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 equilibrium reaction \( NH_4Cl \rightleftharpoons NH_3(g) + HCl(g) \) with \( K_P = 81 \, \text{atm}^2 \), we need to find the value of \( X \) such that the total pressure at equilibrium is \( X \) times the pressure of \( NH_3 \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction involves the dissociation of solid \( NH_4Cl \) into gaseous \( NH_3 \) and \( HCl \). Since \( NH_4Cl \) is a solid, it does not contribute to the pressure.

2. **Setting Up Initial Conditions**:
Assume the initial pressure of the system is \( P \). At equilibrium, let the pressure of \( NH_3 \) be \( P_1 \) and the pressure of \( HCl \) also be \( P_1 \) (since they are produced in a 1:1 ratio).

3. **Calculating Total Pressure at Equilibrium**:
The total pressure at equilibrium can be expressed as:
\[ P_{\text{total}} = P_1 + P_1 = 2P_1 \]

4. **Relating Total Pressure to \( X \)**:
We are given that the total pressure at equilibrium is \( X \) times the pressure of \( NH_3 \):
\[ P_{\text{total}} = X \cdot P_1 \]
Substituting the expression for total pressure:
\[ 2P_1 = X \cdot P_1 \]

5. **Solving for \( X \)**:
We can cancel \( P_1 \) from both sides (assuming \( P_1 \neq 0 \)):
\[ 2 = X \]
Therefore, the value of \( X \) is \( 2 \).

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