If equilibrium constant for the reactions.
` A_2 +B_2 hArr 2AB,`
is K , then the backward reactions.
` AB hArr (1)/(2) A_2 + (1)/(2) B_2.`
its value is 1/K . Is it true or false? If false then write the correct constant.

Answer

To solve the problem, we need to analyze the equilibrium constants for the given reactions step by step.

### Step 1: Identify the Forward Reaction and its Equilibrium Constant
The forward reaction is given as:
\[ A_2 + B_2 \rightleftharpoons 2AB \]
The equilibrium constant \( K \) for this reaction can be expressed as:
\[ K = \frac{[AB]^2}{[A_2][B_2]} \]

### Step 2: Write the Backward Reaction
The backward reaction is:
\[ 2AB \rightleftharpoons A_2 + B_2 \]
For this reaction, the equilibrium constant \( K' \) can be expressed as:
\[ K' = \frac{[A_2][B_2]}{[AB]^2} \]

### Step 3: Relate the Backward Reaction to the Forward Reaction
From the relationship of equilibrium constants for forward and backward reactions, we know:
\[ K' = \frac{1}{K} \]

### Step 4: Consider the Reaction with Coefficients
Now, we need to consider the reaction:
\[ AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2 \]
This reaction is obtained by dividing the backward reaction by 2. When we divide the coefficients of a balanced equation by a factor, the equilibrium constant changes according to the power of that factor.

### Step 5: Calculate the New Equilibrium Constant
When we divide the entire reaction by 2, the new equilibrium constant \( K'' \) is given by:
\[ K'' = K'^{1/2} \]
Substituting \( K' = \frac{1}{K} \):
\[ K'' = \left(\frac{1}{K}\right)^{1/2} = \frac{1}{\sqrt{K}} \]

### Conclusion
The statement in the question is **false**. The correct equilibrium constant for the reaction \( AB \rightleftharpoons \frac{1}{2} A_2 + \frac{1}{2} B_2 \) is:
\[ K'' = \frac{1}{\sqrt{K}} \]