The equilibrium constant for the reaction
`N_(2)(g)+O_(2)(g) hArr 2NO(g)`
at temperature T is `4xx10^(-4)`.
The value of `K_(c)` for the reaction
`NO(g) hArr 1/2 N_(2)(g)+1/2 O_(2)(g)`
at the same temperature is

Options

A. `4xx10^(-4)`
B. 50
C. `2.5xx10^(2)`
D. `0.02`

Answer

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

\[ \text{NO(g)} \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \]

given that the equilibrium constant for the reaction

\[ \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO(g)} \]

is \( K_c = 4 \times 10^{-4} \), we can follow these steps:

### Step 1: Write the given reaction and its equilibrium constant
The first reaction is:

\[ \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO(g)} \]

with \( K_c = 4 \times 10^{-4} \).

### Step 2: Reverse the reaction
To find the equilibrium constant for the reaction we want, we first reverse the given reaction:

\[ 2 \text{NO(g)} \rightleftharpoons \text{N}_2(g) + \text{O}_2(g) \]

When we reverse a reaction, the equilibrium constant becomes the reciprocal:

\[ K_c' = \frac{1}{K_c} = \frac{1}{4 \times 10^{-4}} = 2.5 \times 10^{3} \]

### Step 3: Divide the reversed reaction by 2
Now, we need to divide the entire reversed reaction by 2 to match the desired reaction:

\[ \text{NO(g)} \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \]

When we divide the reaction by 2, the equilibrium constant is raised to the power of \( \frac{1}{2} \):

\[ K_c'' = (K_c')^{\frac{1}{2}} = (2.5 \times 10^{3})^{\frac{1}{2}} \]

### Step 4: Calculate the square root
Now we calculate the square root:

\[ K_c'' = \sqrt{2.5 \times 10^{3}} = \sqrt{2.5} \times \sqrt{10^{3}} = \sqrt{2.5} \times 10^{1.5} \]

Calculating \( \sqrt{2.5} \) gives approximately \( 1.58 \), and \( \sqrt{10^{3}} = 31.62 \).

Thus,

\[ K_c'' \approx 1.58 \times 31.62 \approx 50 \]

### Final Answer
The value of \( K_c \) for the reaction

\[ \text{NO(g)} \rightleftharpoons \frac{1}{2} \text{N}_2(g) + \frac{1}{2} \text{O}_2(g) \]

is approximately \( 50 \).

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