For the reaction `2NO_(2)(g)+(1)/(2)O_(2)(g)hArrN_(2)O_(5)(g)` if the equilibrium constant is `K_(p)`, then the equilibrium constant for the reaction `2N_(2)O_(5)(g)hArr4NO_(2)(g)+O_(2)(g)`would be :

Options

A. `K_(P)^(2)`
B. `(2)/(K_(P))`
C. `(1)/(K_(p)^(2))`
D. `(1)/(sqrt(K_(p)))`

Answer

To solve the problem, we need to determine the equilibrium constant for the reaction:

\[ 2N_2O_5(g) \rightleftharpoons 4NO_2(g) + O_2(g) \]

given that the equilibrium constant for the reaction:

\[ 2NO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons N_2O_5(g) \]

is \( K_p \).

### Step-by-Step Solution:

1. **Identify the Given Reaction and Its Equilibrium Constant**:
- The first reaction is:
\[ 2NO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons N_2O_5(g) \]
- Its equilibrium constant is given as \( K_p \).

2. **Reverse the Given Reaction**:
- To find the equilibrium constant for the desired reaction, we first reverse the given reaction:
\[ N_2O_5(g) \rightleftharpoons 2NO_2(g) + \frac{1}{2}O_2(g) \]
- When a reaction is reversed, the equilibrium constant for the reverse reaction is the reciprocal of the original:
\[ K' = \frac{1}{K_p} \]

3. **Multiply the Reversed Reaction by 2**:
- The next step is to multiply the entire reversed reaction by 2 to match the desired reaction:
\[ 2N_2O_5(g) \rightleftharpoons 4NO_2(g) + O_2(g) \]
- When a reaction is multiplied by a factor, the equilibrium constant for the new reaction is raised to the power of that factor:
\[ K'' = (K')^2 = \left(\frac{1}{K_p}\right)^2 = \frac{1}{K_p^2} \]

4. **Final Result**:
- Thus, the equilibrium constant for the reaction:
\[ 2N_2O_5(g) \rightleftharpoons 4NO_2(g) + O_2(g) \]
- is:
\[ K_{eq} = \frac{1}{K_p^2} \]

### Conclusion:
The equilibrium constant for the reaction \( 2N_2O_5(g) \rightleftharpoons 4NO_2(g) + O_2(g) \) is \( \frac{1}{K_p^2} \).

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