For the reaction,
`I_(2)(aq)hArrI_(2)" (oil), Equilibrium constant is K"_(1).`
`I_(2)"(oil)"hArrI_(2)"(ether), Equilibrium constant is K"_(2).`
`I_(2)(aq)hArrI_(2)"(ether), Equilibrium constant is K"_(3).`
The reaction between `K_(1), K_(2), K_(3)` is

Options

A. `K_3=K_1+K_2`
B. `K_3=K_1K_2`
C. `K_3=K_1//K_2`
D. `K_3=K_2//K_1`

Answer

To solve the problem, we need to establish the relationships between the equilibrium constants \( K_1 \), \( K_2 \), and \( K_3 \) based on the given reactions.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Equilibrium Constants**:
- The first reaction is:
\[ I_2(aq) \rightleftharpoons I_2(oil) \quad \text{with equilibrium constant } K_1 \]
- The second reaction is:
\[ I_2(oil) \rightleftharpoons I_2(ether) \quad \text{with equilibrium constant } K_2 \]
- The third reaction is:
\[ I_2(aq) \rightleftharpoons I_2(ether) \quad \text{with equilibrium constant } K_3 \]

2. **Combine the First Two Reactions**:
- We can add the first and second reactions:
\[ I_2(aq) \rightleftharpoons I_2(oil) \quad (K_1) \]
\[ I_2(oil) \rightleftharpoons I_2(ether) \quad (K_2) \]
- When we add these two reactions, the \( I_2(oil) \) cancels out:
\[ I_2(aq) \rightleftharpoons I_2(ether) \]

3. **Relate the Equilibrium Constants**:
- According to the principle of equilibrium constants, when two reactions are added, their equilibrium constants multiply:
\[ K_3 = K_1 \times K_2 \]

4. **Conclusion**:
- Therefore, the relationship between the equilibrium constants is:
\[ K_3 = K_1 \times K_2 \]

### Final Answer:
The relationship between the equilibrium constants is:
\[ K_3 = K_1 \times K_2 \]