`A(s)hArrB(g)+C(g) K_(p_(1))=36 atm^(2)`
`E(s)hArrB(g)+D(g) K_(p_(2))=64atm^(2)`
Both solids `A "&" E` were taken in a container of constant volume at a give temperature. Total pressure in the container after equilibrium is

Options

A. `6` atm
B. `5` atn
C. `10` atm
D. `20` atm

Answer

To find the total pressure in the container after equilibrium is reached, we can follow these steps:

### Step 1: Write the equilibrium expressions for both reactions.

For the first reaction:
\[ A(s) \rightleftharpoons B(g) + C(g) \]
The equilibrium constant \( K_{p1} \) is given as:
\[ K_{p1} = P_B \cdot P_C = 36 \, \text{atm}^2 \]

For the second reaction:
\[ E(s) \rightleftharpoons B(g) + D(g) \]
The equilibrium constant \( K_{p2} \) is given as:
\[ K_{p2} = P_B \cdot P_D = 64 \, \text{atm}^2 \]

### Step 2: Define the pressures of the gases.

Let:
- \( P_B = P_1 + P_2 \) (where \( P_1 \) is the partial pressure of \( B \) from the first reaction and \( P_2 \) is the partial pressure of \( B \) from the second reaction).

### Step 3: Substitute \( P_B \) in the equilibrium expressions.

From the first reaction:
\[ K_{p1} = P_B \cdot P_C = (P_1 + P_2) \cdot P_1 = 36 \]

From the second reaction:
\[ K_{p2} = P_B \cdot P_D = (P_1 + P_2) \cdot P_2 = 64 \]

### Step 4: Rearrange the equations.

From the first equation:
\[ P_B \cdot P_C = 36 \]
\[ P_C = \frac{36}{P_B} \]

From the second equation:
\[ P_B \cdot P_D = 64 \]
\[ P_D = \frac{64}{P_B} \]

### Step 5: Add the pressures.

The total pressure \( P_{total} \) in the container is:
\[ P_{total} = P_B + P_C + P_D \]
Substituting \( P_C \) and \( P_D \):
\[ P_{total} = P_B + \frac{36}{P_B} + \frac{64}{P_B} \]
\[ P_{total} = P_B + \frac{100}{P_B} \]

### Step 6: Set up the equation using the values of \( K_{p1} \) and \( K_{p2} \).

We know:
\[ K_{p1} + K_{p2} = 100 \]
Thus:
\[ P_B^2 = 100 \]
Taking the square root:
\[ P_B = 10 \]

### Step 7: Calculate the total pressure.

Substituting \( P_B \) back into the total pressure equation:
\[ P_{total} = 10 + \frac{100}{10} \]
\[ P_{total} = 10 + 10 = 20 \, \text{atm} \]

### Final Answer:
The total pressure in the container after equilibrium is:
\[ \boxed{20 \, \text{atm}} \]

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