The reaction `2AX(g)+2B_(2)(g)rarr A_(2)(g)+2B_(2)X(g)` has been studied kinetically and on the baiss of the rate law following mechanism has been proposed.
I. `2A X hArr A_(2)X_(2) " " ("fast and reverse")`
II. `A_(2)X_(2)+B_(2)rarrA_(2)X+B_(2)X`
III. `A_(2)X+B_(2)rarrA_(2)+B_(2)X`
where all the reaction intermediates are gases under ordinary condition.
form the above mechanism in which the steps (elementary) differ conisderably in their rates, the rate law is derived uisng the principle that the slowest step is the rate-determining step (RDS) and the rate of any step varies as the Product of the molar concentrations of each reacting speacting species raised to the power equal to their respective stoichiometric coefficients (law of mass action). If a reacting species is solid or pure liquid, its active mass, i.e., molar concentration is taken to be unity, the standard state. In qrder to find out the final rate law of the reaction, the concentration of any intermediate appearing in the rate law of the RDS is substituted in terms of the concentration of the reactant(s) by means of the law of mass action applied on equilibrium step.
Let the equilibrium constant of Step I be `2xx10^(-3) mol^(-1) L` and the rate constants for the formation of `A_(2)X` and `A_(2)` in Step II and III are `3.0xx10^(-2) mol^(-1) L min^(-1)` and `1xx10^(3) mol^(-1) L min^(-1)` (all data at `25^(@)C)`, then what is the overall rate constant `(mol^(-2) L^(2) min^(-1))` of the consumption of `B_(2)`?

Options

A. `6xx10^(-5)`
B. `1.2xx10^(-4)`
C. `3xx10^(-5)`
D. `1.5xx10^(-5)`

Explanation

To solve the problem, we need to derive the overall rate constant for the consumption of \( B_2 \) based on the given reaction mechanism and the provided equilibrium constant and rate constants. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Rate-Determining Step (RDS)
The slowest step in the given mechanism is the second step:
\[ A_2X_2 + B_2 \rightarrow A_2X + B_2X \]
This step will determine the overall rate of the reaction.

### Step 2: Write the Rate Expression for the RDS
The rate of the reaction for the RDS can be expressed as:
\[ \text{Rate} = k_2 [A_2X_2][B_2] \]
where \( k_2 \) is the rate constant for the second step.

### Step 3: Express the Concentration of the Intermediate \( A_2X_2 \)
Since \( A_2X_2 \) is an intermediate, we need to express its concentration in terms of the reactants using the equilibrium constant from the first step:
\[ 2AX \rightleftharpoons A_2X_2 \]
The equilibrium constant \( K_e \) for this step is given by:
\[ K_e = \frac{[A_2X_2]}{[AX]^2} \]
Given that \( K_e = 2 \times 10^{-3} \, \text{mol}^{-1} \text{L} \), we can rearrange this to find \( [A_2X_2] \):
\[ [A_2X_2] = K_e [AX]^2 \]
Substituting the value of \( K_e \):
\[ [A_2X_2] = (2 \times 10^{-3}) [AX]^2 \]

### Step 4: Substitute \( [A_2X_2] \) into the Rate Expression
Now we substitute \( [A_2X_2] \) back into the rate expression:
\[ \text{Rate} = k_2 (2 \times 10^{-3}) [AX]^2 [B_2] \]

### Step 5: Determine the Overall Rate Constant for \( B_2 \)
Since \( B_2 \) is consumed in the reaction, we need to express the rate in terms of \( [B_2] \):
The overall rate can be expressed as:
\[ \text{Rate} = k_{\text{overall}} [AX]^2 [B_2] \]
Comparing both expressions for the rate, we have:
\[ k_{\text{overall}} = k_2 (2 \times 10^{-3}) \]

### Step 6: Substitute the Value of \( k_2 \)
The rate constant \( k_2 \) for the second step is given as:
\[ k_2 = 3.0 \times 10^{-2} \, \text{mol}^{-1} \text{L min}^{-1} \]
Now substituting this value:
\[ k_{\text{overall}} = (3.0 \times 10^{-2}) (2 \times 10^{-3}) \]
\[ k_{\text{overall}} = 6.0 \times 10^{-5} \, \text{mol}^{-2} \text{L}^2 \text{min}^{-1} \]

### Step 7: Final Calculation
Thus, the overall rate constant for the consumption of \( B_2 \) is:
\[ k_{\text{overall}} = 6.0 \times 10^{-5} \, \text{mol}^{-2} \text{L}^2 \text{min}^{-1} \]