For a reaction, `K = 2 xx 10^(13) e^(-30000//RT)`. When log K(y-axis) is plotted against 1/T (x-axis), slope of line will be……….. Cal

Options

A. `3000/4.6`
B. `-3000/46`
C. `-30000/2.303`
D. `-30000/4.6`

Explanation

To solve the given problem, we need to analyze the equation provided and derive the slope when plotting log K against 1/T.

### Step-by-Step Solution:

1. **Identify the given equation**:
The equation for the reaction is given as:
\[ K = 2 \times 10^{13} e^{-\frac{30000}{RT}} \]

2. **Understand the Arrhenius equation**:
The Arrhenius equation is generally expressed as:
\[ K = A e^{-\frac{E_a}{RT}} \]
where \( A \) is the pre-exponential factor and \( E_a \) is the activation energy.

3. **Compare the equations**:
From the given equation, we can identify:
- Pre-exponential factor \( A = 2 \times 10^{13} \)
- Activation energy \( E_a = 30000 \) (in appropriate units, which we will convert later).

4. **Taking the natural logarithm**:
To plot log K against \( \frac{1}{T} \), we first take the natural logarithm of both sides:
\[ \ln K = \ln(2 \times 10^{13}) - \frac{30000}{RT} \]

5. **Convert to logarithm base 10**:
We need to convert the natural logarithm to base 10 logarithm. Using the conversion:
\[ \log K = \frac{\ln K}{2.303} \]
Thus,
\[ \log K = \frac{\ln(2 \times 10^{13})}{2.303} - \frac{30000}{2.303 RT} \]

6. **Rearranging the equation**:
The equation can be rearranged in the form of a straight line \( y = mx + c \):
\[ \log K = \left(-\frac{30000}{2.303 R}\right) \left(\frac{1}{T}\right) + \frac{\ln(2 \times 10^{13})}{2.303} \]
Here, the slope \( m \) is:
\[ m = -\frac{30000}{2.303 R} \]

7. **Substituting the value of R**:
The value of \( R \) in calories is approximately \( 1.987 \, \text{cal/(mol K)} \). For simplicity, we can round it to \( 2 \, \text{cal/(mol K)} \):
\[ m = -\frac{30000}{2.303 \times 2} \]

8. **Calculating the slope**:
Now we calculate the slope:
\[ m = -\frac{30000}{4.606} \approx -6515.5 \, \text{cal/mol} \]

### Final Answer:
The slope of the line when log K is plotted against \( \frac{1}{T} \) is approximately:
\[ \text{slope} \approx -6515.5 \, \text{cal/mol} \]