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 Reaction and Initial Conditions
The dissociation reaction is given as:
\[ 2AB(s) \rightleftharpoons A_2(g) + B_2(g) \]

Initially, we have:
- Pressure of \( A_2 \) = 0.5 atm
- Pressure of \( B_2 \) = 0 atm (since it is not present initially)
- Solid \( AB \) does not exert pressure.

### Step 2: Define Changes at Equilibrium
Let \( P \) be the change in pressure of \( B_2 \) at equilibrium. According to the stoichiometry of the reaction:
- For every 2 moles of \( AB \) that dissociate, 1 mole of \( A_2 \) and 1 mole of \( B_2 \) are produced.

Thus, at equilibrium:
- Pressure of \( A_2 \) = \( 0.5 + P \)
- Pressure of \( B_2 \) = \( P \)

### Step 3: 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:
\[ K_p = (0.5 + P) \cdot P \]

Given that \( K_p = 0.06 \) atm², we can set up the equation:
\[ (0.5 + P) \cdot P = 0.06 \]

### Step 4: Expand and Rearrange the Equation
Expanding the equation gives:
\[ 0.5P + P^2 = 0.06 \]
Rearranging it leads to:
\[ P^2 + 0.5P - 0.06 = 0 \]

### Step 5: Solve the Quadratic Equation
We can solve this quadratic equation using the quadratic formula:
\[ P = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 0.5, c = -0.06 \).

Calculating the discriminant:
\[ b^2 - 4ac = (0.5)^2 - 4(1)(-0.06) = 0.25 + 0.24 = 0.49 \]

Now substituting into the quadratic formula:
\[ P = \frac{-0.5 \pm \sqrt{0.49}}{2} = \frac{-0.5 \pm 0.7}{2} \]

Calculating the two possible values for \( P \):
1. \( P = \frac{0.2}{2} = 0.1 \) (valid)
2. \( P = \frac{-1.2}{2} = -0.6 \) (not valid since pressure cannot be negative)

Thus, \( P = 0.1 \) atm.

### Step 6: Calculate Total Pressure at Equilibrium
Now we can find the total pressure at equilibrium:
\[ P_{total} = P_{A_2} + P_{B_2} = (0.5 + P) + P = (0.5 + 0.1) + 0.1 = 0.5 + 0.2 = 0.7 \text{ atm} \]

### Final Answer
The total pressure at equilibrium is:
\[ \text{Total Pressure} = 0.7 \text{ atm} \]

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