If a reation follows the Arrhensis equation the plot lnk vs`1/(RT)` gives straight line with a gradient (-y) unti .The energy required to active the reactant is :

Options

A. `-y` unit
B. yR unit
C. y unit
D. y/R unit

Explanation

To solve the problem, we will follow the steps outlined in the video transcript and derive the activation energy from the Arrhenius equation.

### Step-by-Step Solution:

1. **Write the Arrhenius Equation**:
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the 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. **Rearrange the Equation**:
Rearranging the equation gives us:
\[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]
This equation is in the form of \( y = mx + c \), where:
- \( y = \ln k \)
- \( x = \frac{1}{T} \)
- \( m = -\frac{E_a}{R} \) (slope)
- \( c = \ln A \) (y-intercept)

4. **Identify the Slope**:
From the problem statement, we know that the slope of the line (gradient) is given as \( -y \). Therefore, we can equate:
\[ -\frac{E_a}{R} = -y \]

5. **Solve for Activation Energy \( E_a \)**:
By eliminating the negative signs, we get:
\[ \frac{E_a}{R} = y \]
Now, multiplying both sides by \( R \) gives:
\[ E_a = yR \]

6. **Conclusion**:
The energy required to activate the reactant, which is the activation energy \( E_a \), is:
\[ E_a = yR \]

### Final Answer:
The energy required to activate the reactant is \( yR \).