At a particular temperature, the equilibrium constant for
`2NO_2 (g) Leftrightarrow N_2O_4 (g)`
is one. The same reaction is carried out in a container of volume just half of the former. Will the value of `[N_2O_4]/[NO_2]^2` be equal to 1 if no reaction occurred? Will the equilibrium constant change by this change in volume?

Answer

To solve the problem, we need to analyze the equilibrium constant expression and the effect of changing the volume of the container on the equilibrium state of the reaction.

### Step-by-Step Solution:

1. **Understanding the Reaction and Equilibrium Constant:**
The given reaction is:
\[ 2NO_2 (g) \leftrightarrow N_2O_4 (g) \]
The equilibrium constant \( K \) for this reaction is defined as:
\[ K = \frac{[N_2O_4]}{[NO_2]^2} \]
We are told that \( K = 1 \) at a particular temperature.

2. **Effect of Volume Change on Concentrations:**
When the volume of the container is halved, the concentrations of the gases will change. The concentration of a gas is given by:
\[ [C] = \frac{n}{V} \]
If the volume is halved, the concentrations of both \( NO_2 \) and \( N_2O_4 \) will double:
\[ [NO_2] \text{ (new)} = 2[NO_2] \text{ (original)} \]
\[ [N_2O_4] \text{ (new)} = 2[N_2O_4] \text{ (original)} \]

3. **Substituting New Concentrations into the Equilibrium Expression:**
The new expression for the equilibrium constant with the changed concentrations will be:
\[ K' = \frac{[N_2O_4]_{new}}{[NO_2]_{new}^2} = \frac{2[N_2O_4]}{(2[NO_2])^2} \]
Simplifying this gives:
\[ K' = \frac{2[N_2O_4]}{4[NO_2]^2} = \frac{1}{2} \cdot \frac{[N_2O_4]}{[NO_2]^2} \]

4. **Determining the Value of \([N_2O_4]/[NO_2]^2\):**
If no reaction occurs immediately after the volume change, the ratio \(\frac{[N_2O_4]}{[NO_2]^2}\) will still equal 1, as given in the problem. Thus:
\[ K' = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

5. **Conclusion on the Equilibrium Constant:**
The equilibrium constant \( K \) does not change with volume; it is only affected by temperature. Therefore, while the concentrations change due to volume change, the equilibrium constant remains the same at the specified temperature. However, the system will shift to re-establish equilibrium.

### Final Answer:
- The value of \(\frac{[N_2O_4]}{[NO_2]^2}\) will equal 1 if no reaction occurs initially.
- The equilibrium constant \( K \) will not change due to the change in volume; it remains 1 at the specified temperature.