The reaction `A + B hArr C + D` is studied in a one litre vessel at `250^(@)C` . The initial concentration of A was 3n and that of B was n. When equilibrium was attained, equilibrium concentration of C was found to the equal to the equilibrium concentration of B. What is the concentration of D at equilibrium ?

Options

A. `n//2`
B. `(3n-1//2)`
C. `(n-n//3)`
D. n

Explanation

To solve the problem step by step, we will analyze the given reaction and the initial concentrations, then determine the equilibrium concentrations.

### Step 1: Write the reaction and initial concentrations
The reaction is given as:
\[ A + B \rightleftharpoons C + D \]

The initial concentrations are:
- \([A] = 3n\)
- \([B] = n\)
- \([C] = 0\)
- \([D] = 0\)

### Step 2: Define the change in concentrations at equilibrium
Let \( x \) be the amount of \( A \) that reacts at equilibrium. Since the stoichiometry of the reaction is 1:1 for \( A \) and \( B \), \( x \) moles of \( A \) will react with \( x \) moles of \( B \).

At equilibrium, the concentrations will be:
- \([A] = 3n - x\)
- \([B] = n - x\)
- \([C] = x\)
- \([D] = x\)

### Step 3: Use the condition given in the problem
We are told that the equilibrium concentration of \( C \) is equal to the equilibrium concentration of \( B \):
\[ [C] = [B] \]
This translates to:
\[ x = n - x \]

### Step 4: Solve for \( x \)
Now, we can solve the equation:
\[ x + x = n \]
\[ 2x = n \]
\[ x = \frac{n}{2} \]

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

### Conclusion
Thus, the concentration of \( D \) at equilibrium is:
\[ \frac{n}{2} \]