For the equilibrium reaction `NH_4Cl hArr NH_3(g)+HCl(g), K_P=81 "atm"^2`.The total pressure at equilibrium is 'X' times the pressure of `NH_3`.The value of X will be

Explanation

To solve the problem, we need to analyze the equilibrium reaction and the given equilibrium constant \( K_P \).

### Step-by-Step Solution:

1. **Write the Equilibrium Reaction**:
The reaction is:
\[ NH_4Cl (s) \rightleftharpoons NH_3(g) + HCl(g) \]
Here, \( NH_4Cl \) is a solid and does not contribute to the equilibrium expression.

2. **Write the Expression for \( K_P \)**:
The equilibrium constant \( K_P \) for the reaction is given by:
\[ K_P = \frac{P_{NH_3} \cdot P_{HCl}}{1} = P_{NH_3} \cdot P_{HCl} \]
Since \( NH_4Cl \) is a solid, its activity is considered to be 1.

3. **Given Value of \( K_P \)**:
We are given:
\[ K_P = 81 \, \text{atm}^2 \]
Therefore, we can write:
\[ P_{NH_3} \cdot P_{HCl} = 81 \]

4. **Assume Pressures at Equilibrium**:
Let \( P_{NH_3} = P \). Since \( P_{NH_3} \) and \( P_{HCl} \) are produced in a 1:1 ratio, we have:
\[ P_{HCl} = P \]

5. **Substitute into the \( K_P \) Expression**:
Substituting \( P \) into the \( K_P \) expression:
\[ P \cdot P = 81 \implies P^2 = 81 \implies P = 9 \, \text{atm} \]

6. **Calculate Total Pressure at Equilibrium**:
The total pressure \( P_{total} \) at equilibrium is the sum of the pressures of \( NH_3 \) and \( HCl \):
\[ P_{total} = P_{NH_3} + P_{HCl} = P + P = 9 + 9 = 18 \, \text{atm} \]

7. **Relate Total Pressure to \( P_{NH_3} \)**:
We are asked to find \( X \) such that:
\[ P_{total} = X \cdot P_{NH_3} \]
Substituting the values we found:
\[ 18 = X \cdot 9 \]

8. **Solve for \( X \)**:
Dividing both sides by 9 gives:
\[ X = \frac{18}{9} = 2 \]

### Final Answer:
The value of \( X \) is \( 2 \).

---