The temperature dependence of the rate of a chemical reaction is given by arrhenius equation, `k = Ae^(-Ea//RT)`. Which of the following graphs will be a striaght line?

Options

A. In A vs 1/T
B. In A vs `E_(a)`
C. In K vs 1/T
D. In k vs `-E_(a) //R`

Explanation

To determine which graph will be a straight line 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, and
- \( T \) is the temperature in Kelvin.

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

### Step 3: Rearrange the equation
Rearranging the equation, we can express it in the form of \( y = mx + c \):
\[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]

### Step 4: Identify the components of the linear equation
From the rearranged equation:
- The slope \( m \) is \( -\frac{E_a}{R} \)
- The y-intercept \( c \) is \( \ln A \)

### Step 5: Determine the graph that represents a straight line
The equation \( \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \) indicates that if we plot \( \ln k \) (y-axis) against \( \frac{1}{T} \) (x-axis), we will obtain a straight line.

### Conclusion
Thus, the graph that will be a straight line is:
**Option C: \( \ln k \) versus \( \frac{1}{T} \)**.

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