According to Arrhenius equation rate constant K is equal to A `e.^(-E_(a)//RT)` . Which of the following options represents the graph of ln K vs `(1)/(T)`?

Options

A.
B.
C.
D.

Explanation

To solve the question regarding the Arrhenius equation and the graph of ln K versus (1/T), we can follow these steps:

### Step-by-Step Solution:

1. **Write 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 \( k \) and \( T \), we take the natural logarithm of both sides:
\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]

3. **Rearrange the Equation**:
Rearranging the equation gives us:
\[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]
This equation can be compared to the linear equation \( y = mx + c \), where:
- \( y = \ln k \)
- \( x = \frac{1}{T} \)
- \( m = -\frac{E_a}{R} \) (the slope)
- \( c = \ln A \) (the y-intercept)

4. **Identify the Characteristics of the Graph**:
From the rearranged equation:
- The slope \( m \) is negative (\(-\frac{E_a}{R} < 0\)), indicating that the graph will slope downwards.
- The y-intercept \( c \) is positive (\(\ln A > 0\)), indicating that the graph will start above the origin on the y-axis.

5. **Sketch the Graph**:
Based on the information above, the graph of \( \ln k \) versus \( \frac{1}{T} \) will:
- Start from a positive value on the y-axis (due to the positive intercept).
- Decrease as \( \frac{1}{T} \) increases (due to the negative slope).

6. **Select the Correct Option**:
Now, we compare the characteristics of the graph we derived with the provided options. The correct graph will show a downward slope starting from a positive y-intercept.

### Conclusion:
The correct option represents a graph of \( \ln k \) versus \( \frac{1}{T} \) that has a positive intercept and a negative slope.

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