For the reaction, `2NO_(2)(g) rarr 2NO(g) + O_(2)(g), K_(c) = 1.8 xx 10^(-6) "at" 185^(@)C`, the value of `K_(c)` for the reaction `NO(g) + 1/2O_(2)(g) rarr NO_(2)(g)` is

Options

A. `0.9 × 10^(6)`
B. `1.9 × 10^(6)`
C. `7.5 × 10^(2)`
D. `5.7 × 10^(2)`

Explanation

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

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

given the equilibrium constant \( K_c \) for the reaction

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

is \( 1.8 \times 10^{-6} \) at \( 185^\circ C \), we can follow these steps:

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

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

with

\[ K_c = 1.8 \times 10^{-6} \]

### Step 2: Identify the relationship between the two reactions.
We need to find \( K_c \) for the reaction:

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

This reaction is the reverse of the original reaction, and it is also half the stoichiometry of the original reaction.

### Step 3: Apply the equilibrium constant rules.
When a reaction is reversed, the equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant. Therefore, for the reaction:

\[ K_c' = \frac{1}{K_c} \]

### Step 4: Adjust for the change in stoichiometry.
Since the new reaction is half of the original reaction, we take the square root of the equilibrium constant obtained from the previous step:

\[ K_c' = \sqrt{\frac{1}{K_c}} = \frac{1}{\sqrt{K_c}} \]

### Step 5: Substitute the value of \( K_c \).
Now, substituting the value of \( K_c \):

\[ K_c' = \frac{1}{\sqrt{1.8 \times 10^{-6}}} \]

### Step 6: Calculate the value.
Calculating \( \sqrt{1.8 \times 10^{-6}} \):

\[ \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \]

Thus,

\[ K_c' = \frac{1}{1.34 \times 10^{-3}} \approx 7.46 \times 10^{2} \]

### Step 7: Round to significant figures.
Rounding gives us:

\[ K_c' \approx 7.5 \times 10^{2} \]

### Final Answer:
The value of \( K_c \) for the reaction \( \text{NO}(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{NO}_2(g) \) is approximately

\[ \boxed{7.5 \times 10^{2}} \]