The `E_a` of reaction in the presence of catalyst is 4.15 KJ/mol and in absence of catalyst is `8.3KJmol^(-1)` . What is the slope of the plot of lnk vs `1/T` in the absence of catalyst.

Options

A. `+1`
B. `-1`
C. `+1000`
D. `-1000`

Answer

To find the slope of the plot of lnK vs 1/T in the absence of a catalyst, we can use the Arrhenius equation:

1. **Write the Arrhenius equation**:
\[ \ln K = \ln A - \frac{E_a}{RT} \]
where:
- \( K \) is the rate constant,
- \( A \) is the Arrhenius factor,
- \( E_a \) is the activation energy,
- \( R \) is the gas constant,
- \( T \) is the temperature in Kelvin.

2. **Identify the slope of the linear plot**:
The equation can be rearranged to match the linear form \( y = mx + c \):
\[ \ln K = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]
Here, the slope \( m \) of the plot of \( \ln K \) vs \( \frac{1}{T} \) is:
\[ m = -\frac{E_a}{R} \]

3. **Substitute the values**:
- Given \( E_a \) in the absence of a catalyst is \( 8.3 \, \text{kJ/mol} \).
- Convert \( E_a \) to Joules:
\[ E_a = 8.3 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 8300 \, \text{J/mol} \]
- The gas constant \( R \) is \( 8.314 \, \text{J/(mol \cdot K)} \).

4. **Calculate the slope**:
\[ \text{slope} = -\frac{E_a}{R} = -\frac{8300 \, \text{J/mol}}{8.314 \, \text{J/(mol \cdot K)}} \]
Performing the calculation:
\[ \text{slope} = -999.4 \approx -1000 \]

5. **Conclusion**:
The slope of the plot of \( \ln K \) vs \( \frac{1}{T} \) in the absence of a catalyst is approximately \( -1000 \).