Arrhenius studied the effect of temperature on the rate of a reaction and postulated that rate constant varies with temperature exponentially as `k=Ae^(-E_(a)//RT)`. For most of the reactions, it was found that the temperature coefficient of the reaction lies between 2 to 3. The method is generally used for finding the activation energy of a reaction. Keeping temperature constant, the effect of catalyst on the activation energy has also been studied by studying how much the rate of reaction changes in the presence of catalyst. In most of the cases, it is observed that catalyst lowers the activation energy barrier and increases the rate of reaction.
Which of the following plot will be linear ?

Options

A. ln k versus T with - ve slope
B. k versus `1//T` with - ve slope
C. ln k versus `1//T` with - ve slop
D. ln k versus `1//T` with + ve slope.

Explanation

To determine which plot will be linear based on the Arrhenius equation, we can follow these steps:

### Step 1: Write down 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,
- \( T \) is the temperature in Kelvin.

### Step 2: Take the natural logarithm of both sides.
Taking the natural logarithm of the Arrhenius equation gives:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

### Step 3: Rearrange the equation.
Rearranging the equation, we can express it in the form of a linear equation \( y = mx + b \):
\[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]
Here, we identify:
- \( y = \ln k \)
- \( m = -\frac{E_a}{R} \) (the slope)
- \( x = \frac{1}{T} \)
- \( b = \ln A \) (the y-intercept)

### Step 4: Identify the linear relationship.
From the rearranged equation, we see that plotting \( \ln k \) versus \( \frac{1}{T} \) will yield a straight line. The slope of this line will be negative and equal to \(-\frac{E_a}{R}\).

### Conclusion:
The linear plot will be:
\[ \ln k \text{ versus } \frac{1}{T} \]

### Final Answer:
The plot that will be linear is \( \ln k \) versus \( \frac{1}{T} \).

---