Two flasks A and B of equal volume containing 1 mole and 2 moles of `O_(3)` respectively, are heated to the same temperature. When the reaction `2O_(3)=3O_(2)` practically stops, then both the flasks shall have

Options

A. the same ratio: `[O_(2)]//[O_(3)]`
B. the same ratio : `[O_(2)]^(3//2)//[O_(3)]`
C. only `O_(2)`
D. the same time to reach equilibrium

Answer

To solve the problem, we need to analyze the reaction and the conditions in both flasks A and B.

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction given is:
\[ 2O_3 \rightleftharpoons 3O_2 \]
This means that 2 moles of ozone (O3) decompose to form 3 moles of oxygen (O2).

2. **Initial Moles in Each Flask**:
- Flask A contains 1 mole of O3.
- Flask B contains 2 moles of O3.

3. **Volume and Concentration**: Since both flasks have equal volume, the concentration of O3 in each flask can be determined:
- Concentration of O3 in Flask A = \( \frac{1 \text{ mole}}{V} \)
- Concentration of O3 in Flask B = \( \frac{2 \text{ moles}}{V} \)

4. **Equilibrium Constant (Kc)**: The equilibrium constant for the reaction can be expressed as:
\[ K_c = \frac{[O_2]^3}{[O_3]^2} \]
Since both flasks are heated to the same temperature, the value of \( K_c \) will be the same for both reactions.

5. **Setting Up the Equilibrium Expression**:
- Let \( x \) be the amount of O3 that decomposes in Flask A. At equilibrium:
- O3 in Flask A: \( 1 - x \)
- O2 in Flask A: \( \frac{3}{2}x \) (from the stoichiometry of the reaction)
- For Flask B, let \( y \) be the amount of O3 that decomposes:
- O3 in Flask B: \( 2 - y \)
- O2 in Flask B: \( \frac{3}{2}y \)

6. **Equilibrium Concentrations**:
- For Flask A:
\[ K_c = \frac{(\frac{3}{2}x)^3}{(1-x)^2} \]
- For Flask B:
\[ K_c = \frac{(\frac{3}{2}y)^3}{(2-y)^2} \]

7. **Equating the Kc Values**: Since \( K_c \) is the same for both flasks:
\[ \frac{(\frac{3}{2}x)^3}{(1-x)^2} = \frac{(\frac{3}{2}y)^3}{(2-y)^2} \]

8. **Conclusion**: At equilibrium, the concentrations of O2 and O3 will adjust such that the Kc expression holds true for both flasks. Since Flask B starts with more O3, it will produce more O2 until both flasks reach the same Kc value.

Thus, at equilibrium, both flasks will have the same ratio of O2 to O3 based on the Kc value.

### Final Answer:
Both flasks will have the same equilibrium concentrations of O2 and O3, maintaining the same Kc value.