A flask containing `0.5` atm pressure of `A_(2)(g,)`some solid AB added into flask which undergoes dissociation according to `2AB(s)hArrA_(2)(g)+B_(2)(g),K_(p)=0.06atm^(2)`
The total pressure (in atm) at equilibrium is :

Options

A. `0.70`
B. `0.6`
C. `0.10`
D. None of these

Answer

To solve the problem step by step, we will analyze the dissociation of solid AB into gases A₂ and B₂, and calculate the total pressure at equilibrium.

### Step 1: Write the balanced chemical equation
The dissociation reaction is given as:
\[ 2AB(s) \rightleftharpoons A_2(g) + B_2(g) \]

### Step 2: Set up the initial conditions
Initially, we have:
- Pressure of \( A_2 \) = 0.5 atm (given)
- Pressure of \( B_2 \) = 0 atm (since it is not present initially)
- Pressure of \( AB \) = 0 atm (since it is a solid, it does not contribute to pressure)

### Step 3: Define the change in pressure at equilibrium
Let \( x \) be the change in pressure of \( B_2 \) at equilibrium. According to the stoichiometry of the reaction:
- The increase in pressure of \( A_2 \) will be \( x \) (since 1 mole of \( A_2 \) is produced for every mole of \( B_2 \)).
- The pressure of \( A_2 \) at equilibrium will be \( 0.5 + x \).
- The pressure of \( B_2 \) at equilibrium will be \( x \).

### Step 4: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{A_2} \cdot P_{B_2}}{1} = P_{A_2} \cdot P_{B_2} \]
Substituting the pressures at equilibrium:
\[ K_p = (0.5 + x) \cdot x \]
Given that \( K_p = 0.06 \) atm², we can set up the equation:
\[ 0.06 = (0.5 + x) \cdot x \]

### Step 5: Rearranging the equation
Expanding the equation:
\[ 0.06 = 0.5x + x^2 \]
Rearranging gives:
\[ x^2 + 0.5x - 0.06 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 0.5, c = -0.06 \):
- Calculate the discriminant:
\[ D = b^2 - 4ac = (0.5)^2 - 4 \cdot 1 \cdot (-0.06) = 0.25 + 0.24 = 0.49 \]
- Now, substituting into the quadratic formula:
\[ x = \frac{-0.5 \pm \sqrt{0.49}}{2 \cdot 1} = \frac{-0.5 \pm 0.7}{2} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{0.2}{2} = 0.1 \) (valid solution)
2. \( x = \frac{-1.2}{2} = -0.6 \) (not valid as pressure cannot be negative)

### Step 7: Calculate total pressure at equilibrium
Now that we have \( x = 0.1 \):
- The pressure of \( A_2 \) at equilibrium is:
\[ P_{A_2} = 0.5 + 0.1 = 0.6 \, \text{atm} \]
- The pressure of \( B_2 \) at equilibrium is:
\[ P_{B_2} = 0.1 \, \text{atm} \]
- The total pressure \( P_{total} \) at equilibrium is:
\[ P_{total} = P_{A_2} + P_{B_2} = 0.6 + 0.1 = 0.7 \, \text{atm} \]

### Final Answer
The total pressure at equilibrium is **0.7 atm**.

---