Decomposition of ammonium chloride is an endothermic reaction. The equilibrium may be represented as:
`NH_(4)Cl(s) hArr NH_(3)(g)+HCl(g)`
A `6.250 g` sample of `NH_(4)Cl` os placed in an evaculated `4.0 L` container at `27^(@)C`. After equilibrium the total pressure inside the container is `0.820` bar and some solid remains in the container. Answer the followings
The value of `K_(p)` for the reaction at `300 K` is

Options

A. `16.2`
B. `0.168`
C. `1.68`
D. `32.4`

Explanation

To find the value of \( K_p \) for the decomposition of ammonium chloride (\( NH_4Cl \)), we can follow these steps:

### Step 1: Write the equilibrium reaction
The decomposition of ammonium chloride can be represented as:
\[ NH_4Cl(s) \rightleftharpoons NH_3(g) + HCl(g) \]

### Step 2: Understand the equilibrium constant expression
For the reaction, the equilibrium constant \( K_p \) is given by:
\[ K_p = \frac{P_{NH_3} \cdot P_{HCl}}{P_{NH_4Cl}} \]
Since \( NH_4Cl \) is a solid, its activity is considered to be 1, so we can simplify the expression to:
\[ K_p = P_{NH_3} \cdot P_{HCl} \]

### Step 3: Determine the total pressure and partial pressures
Given that the total pressure at equilibrium is \( 0.820 \, \text{bar} \) and that the reaction produces equal moles of \( NH_3 \) and \( HCl \), we can denote the partial pressures of \( NH_3 \) and \( HCl \) as \( P \):
\[ P_{NH_3} = P \quad \text{and} \quad P_{HCl} = P \]
Thus, the total pressure can be expressed as:
\[ P_{total} = P_{NH_3} + P_{HCl} = P + P = 2P \]
Setting this equal to the total pressure:
\[ 2P = 0.820 \, \text{bar} \]

### Step 4: Solve for \( P \)
To find \( P \):
\[ P = \frac{0.820 \, \text{bar}}{2} = 0.410 \, \text{bar} \]

### Step 5: Calculate \( K_p \)
Now we can substitute the values of the partial pressures into the expression for \( K_p \):
\[ K_p = P_{NH_3} \cdot P_{HCl} = P \cdot P = 0.410 \, \text{bar} \cdot 0.410 \, \text{bar} \]
Calculating this gives:
\[ K_p = 0.410^2 = 0.1681 \, \text{bar}^2 \]

### Step 6: Final answer
Thus, the value of \( K_p \) at \( 300 \, K \) is approximately:
\[ K_p \approx 0.168 \]

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