`2A rarr B+C`
The mechanism of the above reaction given as
`2A overset(K_(1))underset(K_(2))hArrX+2B` (fast)
`X+B overset(K_(3))rarrC` (slow)
`E_(1)`= Activation energy for `K_(1)`
`K_(2)`= Activation energy for `K_(2)`
`E_(3)`=Activation energy for `K_(3)`
Calculate `E_("reaction")`
(given : `E_(1) =10 kJ, E_(3)=5 kJ, E_(2) =12 kJ)`

Options

A. 5
B. 6
C. 7
D. 3

Explanation

To calculate the activation energy of the overall reaction given the mechanism, we can follow these steps:

### Step 1: Identify the Rate-Determining Step
The slow step in the provided mechanism is the rate-determining step (RDS). This means that the rate of the overall reaction is determined by this step. In this case, the slow step is:
\[ X + B \overset{K_3}{\rightarrow} C \]

### Step 2: Write the Expression for the Rate
The rate of the reaction can be expressed in terms of the concentration of the reactants involved in the rate-determining step:
\[ \text{Rate} = K_3 [X][B] \]

### Step 3: Use the Equilibrium Expression
The first step of the mechanism is an equilibrium step:
\[ 2A \overset{K_1}{\underset{K_2}{\rightleftharpoons}} X + 2B \]
From this equilibrium, we can express the concentration of the intermediate \( [X] \):
\[ K_{eq} = \frac{[X][B]^2}{[A]^2} = \frac{K_1}{K_2} \]
Thus, rearranging gives:
\[ [X] = \frac{K_1}{K_2} \frac{[A]^2}{[B]^2} \]

### Step 4: Substitute \( [X] \) into the Rate Equation
Substituting the expression for \( [X] \) into the rate equation gives:
\[ \text{Rate} = K_3 \left(\frac{K_1}{K_2} \frac{[A]^2}{[B]^2}\right)[B] \]
This simplifies to:
\[ \text{Rate} = \frac{K_1 K_3}{K_2} [A]^2 [B]^{-1} \]

### Step 5: Determine the Effective Rate Constant
From the above expression, we can identify the effective rate constant \( K_{effective} \):
\[ K_{effective} = \frac{K_1 K_3}{K_2} \]

### Step 6: Calculate the Activation Energy of the Overall Reaction
When combining rate constants, the activation energies behave as follows:
- If rate constants are multiplied, their activation energies add.
- If rate constants are divided, the activation energies subtract.

Thus, the activation energy for the overall reaction \( E_{reaction} \) can be calculated as:
\[ E_{reaction} = E_1 + E_3 - E_2 \]

### Step 7: Substitute the Given Values
Now, substituting the given values:
- \( E_1 = 10 \, \text{kJ} \)
- \( E_2 = 12 \, \text{kJ} \)
- \( E_3 = 5 \, \text{kJ} \)

We get:
\[ E_{reaction} = 10 + 5 - 12 = 3 \, \text{kJ} \]

### Final Answer
Thus, the activation energy for the overall reaction is:
\[ \boxed{3 \, \text{kJ}} \]

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