The graph between the log K versus `1/T` is a straight line. The slope of the line is

Options

A. `-(2.303R)/E_a`
B. `-E_a/(2.303R)`
C. `(2.303R)/E_a`
D. `E_a/(2.303R)`

Answer

To find the slope of the graph between log K versus \( \frac{1}{T} \), we can follow these steps:

### Step-by-Step Solution:

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

2. **Take the Natural Logarithm**:
Taking the natural logarithm of both sides, we get:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

3. **Convert to Logarithm Base 10**:
To convert the natural logarithm to base 10, we use the relationship:
\[ \ln k = 2.303 \log_{10} k \]
Thus, we can rewrite the equation as:
\[ \log_{10} k = \log_{10} A - \frac{E_a}{2.303R} \cdot \frac{1}{T} \]

4. **Identify the Linear Form**:
The equation can be rearranged into the form of a straight line \( y = mx + c \):
\[ \log_{10} k = -\frac{E_a}{2.303R} \cdot \frac{1}{T} + \log_{10} A \]
Here, \( y = \log_{10} k \), \( x = \frac{1}{T} \), \( m = -\frac{E_a}{2.303R} \), and \( c = \log_{10} A \).

5. **Determine the Slope**:
From the linear equation, we can see that the slope \( m \) of the graph of \( \log_{10} k \) versus \( \frac{1}{T} \) is:
\[ m = -\frac{E_a}{2.303R} \]

### Conclusion:
Thus, the slope of the line in the graph of \( \log K \) versus \( \frac{1}{T} \) is:
\[ \text{slope} = -\frac{E_a}{2.303R} \]