The plot of In k versus `1/T` is liner with slope of

Options

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

Answer

To solve the problem regarding the linear relationship between ln k and 1/T, we will 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 pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

**Hint**: Familiarize yourself with the components of the Arrhenius equation and their physical meanings.

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

**Hint**: Remember that taking the logarithm helps in transforming multiplicative relationships into additive ones.

3. **Rearrange the Equation**:
Rearranging the equation gives:
\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]
This equation is now in the form of \( y = mx + c \), where:
- \( y = \ln k \)
- \( m = -\frac{E_a}{R} \)
- \( x = \frac{1}{T} \)
- \( c = \ln A \)

**Hint**: Recognize that this is a linear equation where the slope \( m \) can be directly identified.

4. **Identify the Slope**:
From the rearranged equation, we see that the slope \( m \) is:
\[ m = -\frac{E_a}{R} \]

**Hint**: The slope of a linear graph is crucial for determining relationships between variables.

5. **Select the Correct Option**:
The slope we calculated is \( -\frac{E_a}{R} \). Among the given options, this corresponds to:
- Option 4: \(-\frac{E_a}{R}\)

**Hint**: Always compare your derived results with the provided options carefully.

### Final Answer:
The slope of the plot of ln k versus \( \frac{1}{T} \) is \(-\frac{E_a}{R}\), which corresponds to option 4.