For a first order reaction, the plot of log `k` against 1/T is a straight line. The slope of the line is equal to

Options

A. `(E_(a))/(R)`
B. `(2.303)/(E_(a)xxR)`
C. `(E_(a))/(2.303)`
D. `(-E_(a))/(2.303 R)`

Answer

To solve the question regarding the slope of the plot of log \( k \) against \( \frac{1}{T} \) for a first-order reaction, we will use the Arrhenius equation and the relationship between the rate constant \( k \) and temperature \( T \).

### Step-by-Step Solution:

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

2. **Take the Natural Logarithm**:
To analyze the relationship between \( k \) and \( T \), we take the natural logarithm of both sides:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

3. **Convert to Logarithm Base 10**:
We can convert the natural logarithm to base 10 using the conversion factor \( \log k = \frac{\ln k}{2.303} \):
\[ \log k = \frac{\ln A}{2.303} - \frac{E_a}{2.303RT} \]

4. **Rearranging the Equation**:
Rearranging gives us:
\[ \log k = \left(-\frac{E_a}{2.303R}\right) \left(\frac{1}{T}\right) + \frac{\ln A}{2.303} \]
This is in the form of a linear equation \( y = mx + c \), where:
- \( y = \log k \)
- \( x = \frac{1}{T} \)
- \( m = -\frac{E_a}{2.303R} \) (slope)
- \( c = \frac{\ln A}{2.303} \) (y-intercept)

5. **Identify the Slope**:
From the above equation, we can see that the slope \( m \) of the plot of \( \log k \) against \( \frac{1}{T} \) is:
\[ \text{slope} = -\frac{E_a}{2.303R} \]

### Final Answer:
The slope of the line when plotting \( \log k \) against \( \frac{1}{T} \) for a first-order reaction is:
\[ -\frac{E_a}{2.303R} \]