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

Options

A.
B.
C.
D.

Explanation

To determine the correct graph for the relationship between `log k` and `1/T` as described by the Arrhenius equation, we can follow these steps:

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

### Step 2: Take the Logarithm of the Arrhenius Equation
Taking the logarithm of both sides, we get:
\[ \log k = \log A - \frac{E_A}{2.303R} \cdot \frac{1}{T} \]
This can be rearranged to:
\[ \log k = -\frac{E_A}{2.303R} \cdot \frac{1}{T} + \log A \]

### Step 3: Identify the Linear Form
The equation is now in the form of a straight line \( y = mx + c \), where:
- \( y = \log k \)
- \( m = -\frac{E_A}{2.303R} \) (the slope)
- \( x = \frac{1}{T} \)
- \( c = \log A \) (the y-intercept)

### Step 4: Analyze the Slope
Since the slope \( m \) is negative (because \( E_A \) is always positive), this indicates that as \( \frac{1}{T} \) increases (which means temperature \( T \) decreases), \( \log k \) will decrease.

### Step 5: Determine the Correct Graph
From the analysis, we expect the graph of \( \log k \) versus \( \frac{1}{T} \) to be a straight line with a negative slope. Therefore, the correct graph should show a downward trend as \( \frac{1}{T} \) increases.

### Conclusion
Based on the above analysis, the correct answer is **Option B**, which represents a straight line with a negative slope.

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