For a gaseous reaction `2A+B_(2)to2AB,` the following rate data were obtained at 300 K
`{:("Rate of disappearance of "B_(2),,," ""Concentration",,,),("(""mol lit"^(-1)"min"^(-1)")",,,[A],,,[B_(2)]),((i)1.8xx10^(-3),,,0.015,,,0.15),((ii)1.08xx10^(-2),,,0.09,,,0.15),((iii)5.4xx10^(-3),,,0.015,,,0.45):}`
Calculate the rate constant for the reaction and rate of formation of AB when [A] is 0.02 and `[B_(2)]` is 0.04 "mol lit"^(-1)" at 300 K".`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the rate law expression
The rate law for the reaction \(2A + B_2 \rightarrow 2AB\) can be expressed as:
\[ \text{Rate} = k [A]^{\alpha} [B_2]^{\beta} \]
where \(k\) is the rate constant, and \(\alpha\) and \(\beta\) are the orders of the reaction with respect to \(A\) and \(B_2\), respectively.

### Step 2: Determine the orders of reaction (\(\alpha\) and \(\beta\))
We will use the provided rate data to find the values of \(\alpha\) and \(\beta\).

**Using Experiments (i) and (iii):**
\[ \frac{R_1}{R_3} = \frac{k [A]_1^{\alpha} [B_2]_1^{\beta}}{k [A]_3^{\alpha} [B_2]_3^{\beta}} \]
Substituting the values:
\[ \frac{1.8 \times 10^{-3}}{5.4 \times 10^{-3}} = \frac{0.015^{\alpha} \cdot 0.15^{\beta}}{0.015^{\alpha} \cdot 0.45^{\beta}} \]
This simplifies to:
\[ \frac{1.8}{5.4} = \frac{0.15^{\beta}}{0.45^{\beta}} \]
\[ \frac{1}{3} = \left(\frac{0.15}{0.45}\right)^{\beta} \]
\[ \frac{1}{3} = \left(\frac{1}{3}\right)^{\beta} \]
Thus, \(\beta = 1\).

**Using Experiments (i) and (ii):**
\[ \frac{R_1}{R_2} = \frac{k [A]_1^{\alpha} [B_2]_1^{\beta}}{k [A]_2^{\alpha} [B_2]_2^{\beta}} \]
Substituting the values:
\[ \frac{1.8 \times 10^{-3}}{1.08 \times 10^{-2}} = \frac{0.015^{\alpha} \cdot 0.15^{\beta}}{0.09^{\alpha} \cdot 0.15^{\beta}} \]
This simplifies to:
\[ \frac{1.8}{10.8} = \frac{0.015^{\alpha}}{0.09^{\alpha}} \]
\[ \frac{1}{6} = \left(\frac{0.015}{0.09}\right)^{\alpha} \]
\[ \frac{1}{6} = \left(\frac{1}{6}\right)^{\alpha} \]
Thus, \(\alpha = 1\).

### Step 3: Write the complete rate law
Now we can express the rate law as:
\[ \text{Rate} = k [A][B_2] \]

### Step 4: Calculate the rate constant \(k\)
Using the first experiment data:
\[ 1.8 \times 10^{-3} = k (0.015)(0.15) \]
Solving for \(k\):
\[ k = \frac{1.8 \times 10^{-3}}{0.015 \times 0.15} \]
\[ k = \frac{1.8 \times 10^{-3}}{0.00225} = 0.8 \, \text{mol}^{-1} \text{lit} \text{min}^{-1} \]

### Step 5: Calculate the rate of formation of \(AB\)
Using the rate law:
\[ \text{Rate} = k [A][B_2] \]
Substituting \(k\), \([A] = 0.02\), and \([B_2] = 0.04\):
\[ \text{Rate} = 0.8 \times 0.02 \times 0.04 \]
\[ \text{Rate} = 0.8 \times 0.0008 = 0.00064 \, \text{mol lit}^{-1} \text{min}^{-1} \]
Since the formation of \(AB\) is twice the rate of disappearance of \(B_2\):
\[ \text{Rate of formation of } AB = 2 \times 0.00064 = 0.00128 \, \text{mol lit}^{-1} \text{min}^{-1} \]

### Final Answers:
- Rate constant \(k = 0.8 \, \text{mol}^{-1} \text{lit} \text{min}^{-1}\)
- Rate of formation of \(AB = 0.00128 \, \text{mol lit}^{-1} \text{min}^{-1}\)