Write the relationship between equilibrium constants for the following reactions:
`2NO(g)+O_2(g) hArr 2NO_2(g)`
`4NO_2(g) hArr 4NO(g) + 2O_2(g)`

Explanation

To find the relationship between the equilibrium constants for the given reactions, we will analyze each reaction step by step.

### Step 1: Write the equilibrium constant expressions for both reactions.

For the first reaction:
\[ 2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g) \]

The equilibrium constant \( K_1 \) is given by:
\[ K_1 = \frac{[NO_2]^2}{[NO]^2[O_2]} \]

For the second reaction:
\[ 4NO_2(g) \rightleftharpoons 4NO(g) + 2O_2(g) \]

The equilibrium constant \( K_2 \) is given by:
\[ K_2 = \frac{[NO]^4[O_2]^2}{[NO_2]^4} \]

### Step 2: Relate the two reactions.

Notice that the second reaction is essentially the reverse of the first reaction, multiplied by 2. To express this mathematically, we can reverse the first reaction and multiply it by 2:

Reversing the first reaction gives:
\[ 2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g) \]

The equilibrium constant for this reversed reaction, \( K' \), is:
\[ K' = \frac{[NO]^2[O_2]}{[NO_2]^2} \]
Since reversing the reaction inverts the equilibrium constant:
\[ K' = \frac{1}{K_1} \]

### Step 3: Multiply the reversed reaction by 2.

Now, we multiply the reversed reaction by 2:
\[ 4NO_2(g) \rightleftharpoons 4NO(g) + 2O_2(g) \]

When we multiply a reaction by a factor of \( n \), the equilibrium constant is raised to the power of \( n \):
\[ K_2 = (K')^2 = \left(\frac{1}{K_1}\right)^2 = \frac{1}{K_1^2} \]

### Conclusion: Relationship between the equilibrium constants.

Thus, the relationship between the equilibrium constants for the two reactions is:
\[ K_2 = \frac{1}{K_1^2} \]