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, we need to analyze the Arrhenius equation and derive the relationship between the natural logarithm of the rate constant (ln k) and the reciprocal of temperature (1/T). Here’s a step-by-step solution:

### Step 1: Write the Arrhenius Equation
The Arrhenius equation is given as:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Take the Natural Logarithm of Both Sides
To analyze the relationship, we take the natural logarithm of both sides of the Arrhenius equation:
\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]

### Step 3: Rearrange the Equation
We can rearrange the equation to match the form of a straight line \( y = mx + c \):
\[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]
Here, we can identify:
- \( y = \ln k \)
- \( m = -\frac{E_a}{R} \) (slope)
- \( x = \frac{1}{T} \)
- \( c = \ln A \) (y-intercept)

### Step 4: Identify the Graph Characteristics
From the rearranged equation, we can conclude:
- The graph of \( \ln k \) versus \( \frac{1}{T} \) will be a straight line.
- The slope of the line will be negative (since \( -\frac{E_a}{R} \) is negative).
- The y-intercept of the line will be \( \ln A \).

### Step 5: Conclusion
Thus, the graph of \( \ln k \) versus \( \frac{1}{T} \) is a straight line with a negative slope and a y-intercept of \( \ln A \).

Now, we can determine which option from the given choices represents this graph. The correct option will show a straight line that slopes downwards from left to right.