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.
How many times the rate of formation of `A_(2)` will increase if concentrations of `AX` is doubled and that of `B_(2)` is increased theee fold?

Options

A. `36`
B. `12`
C. `6`
D. `8`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the Rate Law for the Rate Determining Step (RDS)
The rate law for the second step (which is the slowest step and thus 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 2: Substitute the Equilibrium Expression from Step 1
From the first step, which is an equilibrium reaction:
\[ 2AX \rightleftharpoons A_2X_2 \]
The equilibrium constant \( K_{eq} \) for this step can be expressed as:
\[ K_{eq} = \frac{[A_2X_2]}{[AX]^2} \]
Thus, we can express \( [A_2X_2] \) in terms of \( [AX] \):
\[ [A_2X_2] = K_{eq} [AX]^2 \]

### Step 3: Substitute into the Rate Law
Now, substituting \( [A_2X_2] \) into the rate law:
\[ \text{Rate} = k_2 \cdot K_{eq} [AX]^2 [B_2] \]
Let’s denote \( k = k_2 \cdot K_{eq} \) for simplicity:
\[ \text{Rate} = k [AX]^2 [B_2] \]

### Step 4: Calculate the New Rate with Changed Concentrations
If the concentration of \( AX \) is doubled and the concentration of \( B_2 \) is increased threefold, we have:
- New concentration of \( AX = 2[AX] \)
- New concentration of \( B_2 = 3[B_2] \)

Substituting these into the rate equation:
\[ \text{New Rate} = k (2[AX])^2 (3[B_2]) \]
\[ = k \cdot 4[AX]^2 \cdot 3[B_2] \]
\[ = 12k [AX]^2 [B_2] \]

### Step 5: Relate New Rate to Original Rate
The original rate was:
\[ \text{Original Rate} = k [AX]^2 [B_2] \]
Thus, the new rate can be expressed as:
\[ \text{New Rate} = 12 \times \text{Original Rate} \]

### Conclusion
The rate of formation of \( A_2 \) will increase by a factor of 12 when the concentration of \( AX \) is doubled and the concentration of \( B_2 \) is increased threefold.

### Final Answer
The rate of formation of \( A_2 \) will increase by **12 times**.

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