A graph plotted between `log k` versus `1//T` for calculating activation energy is shown by

Options

A.
B.
C.
D.

Explanation

To solve the question regarding the graph plotted between `log k` versus `1/T` for calculating activation energy, we can follow these steps:

### Step 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,
- \( T \) is the temperature in Kelvin.

### Step 2: Take the Logarithm of the Arrhenius Equation
To analyze the relationship between `log k` and `1/T`, we take the logarithm of both sides:
\[ \log k = \log A - \frac{E_A}{2.303 R} \cdot \frac{1}{T} \]

### Step 3: Rearrange the Equation
This can be rearranged to fit the linear equation format \( y = mx + c \):
\[ \log k = -\frac{E_A}{2.303 R} \cdot \frac{1}{T} + \log A \]
Here:
- \( y = \log k \)
- \( x = \frac{1}{T} \)
- The slope \( m = -\frac{E_A}{2.303 R} \)
- The intercept \( c = \log A \)

### Step 4: Identify the Nature of the Graph
Since the slope \( m \) is negative, the graph of `log k` versus `1/T` will be a straight line that decreases as `1/T` increases. This indicates that as temperature increases, the rate constant \( k \) increases, leading to a decrease in `log k`.

### Step 5: Analyze the Options
Now, we need to analyze the provided options for the graph:
- A graph with an increasing slope (positive slope) is incorrect.
- A graph with a decreasing slope (negative slope) is correct.

### Conclusion
The correct answer is the option that shows a straight line with a negative slope.

### Final Answer
The correct answer is **Option B**.

---