The `E_a` of reaction in the presence of catalyst is `5.25 kJ//mol` in the absence of catalyst is `8.314 kJ mol^(-1)`. What is the slope of the plot of Ink vs `1/T` in the absence of catalyst.
`(R = 8.314 J k^(-1) mol^(-1))`

Options

A. 100
B. `-100`
C. `-1000`
D. `+1000`

Answer

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

\[ k = A e^{-\frac{E_a}{RT}} \]

Taking the natural logarithm of both sides gives us:

\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]

This can be rearranged to match the form of a linear equation \( y = mx + c \):

\[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]

Where:
- \( y = \ln k \)
- \( m = -\frac{E_a}{R} \) (the slope)
- \( x = \frac{1}{T} \)
- \( c = \ln A \)

### Step-by-Step Solution:

1. **Identify the Activation Energies**:
- Activation energy in the presence of a catalyst, \( E_a = 5.25 \, \text{kJ/mol} \)
- Activation energy in the absence of a catalyst, \( E_a = 8.314 \, \text{kJ/mol} \)

2. **Convert Activation Energy to Joules**:
- Since \( R \) is given in J, we need to convert \( E_a \) from kJ to J:
\[ E_a = 8.314 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 8314 \, \text{J/mol} \]

3. **Use the Arrhenius Equation to Find the Slope**:
- The slope \( m \) of the plot of \( \ln k \) versus \( \frac{1}{T} \) is given by:
\[ m = -\frac{E_a}{R} \]

4. **Substitute the Values**:
- Substituting the values of \( E_a \) and \( R \):
\[ m = -\frac{8314 \, \text{J/mol}}{8.314 \, \text{J/(mol K)}} = -1000 \]

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

### Final Answer:
The slope of the plot of \( \ln k \) vs \( \frac{1}{T} \) in the absence of a catalyst is \( -1000 \).

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