According to Arrhenius equation rate constant k is equal to `Ae^(-E_(a)//RT)`. Which of the following option. Represents the graph of ln k us `(1)/(T)`?

Options

A.
B.
C.
D.

Answer

To solve the problem regarding the Arrhenius equation and the graph of ln k versus (1/T), we will 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 Arrhenius constant (pre-exponential factor),
- \( E_a \) is the activation energy,
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin.

2. **Take the Natural Logarithm**:
To find the relationship between ln k and (1/T), we take the natural logarithm of both sides:
\[ \ln k = \ln A + \ln\left(e^{-\frac{E_a}{RT}}\right) \]
Using the property of logarithms, this simplifies to:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

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

4. **Identifying the Graph**:
From the equation \( \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \), we can conclude:
- The graph of \( \ln k \) versus \( \frac{1}{T} \) is a straight line.
- The slope of the line is negative (since \(-\frac{E_a}{R}\) is negative).
- The y-intercept is positive (\(\ln A\)).

5. **Choosing the Correct Option**:
Based on the characteristics of the graph:
- The graph should be a straight line with a negative slope and a positive y-intercept.
- Therefore, the correct option that represents this graph is option **A**.