`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g), DeltaH^(ɵ)=-22.4 kJ`
The pressure inside the chamber is `100` atm and temperature at `300 K`
If `K_(p)` for the reaction is `1.44xx10^(-5)`, then the value of `K_(p)` for the decomposition of `NH_(3)`
`2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g)`
will be:

Options

A. (a) `sqrt(1.44xx10^(-5))`
B. (b) `(1.44xx10^(-5))^(4)`
C. (c) `(1)/(1.44xx10^(-5))`
D. (d) `1.00xx10^(-3)`

Explanation

To find the value of \( K_p \) for the decomposition of \( NH_3 \) given the reaction:

\[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \]

we can follow these steps:

### Step 1: Identify the Given Information
We know the following:
- The forward reaction is:
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]
- The equilibrium constant \( K_{p1} \) for this reaction is given as:
\[ K_{p1} = 1.44 \times 10^{-5} \]
- We need to find \( K_{p2} \) for the reverse reaction:
\[ 2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g) \]

### Step 2: Understand the Relationship Between \( K_{p1} \) and \( K_{p2} \)
When a reaction is reversed, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. Therefore, we can express this relationship as:

\[ K_{p2} = \frac{1}{K_{p1}} \]

### Step 3: Calculate \( K_{p2} \)
Using the value of \( K_{p1} \):

\[ K_{p2} = \frac{1}{1.44 \times 10^{-5}} \]

Calculating this gives:

\[ K_{p2} = \frac{1}{1.44} \times 10^{5} = 0.6944 \times 10^{5} = 6.944 \times 10^{4} \]

### Step 4: Final Answer
Thus, the value of \( K_{p2} \) for the decomposition of \( NH_3 \) is:

\[ K_{p2} \approx 6.94 \times 10^{4} \]