A reaction rate constant is given by
`k = 1.2 xx 10^(14) e^((-2500)/(RT))s^(-1)`. It means

Options

A. `log k` versus ` log T` will give a straight line with a slope as `25000`.
B. `log k` versus `log T` will give a straight line with a slope as `-25000`.
C. `log k` versus `T` will give a straight line with a slope as `-25000`.
D. `log k` versus `1//T` will give a straight line.

Answer

To analyze the given reaction rate constant equation \( k = 1.2 \times 10^{14} e^{\frac{-2500}{RT}} \, s^{-1} \), we can follow these steps:

### Step 1: Identify the Form of the Equation
The equation for the rate constant \( k \) resembles the Arrhenius equation, which is given by:
\[ k = A e^{\frac{-E_a}{RT}} \]
where:
- \( A \) is the pre-exponential factor (frequency factor),
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 2: Compare the Given Equation with the Arrhenius Equation
From the given equation, we can identify:
- \( A = 1.2 \times 10^{14} \)
- \( E_a = 2500 \, \text{J/mol} \) (since it is in the exponent as \(-\frac{E_a}{RT}\))

### Step 3: Take the Logarithm of Both Sides
To analyze the relationship between \( k \) and \( T \), we take the logarithm of both sides:
\[ \log k = \log A + \frac{-E_a}{2.303RT} \]
This can be rearranged to:
\[ \log k = \log A - \frac{E_a}{2.303R} \cdot \frac{1}{T} \]

### Step 4: Identify the Linear Relationship
This equation is in the form of \( y = mx + c \), where:
- \( y = \log k \)
- \( x = \frac{1}{T} \)
- The slope \( m = -\frac{E_a}{2.303R} \)

### Step 5: Determine the Slope
The slope of the line when plotting \( \log k \) versus \( \frac{1}{T} \) will be negative, indicating that as temperature increases, the rate constant \( k \) increases.

### Step 6: Analyze the Options
Now, we can analyze the options provided:
- **Option A**: Incorrect, as we cannot plot \( \log k \) versus \( \log T \).
- **Option B**: Incorrect for the same reason as option A.
- **Option C**: Incorrect, as we do not plot \( \log k \) versus \( T \).
- **Option D**: Correct, as we can plot \( \log k \) versus \( \frac{1}{T} \) and it will yield a straight line.

### Conclusion
The correct interpretation of the given rate constant equation is that it follows the Arrhenius equation, and plotting \( \log k \) versus \( \frac{1}{T} \) will yield a straight line.

---