`A_(2)B_((g))` is introduced in a vessel at `1000K`. If partial pressure of `A_(2)B(g) "is" 1` atm initially and `K_(P)` for reaction `A_(2)B_((g))hArr2A(g)+B(g) "is" 81xx10^(-6)` then calculate percentage of dissociation of `A_(2)B`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium reaction
The equilibrium reaction is given as:
\[ A_2B_{(g)} \rightleftharpoons 2A_{(g)} + B_{(g)} \]

### Step 2: Set up the initial conditions
Initially, the partial pressure of \( A_2B \) is given as 1 atm. Therefore, we can summarize the initial conditions as:
- \( P_{A_2B} = 1 \, \text{atm} \)
- \( P_A = 0 \, \text{atm} \)
- \( P_B = 0 \, \text{atm} \)

### Step 3: Define the change in pressure due to dissociation
Let \( \alpha \) be the degree of dissociation of \( A_2B \). At equilibrium, the changes in pressure can be expressed as:
- The pressure of \( A_2B \) will decrease by \( \alpha \) atm, so:
\[ P_{A_2B} = 1 - \alpha \]
- The pressure of \( A \) will increase by \( 2\alpha \) atm (since 2 moles of \( A \) are produced for every mole of \( A_2B \)):
\[ P_A = 2\alpha \]
- The pressure of \( B \) will increase by \( \alpha \) atm:
\[ P_B = \alpha \]

### Step 4: Write the expression for \( K_P \)
The equilibrium constant \( K_P \) for the reaction is given by:
\[ K_P = \frac{(P_A)^2 \cdot (P_B)}{P_{A_2B}} \]
Substituting the expressions from Step 3 into this equation gives:
\[ K_P = \frac{(2\alpha)^2 \cdot (\alpha)}{(1 - \alpha)} \]
\[ K_P = \frac{4\alpha^2 \cdot \alpha}{(1 - \alpha)} \]
\[ K_P = \frac{4\alpha^3}{(1 - \alpha)} \]

### Step 5: Substitute the known value of \( K_P \)
We know that \( K_P = 81 \times 10^{-6} \). Therefore, we can set up the equation:
\[ \frac{4\alpha^3}{(1 - \alpha)} = 81 \times 10^{-6} \]

### Step 6: Make an assumption for small \( \alpha \)
Assuming \( \alpha \) is small compared to 1, we can approximate \( (1 - \alpha) \approx 1 \):
\[ 4\alpha^3 \approx 81 \times 10^{-6} \]
\[ \alpha^3 \approx \frac{81 \times 10^{-6}}{4} \]
\[ \alpha^3 \approx 20.25 \times 10^{-6} \]

### Step 7: Solve for \( \alpha \)
Taking the cube root of both sides:
\[ \alpha \approx \sqrt[3]{20.25 \times 10^{-6}} \]
Calculating this gives:
\[ \alpha \approx 0.027 \]

### Step 8: Calculate the percentage of dissociation
To find the percentage of dissociation, we multiply \( \alpha \) by 100:
\[ \text{Percentage of dissociation} = \alpha \times 100 \approx 0.027 \times 100 = 2.7\% \]

### Final Answer
The percentage of dissociation of \( A_2B \) is approximately **2.7%**.

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