a))/("R ln A")`
C. `0, log A`
D. None of these.

Explanation

To solve the problem, we will use the Arrhenius equation and analyze the graph of ln k versus 1/T.

### Step-by-Step Solution:

1. **Understand the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = Arrhenius constant (pre-exponential factor)
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

2. **Take the Natural Logarithm**:
Taking the natural logarithm of both sides gives:
\[ \ln k = \ln A - \frac{E_a}{RT} \]
This can be rearranged to:
\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]

3. **Identify the Linear Form**:
The equation can be compared to the linear form \( y = mx + c \):
- Let \( y = \ln k \)
- Let \( x = \frac{1}{T} \)
- The slope \( m = -\frac{E_a}{R} \)
- The y-intercept \( c = \ln A \)

4. **Determine the Intercepts**:
- **Y-Intercept**: When \( \frac{1}{T} = 0 \) (i.e., \( T \to \infty \)), the value of \( \ln k \) at this point is \( \ln A \). Thus, the y-intercept corresponds to \( T = \infty \).
- **X-Intercept**: When \( \ln k = 0 \), we can set the equation to zero:
\[ 0 = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]
Rearranging gives:
\[ \frac{E_a}{R} \cdot \frac{1}{T} = \ln A \]
Therefore, solving for \( T \):
\[ T = \frac{E_a}{R \ln A} \]

5. **Final Results**:
- The y-intercept corresponds to \( T \to \infty \).
- The x-intercept corresponds to \( T = \frac{E_a}{R \ln A} \).

### Summary:
- The point where the line cuts the y-axis corresponds to \( T = \infty \).
- The point where the line cuts the x-axis corresponds to \( T = \frac{E_a}{R \ln A} \).