In the reaction, `A+BhArrC+D` in a one - litre container, concentration of B at fixed temperature was n mole and initial concentration of A was 3n mole. If the concentration of C at equilibrium is equivalent to that of B, the concentration of D will be

Options

A. `(n)/(4)`
B. n
C. `(n)/(3)`
D. `(n)/(2)`

Answer

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

**Reaction:**
\[ A + B \rightleftharpoons C + D \]

**Initial Concentrations:**
- Concentration of A = \( 3n \) moles
- Concentration of B = \( n \) moles

Since the reaction occurs in a one-litre container, the concentrations are equal to the number of moles.

### Step 1: Write the Initial Concentrations
At time \( t = 0 \):
- \[ [A] = 3n \]
- \[ [B] = n \]
- \[ [C] = 0 \]
- \[ [D] = 0 \]

### Step 2: Define the Change in Concentrations
Let \( x \) be the amount of A and B that react at equilibrium. Therefore, at equilibrium, the concentrations will be:
- \[ [A] = 3n - x \]
- \[ [B] = n - x \]
- \[ [C] = x \]
- \[ [D] = x \]

### Step 3: Use the Given Condition
We know that at equilibrium, the concentration of C is equal to the concentration of B:
\[ [C] = [B] \]
Thus,
\[ x = n - x \]

### Step 4: Solve for x
Rearranging the equation:
\[ x + x = n \]
\[ 2x = n \]
\[ x = \frac{n}{2} \]

### Step 5: Find the Concentration of D
Since the concentration of D is equal to \( x \):
\[ [D] = x = \frac{n}{2} \]

### Conclusion
The concentration of D at equilibrium is:
\[ \frac{n}{2} \]

### Final Answer
The concentration of D will be \( \frac{n}{2} \).

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