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

Options

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

Answer

To solve the question regarding the plot of ln k versus 1/T and its slope, we can use the Arrhenius equation, which 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,
- \( T \) is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Take the Natural Logarithm of Both Sides:**
Start by taking the natural logarithm of the Arrhenius equation:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

2. **Rearrange the Equation:**
Rearranging the equation gives:
\[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]

3. **Identify the Linear Form:**
The equation \(\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A\) can be recognized as the equation of a straight line in the form \(y = mx + c\), where:
- \(y\) is \(\ln k\),
- \(m\) (the slope) is \(-\frac{E_a}{R}\),
- \(x\) is \(\frac{1}{T}\),
- \(c\) is \(\ln A\).

4. **Determine the Slope:**
From the linear equation, we can conclude that the slope of the plot of \(\ln k\) versus \(\frac{1}{T}\) is:
\[ \text{slope} = -\frac{E_a}{R} \]

5. **Conclusion:**
Therefore, the slope of the plot of \(\ln k\) versus \(\frac{1}{T}\) is \(-\frac{E_a}{R}\), which corresponds to the activation energy divided by the gas constant, with a negative sign.

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

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