The equilibrium constant for the reaction `NH_(4)NO_(2)(s)hArr N_(2)(g)+2H_(2)O(g)`, is given by

Options

A. `([NH_(4)NO_(2)])/([N_(2)][H_(2)O]^(2))`
B. `[N_(2)][H_(2)O]^(2)`
C. `([N_(2)][H_(2)O])/([NH_(4)NO_(2)])`
D. `([N_(2)][2H_(2)O]^(2))/([NH_(4)NO_(2)]^(2))`

Explanation

To determine the equilibrium constant \( K_c \) for the reaction

\[ NH_4NO_2(s) \rightleftharpoons N_2(g) + 2H_2O(g), \]

we will follow these steps:

### Step 1: Identify the products and reactants
In the given reaction, the reactant is \( NH_4NO_2(s) \) and the products are \( N_2(g) \) and \( H_2O(g) \).

### Step 2: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) is defined as the ratio of the concentrations of the products raised to the power of their coefficients to the concentrations of the reactants raised to the power of their coefficients.

For the reaction:

\[ K_c = \frac{[N_2][H_2O]^2}{[NH_4NO_2]} \]

### Step 3: Consider the states of the substances
In this reaction, \( NH_4NO_2 \) is a solid. The concentration of pure solids and pure liquids is not included in the equilibrium expression. Therefore, we treat the concentration of \( NH_4NO_2 \) as 1.

### Step 4: Simplify the expression
Since the concentration of the solid \( NH_4NO_2 \) is considered to be 1, the expression for \( K_c \) simplifies to:

\[ K_c = [N_2][H_2O]^2 \]

### Step 5: Final expression for the equilibrium constant
Thus, the final expression for the equilibrium constant \( K_c \) for the reaction is:

\[ K_c = [N_2][H_2O]^2 \]

### Conclusion
The equilibrium constant for the reaction \( NH_4NO_2(s) \rightleftharpoons N_2(g) + 2H_2O(g) \) is given by:

\[ K_c = [N_2][H_2O]^2 \]

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