The equilibrium constant for the given reaction is 100.
`N_(2)(g)+2O_(2)(g) hArr 2NO_(2)(g)`
What is the equilibrium constant for the reaction ?
`NO_(2)(g) hArr 1//2 N_(2)(g) +O_(2)(g)`

Options

A. 10
B. 1
C. 0.1
D. 0.001

Answer

To find the equilibrium constant for the reaction:

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

we start with the given reaction:

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

and its equilibrium constant \( K_c = 100 \).

### Step 1: Reverse the Reaction
When we reverse a chemical reaction, the equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant.

For the reaction:

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

the equilibrium constant \( K_c' \) becomes:

\[ K_c' = \frac{1}{K_c} = \frac{1}{100} = 0.01 \]

### Step 2: Adjust the Reaction Coefficients
Next, we need to adjust the coefficients of the reversed reaction to match the desired reaction. The desired reaction is:

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

To obtain this reaction from the reversed reaction, we need to divide the entire equation by 2.

### Step 3: Apply the Equilibrium Constant Rule
When we multiply or divide a reaction by a factor, we raise the equilibrium constant to the power of that factor. Since we are dividing the reaction by 2, we take the square root of the equilibrium constant we found in Step 1.

Thus, the new equilibrium constant \( K_c'' \) for the desired reaction is:

\[ K_c'' = (K_c')^{\frac{1}{2}} = (0.01)^{\frac{1}{2}} = 0.1 \]

### Final Answer
The equilibrium constant for the reaction

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

is

\[ K_c = 0.1 \]

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