The value of equilibrium constant for the reaction `[N_2O_5 (g) hArr 2NO_2(g) + 1/2 O_2(g)]` is 0.5. The equilibruim constant for the reaction `[4NO_2(g) + O_2(g) hArr 2N_2O_5(g)]` is

Options

A. 0.4
B. 0.5
C. 2
D. 4

Answer

To solve the problem, we need to find the equilibrium constant for the reaction:

\[ 4NO_2(g) + O_2(g) \rightleftharpoons 2N_2O_5(g) \]

given that the equilibrium constant for the reaction:

\[ N_2O_5(g) \rightleftharpoons 2NO_2(g) + \frac{1}{2} O_2(g) \]

is \( K_1 = 0.5 \).

### Step-by-Step Solution:

1. **Write the expression for the first equilibrium constant (K1)**:
The equilibrium constant \( K_1 \) for the first reaction can be expressed as:
\[ K_1 = \frac{[NO_2]^2 \cdot [O_2]^{1/2}}{[N_2O_5]} \]
Given that \( K_1 = 0.5 \).

2. **Reverse the first reaction**:
To find \( K_2 \), we need to reverse the first reaction. When we reverse a reaction, the equilibrium constant becomes the reciprocal:
\[ N_2O_5(g) \leftarrow 2NO_2(g) + \frac{1}{2} O_2(g) \]
The equilibrium constant for the reverse reaction \( K_{rev} \) is:
\[ K_{rev} = \frac{1}{K_1} = \frac{1}{0.5} = 2 \]

3. **Square the reversed reaction**:
The next step is to square the reversed reaction to match the stoichiometry of the second reaction:
\[ 2N_2O_5(g) \rightleftharpoons 4NO_2(g) + O_2(g) \]
When we square the equilibrium constant, we square the value we found:
\[ K_2 = (K_{rev})^2 = (2)^2 = 4 \]

4. **Conclusion**:
Therefore, the equilibrium constant \( K_2 \) for the reaction \( 4NO_2(g) + O_2(g) \rightleftharpoons 2N_2O_5(g) \) is:
\[ K_2 = 4 \]

### Final Answer:
The equilibrium constant for the reaction \( 4NO_2(g) + O_2(g) \rightleftharpoons 2N_2O_5(g) \) is **4**.

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