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_2/2-K_3`
B. `K_1+(K_2)^(1//2)-K_3`
C. `(K_1K_2)/(2K_3)`
D. `K_1(K_2)^(1//2)//K_3`

Answer

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

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

in terms of \( K_1, K_2, \) and \( K_3 \), we will manipulate the given reactions and their equilibrium constants.

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

1. **Reaction 1**:
\[ \frac{1}{2} N_2 + O_2 \rightleftharpoons NO_2 \quad (K_1) \]

2. **Reaction 2**:
\[ 2 NO_2 \rightleftharpoons 2 NO + O_2 \quad (K_2) \]

3. **Reaction 3**:
\[ NOBr \rightleftharpoons NO + \frac{1}{2} Br_2 \quad (K_3) \]

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

To obtain the desired reaction, we will:

- Use **Reaction 1** as it is.
- Use **Reaction 2** but divide the entire reaction by 2.
- Reverse **Reaction 3**.

### Step 3: Adjust Reaction 2

Dividing Reaction 2 by 2 gives:

\[ NO_2 \rightleftharpoons NO + \frac{1}{2} O_2 \quad (K_2^{1/2} = \sqrt{K_2}) \]

### Step 4: Reverse Reaction 3

Reversing Reaction 3 gives:

\[ NO + \frac{1}{2} Br_2 \rightleftharpoons NOBr \quad \left(\frac{1}{K_3}\right) \]

### Step 5: Combine the reactions

Now we can combine the adjusted reactions:

1. From Reaction 1:
\[ \frac{1}{2} N_2 + O_2 \rightleftharpoons NO_2 \quad (K_1) \]

2. From the adjusted Reaction 2:
\[ NO_2 \rightleftharpoons NO + \frac{1}{2} O_2 \quad (\sqrt{K_2}) \]

3. From the reversed Reaction 3:
\[ NO + \frac{1}{2} Br_2 \rightleftharpoons NOBr \quad \left(\frac{1}{K_3}\right) \]

Adding these reactions together:

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

This simplifies to:

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

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

The overall equilibrium constant \( K \) can be expressed as:

\[ K = K_1 \cdot K_2^{1/2} \cdot \frac{1}{K_3} \]

This simplifies to:

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

### Final Answer

Thus, the equilibrium constant \( K \) for the reaction

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

is given by:

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