According to Arrhenius equation, rate constant k is equal to `A e^(- E_a //RT)` Which of the following options represents the graph of In k vs`1/T` ?

Options

A.
B.
C.
D.

Explanation

To solve the question regarding the Arrhenius equation and the graph of ln 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 pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

2. **Take the Natural Logarithm**:
To analyze the relationship between ln k and 1/T, we take the natural logarithm of both sides:
\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]

3. **Rearrange the Equation**:
This equation can be rearranged to fit the linear form \( y = mx + c \):
\[ \ln k = -\frac{E_a}{R} \cdot \left(\frac{1}{T}\right) + \ln A \]
Here, we can identify:
- \( y = \ln k \)
- \( x = \frac{1}{T} \)
- \( m = -\frac{E_a}{R} \) (the slope)
- \( c = \ln A \) (the y-intercept)

4. **Determine the Nature of the Graph**:
Since the slope \( m \) is negative (because \( E_a \) and \( R \) are positive), the graph of ln k vs 1/T will be a straight line with a negative slope.

5. **Identify the Correct Graph**:
Among the given options, we need to select the graph that shows a straight line with a negative slope. This means as \( \frac{1}{T} \) increases, \( \ln k \) decreases.

### Conclusion:
The correct option will be the one that represents a straight line with a negative slope.

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