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 lnK versus 1/T in the absence of a catalyst, we will use the Arrhenius equation:

### Step-by-Step Solution:

1. **Understand the Arrhenius Equation**:
The Arrhenius equation can be expressed as:
\[ \ln K = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]
where:
- \( K \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

2. **Identify the Slope**:
From the equation, we can see that the plot of \(\ln K\) versus \(\frac{1}{T}\) is a straight line of the form:
\[ y = c - mx \]
where:
- \( y = \ln K \)
- \( x = \frac{1}{T} \)
- The slope \( m = -\frac{E_a}{R} \)

3. **Substituting the Values**:
Given:
- The activation energy in the absence of catalyst \( E_a = 8.314 \, \text{kJ/mol} \)
- Convert \( E_a \) from kJ to J:
\[ E_a = 8.314 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 8314 \, \text{J/mol} \]
- The gas constant \( R = 8.314 \, \text{J/(K mol)} \)

4. **Calculate the Slope**:
Now, substituting the values into the slope formula:
\[ \text{slope} = -\frac{E_a}{R} = -\frac{8314 \, \text{J/mol}}{8.314 \, \text{J/(K mol)}} \]
\[ \text{slope} = -1000 \]

5. **Final Answer**:
Therefore, the slope of the plot of \(\ln K\) versus \(\frac{1}{T}\) in the absence of catalyst is:
\[ \text{slope} = -1000 \]