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 need to derive the activation energy (Ea) from the Arrhenius equation and analyze the given information about the plot of ln k versus 1/(RT).

### 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
- \( A \) = frequency factor (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, we get:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

3. **Rearrange the Equation**:
Rearranging the equation gives:
\[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]
This is in the form of the straight-line equation \( y = mx + c \), where:
- \( y = \ln k \)
- \( m = -\frac{E_a}{R} \)
- \( x = \frac{1}{T} \)
- \( c = \ln A \)

4. **Identify the Slope**:
From the equation, we can see that the slope \( m \) is given by:
\[ m = -\frac{E_a}{R} \]

5. **Given Information**:
The problem states that the slope of the plot is \(-y\). Therefore, we can equate:
\[ -\frac{E_a}{R} = -y \]

6. **Solve for Activation Energy (Ea)**:
Rearranging the equation gives:
\[ \frac{E_a}{R} = y \]
Thus, we can express the activation energy as:
\[ E_a = y \cdot R \]

### Final Answer:
The energy required to activate the reactant is:
\[ E_a = y \cdot R \]