For the reaction:
`2A(g)+B(g) hArr 3C(g)+D(g)`
Two moles each of `A` and `B` were taken into a flask. The following must always be true when the system attained equilibrium

Options

A. `[A] = [B]`
B. `[A] lt [B]`
C. `[B]=[C]`
D. `[A] gt [B]`

Explanation

To solve the problem, we need to analyze the given reaction and the conditions at equilibrium. The reaction is:

\[ 2A(g) + B(g) \rightleftharpoons 3C(g) + D(g) \]

Initially, we have:
- 2 moles of \( A \)
- 2 moles of \( B \)
- 0 moles of \( C \)
- 0 moles of \( D \)

### Step-by-Step Solution:

1. **Set Up the Initial Conditions:**
- At time \( t = 0 \):
- Moles of \( A = 2 \)
- Moles of \( B = 2 \)
- Moles of \( C = 0 \)
- Moles of \( D = 0 \)

2. **Define the Change in Moles at Equilibrium:**
- Let \( x \) be the moles of \( B \) that are consumed at equilibrium.
- According to the stoichiometry of the reaction:
- For every 1 mole of \( B \) consumed, 2 moles of \( A \) are consumed.
- Therefore, the moles of \( A \) consumed will be \( 2x \).
- The moles of \( C \) formed will be \( 3x \).
- The moles of \( D \) formed will be \( x \).

3. **Calculate Moles at Equilibrium:**
- Moles of \( A \) at equilibrium:
\[ 2 - 2x \]
- Moles of \( B \) at equilibrium:
\[ 2 - x \]
- Moles of \( C \) at equilibrium:
\[ 3x \]
- Moles of \( D \) at equilibrium:
\[ x \]

4. **Analyze the Given Options:**
- **Option A:** At equilibrium, the concentration of \( A \) is equal to the concentration of \( B \).
- This implies \( 2 - 2x = 2 - x \). Solving this gives \( x = 0 \), which means no reaction has occurred, hence this option is incorrect.

- **Option B:** The concentration of \( A \) is less than the concentration of \( B \).
- This implies \( 2 - 2x < 2 - x \). This is true for \( x > 0 \), so this option is correct.

- **Option C:** The concentration of \( B \) is equal to the concentration of \( C \).
- This implies \( 2 - x = 3x \). Solving gives \( 2 = 4x \) or \( x = \frac{1}{2} \). This is a specific condition and not always true, hence this option is incorrect.

- **Option D:** The concentration of \( C \) is equal to the concentration of \( D \).
- This implies \( 3x = x \), which is only true if \( x = 0 \), hence this option is also incorrect.

5. **Conclusion:**
- The only option that must always be true when the system attains equilibrium is **Option B**: The concentration of \( A \) is less than the concentration of \( B \).

### Final Answer:
The correct option is **B**: The concentration of \( A \) is less than the concentration of \( B \).