A reaction takes place in theee steps: the rate constant are `k_(1), k_(2), and k_(3)`. The overall rate constant `k=(k_(1)k_(3))/(k_(2))`. If `E_(1), E_(2)`, and `E_(3)` (energy of activation) are `60, 30` and `10kJ`, respectively, the overall energy of activation is

Options

A. `40`
B. `30`
C. `400`
D. `60`

Explanation

To find the overall energy of activation for the reaction that takes place in three steps, we can follow these steps:

### Step 1: Understand the given information
We have three steps in the reaction with activation energies:
- \( E_1 = 60 \, \text{kJ} \)
- \( E_2 = 30 \, \text{kJ} \)
- \( E_3 = 10 \, \text{kJ} \)

The overall rate constant is given by the equation:
\[ k = \frac{k_1 k_3}{k_2} \]

### Step 2: Write the expression for the natural logarithm of the rate constant
Taking the natural logarithm of both sides, we have:
\[ \ln k = \ln k_1 + \ln k_3 - \ln k_2 \]

### Step 3: Substitute the Arrhenius equation
According to the Arrhenius equation, we can express the natural logarithm of the rate constants in terms of activation energy:
\[ \ln k = -\frac{E_a}{RT} + \ln A \]
where \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature.

### Step 4: Write the expressions for each rate constant
For each rate constant, we can write:
- For \( k_1 \):
\[ \ln k_1 = -\frac{E_1}{RT} + \ln A \]
- For \( k_2 \):
\[ \ln k_2 = -\frac{E_2}{RT} + \ln A \]
- For \( k_3 \):
\[ \ln k_3 = -\frac{E_3}{RT} + \ln A \]

### Step 5: Substitute these expressions into the equation for \( \ln k \)
Substituting these into the expression for \( \ln k \), we get:
\[ \ln k = \left(-\frac{E_1}{RT} + \ln A\right) + \left(-\frac{E_3}{RT} + \ln A\right) - \left(-\frac{E_2}{RT} + \ln A\right) \]

### Step 6: Simplify the equation
Combining the terms, we have:
\[ \ln k = -\frac{E_1 + E_3 - E_2}{RT} + 3 \ln A \]
The \( \ln A \) terms will cancel out when we equate both sides.

### Step 7: Rearranging to find overall activation energy
From this, we can see that the overall activation energy \( E_a \) can be expressed as:
\[ E_a = E_1 + E_3 - E_2 \]

### Step 8: Substitute the values of activation energies
Now substituting the values:
\[ E_a = 60 \, \text{kJ} + 10 \, \text{kJ} - 30 \, \text{kJ} \]
\[ E_a = 40 \, \text{kJ} \]

### Conclusion
The overall energy of activation for the reaction is:
\[ \boxed{40 \, \text{kJ}} \]