The reaction ,` 2AB(g) + 2C(g) to A_(2(g)) + 2BC_((g))` proceeds according to the mechanism .
I. `2AB hArr A_(2)B_(2)` (fast)
II. `A_(2)B_(2)+C to A_(2)B + BC ` (slow )
III. `A_(2)B+C to A_(2)+BC` (fast)
what will be the initial rate taking [AB] = 0.2 M and [C] = 0.5 M ? The `K_(c)` for the step I is ` 10^(2) M^(-1)` and rate constant for the step II is ` 3.0 xx 10^(-3) mol^(-1) min^(-1)`

Options

A. `0.0716 M min^(-1)`
B. `0.0891 M min^(-1)`
C. `0.006 M min^(-1)`
D. `0.0257 M min^(-1)`

Answer

To solve the given problem, we will follow these steps:

### Step 1: Identify the rate-determining step
The rate of a reaction mechanism is determined by the slowest step, which is step II in this case. Therefore, we will focus on this step to find the rate of the reaction.

### Step 2: Write the rate law for the slow step
The rate law for step II can be expressed as:
\[ \text{Rate} = k \cdot [A_2B_2] \cdot [C] \]
where \( k \) is the rate constant for step II.

### Step 3: Determine the concentration of \( A_2B_2 \)
Since step I is an equilibrium step, we can use the equilibrium constant \( K_c \) to relate the concentrations of the species involved. The equilibrium constant for step I is given by:
\[ K_c = \frac{[A_2B_2]}{[AB]^2} \]
From this, we can express the concentration of \( A_2B_2 \):
\[ [A_2B_2] = K_c \cdot [AB]^2 \]

### Step 4: Substitute the values into the equation
Now we can substitute the values provided in the question:
- \( K_c = 10^2 \, \text{M}^{-1} \)
- \( [AB] = 0.2 \, \text{M} \)
- \( [C] = 0.5 \, \text{M} \)
- \( k = 3.0 \times 10^{-3} \, \text{mol}^{-1} \, \text{min}^{-1} \)

Calculating \( [A_2B_2] \):
\[ [A_2B_2] = 10^2 \cdot (0.2)^2 = 100 \cdot 0.04 = 4 \, \text{M} \]

### Step 5: Calculate the rate of the reaction
Now we can substitute \( [A_2B_2] \) into the rate law:
\[ \text{Rate} = k \cdot [A_2B_2] \cdot [C] \]
\[ \text{Rate} = (3.0 \times 10^{-3}) \cdot (4) \cdot (0.5) \]
\[ \text{Rate} = (3.0 \times 10^{-3}) \cdot 2 = 6.0 \times 10^{-3} \, \text{mol} \, \text{min}^{-1} \]

### Final Answer
The initial rate of the reaction is:
\[ \text{Rate} = 0.006 \, \text{mol} \, \text{min}^{-1} \]