In Arrehenius equation if a graph is plotted between logK and 1/T, the slope of the curve will be:

Options

A. `(-E_a)/R`
B. `(-E_a)/(2.303R)`
C. `E_a/R`
D. `E_a/(2.303R)`

Answer

To find the slope of the graph plotted between log K and 1/T using the Arrhenius equation, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

2. **Take the Natural Logarithm**:
Taking the natural logarithm of both sides gives:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

3. **Convert Natural Logarithm to Common Logarithm**:
To convert from natural logarithm to common logarithm (base 10), we use the conversion factor \( 2.303 \):
\[ \ln k = 2.303 \log k \]
Thus, we can rewrite the equation as:
\[ 2.303 \log k = \ln A - \frac{E_a}{RT} \]
Rearranging gives:
\[ \log k = \log A - \frac{E_a}{2.303R} \cdot \frac{1}{T} \]

4. **Identify the Linear Form**:
The equation can be compared to the linear form \( y = mx + c \):
- Let \( y = \log k \)
- Let \( x = \frac{1}{T} \)
- The slope \( m \) is given by:
\[ m = -\frac{E_a}{2.303R} \]

5. **Conclusion**:
Therefore, the slope of the curve when plotting log K against \( \frac{1}{T} \) is:
\[ \text{slope} = -\frac{E_a}{2.303R} \]

### Final Answer:
The slope of the curve will be \( -\frac{E_a}{2.303R} \).

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