`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 given reaction is `1.44xx10^(-5)`, then the value of `K_(c)` will be:

Options

A. `(1.44xx10^(-5))/((0.082xx500)^(-2)) mol L^(-1)`
B. `(1.44xx10^(-5))/((8.314xx200)^(-2)) mol L^(-1)`
C. `(1.44xx10^(-5))/((0.082xx700)^(2)) mol L^(-1)`
D. `(1.44xx10^(-5))/((0.082xx300)^(-2)) mol L^(-1)`

Explanation

To find the value of \( K_c \) from the given \( K_p \) for the reaction:

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

we will use the relationship between \( K_p \) and \( K_c \):

\[ K_p = K_c \cdot R T^{\Delta n_g} \]

### Step 1: Calculate \( \Delta n_g \)

First, we need to determine \( \Delta n_g \), which is the change in the number of moles of gas during the reaction.

- Moles of products: \( 2 \) (from \( 2NH_3 \))
- Moles of reactants: \( 1 + 3 = 4 \) (from \( N_2 + 3H_2 \))

Now, we calculate \( \Delta n_g \):

\[ \Delta n_g = \text{Moles of products} - \text{Moles of reactants} = 2 - 4 = -2 \]

### Step 2: Use the given values

Given:
- \( K_p = 1.44 \times 10^{-5} \)
- \( R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 300 \, \text{K} \)

### Step 3: Substitute values into the equation

Now we substitute the values into the equation:

\[ K_p = K_c \cdot R T^{\Delta n_g} \]

Rearranging gives us:

\[ K_c = \frac{K_p}{R T^{\Delta n_g}} \]

Substituting the known values:

\[ K_c = \frac{1.44 \times 10^{-5}}{0.082 \times 300^{-2}} \]

### Step 4: Calculate \( K_c \)

Calculating \( T^{\Delta n_g} \):

\[ T^{-2} = (300)^{-2} = \frac{1}{90000} \approx 1.11 \times 10^{-5} \]

Now substituting this back into the equation:

\[ K_c = \frac{1.44 \times 10^{-5}}{0.082 \times 1.11 \times 10^{-5}} \]

Calculating the denominator:

\[ 0.082 \times 1.11 \times 10^{-5} \approx 9.1 \times 10^{-7} \]

Now calculating \( K_c \):

\[ K_c \approx \frac{1.44 \times 10^{-5}}{9.1 \times 10^{-7}} \approx 15.8 \]

### Final Answer

Thus, the value of \( K_c \) is approximately \( 15.8 \).

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