If two gases `AB_(2)` and `B_(2)C` are mixed, following equilibria are readily established:
`AB_(2)(g)+B_(2)C(g)rarr AB_(3)(g)+BC(g)`,
`BC(g)+B_(2)C(g) rarr B_(3)C_(2)(g)`
If the reaction started only with `AB_(2)` with `B_(2)C`, then which of the following is necessarily true at equilibrium?

Options

A. `[AB_(3)]_(eq)=[BC]_(eq)`
B. `[AB_(2)]_(eq)=[B_(2)C]_(eq)`
C. `[AB_(3)]_(eq)gt [B_(3)C_(2)]_(eq)`
D. `[AB_(3)]_(eq)gt [BC]_(eq)`

Explanation

To solve the problem, we need to analyze the two equilibria established when the gases \( AB_2 \) and \( B_2C \) are mixed:

1. **First Equilibrium**:
\[ AB_2(g) + B_2C(g) \rightleftharpoons AB_3(g) + BC(g) \]

2. **Second Equilibrium**:
\[ BC(g) + B_2C(g) \rightleftharpoons B_3C_2(g) \]

### Step-by-step Solution:

**Step 1: Initial Conditions**
- Let’s assume we start with \( x \) moles of \( AB_2 \) and \( y \) moles of \( B_2C \). Initially, we have:
- \( [AB_2] = x \)
- \( [B_2C] = y \)
- \( [AB_3] = 0 \)
- \( [BC] = 0 \)
- \( [B_3C_2] = 0 \)

**Step 2: Change in Concentrations**
- Let \( A \) moles of \( AB_2 \) react with \( B_2C \) to form \( AB_3 \) and \( BC \). At equilibrium, we have:
- \( [AB_2] = x - A \)
- \( [B_2C] = y - A \)
- \( [AB_3] = A \)
- \( [BC] = A \)

**Step 3: Second Reaction**
- The \( BC \) produced can react with \( B_2C \) to form \( B_3C_2 \). Let \( Z \) moles of \( BC \) react:
- At equilibrium, we have:
- \( [BC] = A - Z \)
- \( [B_2C] = y - A - Z \)
- \( [B_3C_2] = Z \)

**Step 4: Establishing Relationships**
- From the above, we can summarize the concentrations at equilibrium:
- \( [AB_2] = x - A \)
- \( [B_2C] = y - A - Z \)
- \( [AB_3] = A \)
- \( [BC] = A - Z \)
- \( [B_3C_2] = Z \)

**Step 5: Analyzing the Options**
1. **Option A**: \( [AB_3] = [BC] \)
- \( A \neq A - Z \) (since \( Z > 0 \)), so this is **false**.

2. **Option B**: \( [AB_2] = [B_2C] \)
- \( x - A \) and \( y - A - Z \) cannot be determined to be equal without knowing \( x \) and \( y \), so this is **false**.

3. **Option C**: \( [AB_3] > [B_3C_2] \)
- \( A > Z \) (since \( Z < A \)), so this is **true**.

4. **Option D**: \( [AB_3] > [BC] \)
- \( A > A - Z \) (since \( Z > 0 \)), so this is **true**.

### Conclusion:
At equilibrium, the necessarily true statements are:
- \( [AB_3] > [B_3C_2] \)
- \( [AB_3] > [BC] \)