If `K_(1),K_(2),K_(3)` are equilibrium constant for formation of `AD,AD_(2),AD_(3)` respectively as follows `A+DhArrD,AD+DhArrAD_(2),AD_(2)+DhArrAD_(3)`. Then equilibrium constant `'K' "for" A+3DhArrAD_(3)` is related as

Options

A. `K_(1)+K_(2)+K_(3)=K`
B. `logK_(1)+logK_(2)+logK_(3)=logK`
C. `K_(1)+K_(2)=K_(3)+K`
D. `logK_(1)+logK_(2)=logK_(3)+logK`

Explanation

To solve the problem, we need to analyze the given reactions and their equilibrium constants. Let's go through the steps systematically.

### Step 1: Write down the reactions and their equilibrium constants.
1. **Reaction 1:**
\[ A + D \rightleftharpoons AD \quad (K_1) \]
2. **Reaction 2:**
\[ AD + D \rightleftharpoons AD_2 \quad (K_2) \]
3. **Reaction 3:**
\[ AD_2 + D \rightleftharpoons AD_3 \quad (K_3) \]

### Step 2: Add the reactions together.
To find the equilibrium constant for the overall reaction \( A + 3D \rightleftharpoons AD_3 \), we can add the three reactions together.

When we add the reactions:
- From Reaction 1, we have \( A + D \rightleftharpoons AD \).
- From Reaction 2, we have \( AD + D \rightleftharpoons AD_2 \).
- From Reaction 3, we have \( AD_2 + D \rightleftharpoons AD_3 \).

### Step 3: Combine the reactions.
When we combine these reactions, we can cancel out the intermediate species:
- The \( AD \) from Reaction 1 cancels with \( AD \) from Reaction 2.
- The \( AD_2 \) from Reaction 2 cancels with \( AD_2 \) from Reaction 3.

The overall reaction simplifies to:
\[ A + 3D \rightleftharpoons AD_3 \]

### Step 4: Relate the equilibrium constants.
According to the principle of equilibrium constants, when reactions are added, their equilibrium constants multiply. Therefore, the equilibrium constant \( K \) for the overall reaction is given by:
\[ K = K_1 \times K_2 \times K_3 \]

### Step 5: Express in logarithmic form.
If we take the logarithm of both sides, we can express it as:
\[ \log K = \log(K_1 \times K_2 \times K_3) \]
Using the property of logarithms that states \( \log(a \times b) = \log a + \log b \), we can rewrite this as:
\[ \log K = \log K_1 + \log K_2 + \log K_3 \]

### Final Answer:
Thus, the relationship between the equilibrium constants is:
\[ \log K = \log K_1 + \log K_2 + \log K_3 \]