The equilibrium constant for the reaction
`A_(2)(g)+B_(2)(g) hArr 2AB(g)`
is `20` at `500K`. The equilibrium constant for the reaction `2AB(g) hArr A_(2)(g)+B_(2)(g)` would be

Options

A. `20`
B. `0.5`
C. `0.05`
D. `10`

Explanation

To solve the problem, we need to determine the equilibrium constant for the reverse reaction given the equilibrium constant for the forward reaction.

### Step-by-Step Solution:

1. **Identify the Forward Reaction and its Equilibrium Constant**:
The forward reaction is:
\[ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \]
The equilibrium constant \( K_f \) for this reaction is given as \( 20 \) at \( 500 \, K \).

2. **Write the Reverse Reaction**:
The reverse reaction is:
\[ 2AB(g) \rightleftharpoons A_2(g) + B_2(g) \]
We need to find the equilibrium constant \( K_r \) for this reaction.

3. **Relate the Equilibrium Constants**:
The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. This is a fundamental property of equilibrium constants:
\[ K_r = \frac{1}{K_f} \]

4. **Calculate the Equilibrium Constant for the Reverse Reaction**:
Substituting the value of \( K_f \):
\[ K_r = \frac{1}{20} = 0.05 \]

5. **Conclusion**:
The equilibrium constant for the reaction \( 2AB(g) \rightleftharpoons A_2(g) + B_2(g) \) is \( 0.05 \).

### Final Answer:
The equilibrium constant for the reaction \( 2AB(g) \rightleftharpoons A_2(g) + B_2(g) \) is \( 0.05 \).

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