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_(1)K_(2)`
C. `K_(3)=K_(1)//K_(2)`
D. `K_(3)=K_(2)//K_(1)`

Answer

To find the relationship between the equilibrium constants \( K_1 \), \( K_2 \), and \( K_3 \) for the given reactions, we can follow these steps:

### Step 1: Write down the reactions and their equilibrium constants
1. **Reaction 1**: \( I_2(aq) \rightleftharpoons I_2(oil) \) with equilibrium constant \( K_1 \)
2. **Reaction 2**: \( I_2(oil) \rightleftharpoons I_2(ether) \) with equilibrium constant \( K_2 \)
3. **Reaction 3**: \( I_2(aq) \rightleftharpoons I_2(ether) \) with equilibrium constant \( K_3 \)

### Step 2: Combine the first two reactions
By adding Reaction 1 and Reaction 2, we can eliminate \( I_2(oil) \):
- From Reaction 1: \( I_2(aq) \rightleftharpoons I_2(oil) \)
- From Reaction 2: \( I_2(oil) \rightleftharpoons I_2(ether) \)

When we add these two reactions, \( I_2(oil) \) cancels out, resulting in:
\[ I_2(aq) \rightleftharpoons I_2(ether) \]

### Step 3: Relate the equilibrium constants
According to the principle of equilibrium constants, when two reactions are added, the equilibrium constants are multiplied:
\[ K_3 = K_1 \times K_2 \]

### Conclusion
Thus, the relationship between the equilibrium constants is:
\[ K_3 = K_1 \times K_2 \]