For the reaction ` N_2(g) +O_2( g) hArr 2NO(g), ` the equilibrium constant is K1 and for reaction `2NO(g) +O_2(g) hArr 2NO_2(g) .` equilibrium constant is K2 What is K for the reaction `NO_2(g) hArr (1)/(2) N_2(g) + O_2(g)`

Options

A. ` 1//4( 4K_1K_2)`
B. ` (1//K_1K_2) ^(1//2)`
C. ` 1//(K_1K_2)`
D. ` 1//(2K_1K_2)`

Answer

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

\[ NO_2(g) \rightleftharpoons \frac{1}{2} N_2(g) + O_2(g), \]

we start with the given reactions and their equilibrium constants.

1. **Write the equilibrium expressions for the given reactions:**

For the reaction
\[ N_2(g) + O_2(g) \rightleftharpoons 2NO(g), \]
the equilibrium constant \( K_1 \) is given by:
\[ K_1 = \frac{[NO]^2}{[N_2][O_2]}. \]

For the reaction
\[ 2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g), \]
the equilibrium constant \( K_2 \) is given by:
\[ K_2 = \frac{[NO_2]^2}{[NO]^2[O_2]}. \]

2. **Express the equilibrium constant \( K \) for the desired reaction:**

For the reaction
\[ NO_2(g) \rightleftharpoons \frac{1}{2} N_2(g) + O_2(g), \]
the equilibrium constant \( K \) can be expressed as:
\[ K = \frac{[N_2]^{1/2}[O_2]}{[NO_2]}. \]

3. **Relate \( K \) to \( K_1 \) and \( K_2 \):**

We can express \( K \) in terms of \( K_1 \) and \( K_2 \). We assume:
\[ K = K_1^a \cdot K_2^b. \]

To find \( a \) and \( b \), we will compare the exponents of the species in the equilibrium expressions.

4. **Set up the equations based on the stoichiometry:**

- For \( O_2 \):
\[ a + b - 1 = 1 \quad \text{(from K)} \]
- For \( NO \):
\[ 2a - b = 0 \quad \text{(from K)} \]

5. **Solve the equations:**

From the first equation:
\[ a + b = 2. \]

From the second equation:
\[ 2a = b \implies b = 2a. \]

Substituting \( b = 2a \) into \( a + b = 2 \):
\[ a + 2a = 2 \implies 3a = 2 \implies a = \frac{2}{3}, \quad b = \frac{4}{3}. \]

6. **Substitute back to find \( K \):**

Now substituting \( a \) and \( b \) back into the expression for \( K \):
\[ K = K_1^{\frac{2}{3}} \cdot K_2^{\frac{4}{3}}. \]

7. **Final expression:**

Rearranging gives:
\[ K = \frac{K_2^{\frac{4}{3}}}{K_1^{\frac{2}{3}}}. \]

**Final Result:**
\[ K = \frac{K_2^{\frac{4}{3}}}{K_1^{\frac{2}{3}}}. \]