Consider these reactions and their corresponding `K_s`.
`{:(1/2N_2+O_2toNO_2,K_1),(2NO_2to2NO+O_2,K_2),(NOBrtoNO+1/2Br_2,K_3):}`
Express the K value for the reaction below in terms of `K_1,K_2 and K_3`
`1/2N_2+1/2O_2+1/2Br_2toNOBrK=?`

Options

A. `K_1 K_1 K_3`
B. `(K_1K_2)/K_3`
C. `(K_1K_3^2)/K_2`
D. `(K_2K_3^3)/K_1`

Answer

To express the equilibrium constant \( K \) for the reaction

\[ \frac{1}{2} N_2 + \frac{1}{2} O_2 + \frac{1}{2} Br_2 \rightarrow NOBr \]

in terms of \( K_1, K_2, \) and \( K_3 \), we will manipulate the given reactions step by step.

### Step 1: Write down the given reactions and their equilibrium constants.

1. \(\frac{1}{2} N_2 + O_2 \rightarrow NO_2, \quad K = K_1\)
2. \(2 NO_2 \rightarrow 2 NO + O_2, \quad K = K_2\)
3. \(NOBr \rightarrow NO + \frac{1}{2} Br_2, \quad K = K_3\)

### Step 2: Manipulate the reactions to derive the desired reaction.

**For the first reaction**:
- We will use it as is:
\[ \frac{1}{2} N_2 + O_2 \rightarrow NO_2 \quad (K_1) \]

**For the second reaction**:
- We need to divide the entire reaction by 2 to adjust the stoichiometry:
\[ NO_2 \rightarrow NO + \frac{1}{2} O_2 \quad (K = K_2^{1/2} = \sqrt{K_2}) \]

**For the third reaction**:
- We need to reverse it to have \( NOBr \) on the reactant side:
\[ NO + \frac{1}{2} Br_2 \rightarrow NOBr \quad (K = \frac{1}{K_3}) \]

### Step 3: Combine the manipulated reactions.

Now we can add the reactions together:

1. \(\frac{1}{2} N_2 + O_2 \rightarrow NO_2 \quad (K_1)\)
2. \(NO_2 \rightarrow NO + \frac{1}{2} O_2 \quad (K = \sqrt{K_2})\)
3. \(NO + \frac{1}{2} Br_2 \rightarrow NOBr \quad (K = \frac{1}{K_3})\)

### Step 4: Write the overall reaction.

When we add these reactions, we get:

\[ \frac{1}{2} N_2 + O_2 + NO_2 + \frac{1}{2} Br_2 \rightarrow NO_2 + NO + \frac{1}{2} O_2 + NOBr \]

### Step 5: Cancel out the common terms.

- \( NO_2 \) cancels out on both sides.
- \( O_2 \) cancels out as well.

This results in:

\[ \frac{1}{2} N_2 + \frac{1}{2} Br_2 \rightarrow NOBr \]

### Step 6: Calculate the overall equilibrium constant \( K \).

The overall equilibrium constant \( K \) for the reaction is given by the product of the constants of the individual reactions:

\[ K = K_1 \cdot \sqrt{K_2} \cdot \frac{1}{K_3} \]

Thus, we can express \( K \) as:

\[ K = \frac{K_1 \sqrt{K_2}}{K_3} \]

### Final Answer

\[ K = \frac{K_1 \sqrt{K_2}}{K_3} \]