The equilibrium `A(g) +4B(g) hArr AB_4(g)` is attained by mixing equal moles of A and B in a one litre vessel Then at equilibrium

Options

A. [A]=[B]
B. `[A] gt [B]`
C. `[A] lt [B]`
D. `[AB_4] gt [A]`

Answer

To solve the problem, we need to analyze the equilibrium reaction given:

\[ A(g) + 4B(g) \rightleftharpoons AB_4(g) \]

### Step 1: Initial Moles
We start with equal moles of A and B in a 1-liter vessel. Let's assume we have 1 mole of A and 1 mole of B initially.

- Initial moles of A = 1 mole
- Initial moles of B = 1 mole

### Step 2: Change in Moles at Equilibrium
At equilibrium, let \( x \) be the amount of A that reacts. According to the stoichiometry of the reaction, for every mole of A that reacts, 4 moles of B will also react. Therefore, the changes in moles can be expressed as:

- Change in moles of A = \( -x \)
- Change in moles of B = \( -4x \)

### Step 3: Moles at Equilibrium
Now, we can express the moles of A and B at equilibrium:

- Moles of A at equilibrium = \( 1 - x \)
- Moles of B at equilibrium = \( 1 - 4x \)

### Step 4: Analyzing the Equilibrium Condition
Since we started with 1 mole of A and 1 mole of B, we need to consider the limits of \( x \). The maximum value of \( x \) can be 0.25 (since 4 moles of B are required for every mole of A). If \( x \) were to exceed 0.25, we would end up with negative moles of B, which is not possible.

### Step 5: Comparing Moles of A and B
At equilibrium, we can see that:

- Moles of A = \( 1 - x \)
- Moles of B = \( 1 - 4x \)

Since \( x \) can be at most 0.25, we can substitute this into the equations:

- If \( x = 0.25 \):
- Moles of A = \( 1 - 0.25 = 0.75 \)
- Moles of B = \( 1 - 4(0.25) = 1 - 1 = 0 \)

Thus, as \( x \) increases from 0 to 0.25, the moles of B decrease at a faster rate than the moles of A. Therefore, at equilibrium, the moles of B will always be less than the moles of A.

### Conclusion
At equilibrium, the number of moles of B will be less than the number of moles of A.

### Final Answer
The correct option is that the number of moles of B becomes less than A.

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