The slope of Arrhenius plot `(l_(n)k " vs " 1//J)` of a first order reaction is `-5 xx 10^(3)`. The value of `E_(a)` of the reaction is `(R = 8.314 J K^(-1)mol^(-1))`

Options

A. `41.5 kJmol^(-1)`
B. `83 kJmol^(-1)`
C. `-41.5 kJmol^(-1)`
D. `-83 kJ mol^(-1)`

Explanation

To find the activation energy (Ea) of a first-order reaction using the slope of the Arrhenius plot, we can use the Arrhenius equation:

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

Taking the natural logarithm of both sides gives us:

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

This equation can be rearranged to the form:

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

From this equation, we can see that if we plot \(\ln k\) against \(\frac{1}{T}\), the slope of the line will be:

\[ \text{slope} = -\frac{E_a}{R} \]

Given that the slope of the Arrhenius plot is \(-5 \times 10^3\), we can set this equal to \(-\frac{E_a}{R}\):

1. **Set the equation for the slope**:
\[ -5 \times 10^3 = -\frac{E_a}{R} \]

2. **Substituting the value of R**:
The value of R is given as \(8.314 \, \text{J K}^{-1} \text{mol}^{-1}\).
\[ -5 \times 10^3 = -\frac{E_a}{8.314} \]

3. **Rearranging to solve for \(E_a\)**:
\[ E_a = 5 \times 10^3 \times 8.314 \]

4. **Calculating \(E_a\)**:
\[ E_a = 41657 \, \text{J/mol} \]

5. **Converting to kJ/mol**:
\[ E_a = 41.657 \, \text{kJ/mol} \]

Thus, the activation energy \(E_a\) of the reaction is approximately **41.66 kJ/mol**.