In a plot of log k vs 1/T, the slope is

Options

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

Explanation

To solve the question regarding the slope of the plot of log k vs. 1/T, we can follow these steps:

### Step-by-Step Solution:

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

2. **Take the Natural Logarithm**:
To analyze the equation, we take the natural logarithm (ln) of both sides:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

3. **Convert to Logarithm Base 10**:
We can convert the natural logarithm to base 10 logarithm using the relationship \( \ln x = \log_{10} x \cdot 2.303 \):
\[ \ln k = \log_{10} A \cdot 2.303 - \frac{E_a}{R} \cdot \frac{1}{T} \]
This can be rewritten as:
\[ \log_{10} k = \log_{10} A - \frac{E_a}{2.303 R} \cdot \frac{1}{T} \]

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

5. **Determine the Slope**:
From the linear equation, we can see that the slope \( m \) of the plot of log k vs. 1/T is:
\[ \text{slope} = -\frac{E_a}{2.303 R} \]

### Final Answer:
In a plot of log k vs. 1/T, the slope is given by:
\[ \text{slope} = -\frac{E_a}{2.303 R} \]