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 given reaction and its equilibrium constant**:
The forward reaction is:
\[ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \]
The equilibrium constant for this reaction, denoted as \( K_1 \), is given as \( 20 \) at \( 500K \).

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 for this reaction, which we will denote as \( K_2 \).

3. **Relate the equilibrium constants**:
The equilibrium constant for a reaction is related to the equilibrium constant of its reverse reaction. Specifically, if you reverse a reaction, the equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction:
\[ K_2 = \frac{1}{K_1} \]

4. **Substitute the known value of \( K_1 \)**:
Since \( K_1 = 20 \):
\[ K_2 = \frac{1}{20} \]

5. **Calculate \( K_2 \)**:
Performing the calculation gives:
\[ K_2 = 0.05 \]

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