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

Answer

To solve the problem step by step, we will analyze the given reaction and the information provided.

### Step 1: Write the balanced chemical equation
The reaction is given as:
\[ A + B \rightleftharpoons C + D \]

### Step 2: Identify initial concentrations
From the problem, we know:
- Initial concentration of A = \( 3n \)
- Initial concentration of B = \( n \)
- Initial concentrations of C and D = \( 0 \) (since they are products and initially not present)

### Step 3: Set up the change in concentrations at equilibrium
Let \( x \) be the amount of A and B that react to reach equilibrium. Therefore, at equilibrium, the concentrations will be:
- Concentration of A = \( 3n - x \)
- Concentration of B = \( n - x \)
- Concentration of C = \( x \)
- Concentration of D = \( x \)

### Step 4: Use the information given about equilibrium concentrations
It is given that the equilibrium concentration of C is equal to the equilibrium concentration of B:
\[ x = n - x \]

### Step 5: Solve for \( x \)
Rearranging the equation:
\[ x + x = n \]
\[ 2x = n \]
\[ x = \frac{n}{2} \]

### Step 6: Find the equilibrium concentrations
Now we can substitute \( x \) back into the expressions for the equilibrium concentrations:
- Concentration of C = \( x = \frac{n}{2} \)
- Concentration of D = \( x = \frac{n}{2} \)

### Step 7: Conclusion
The concentration of D at equilibrium is:
\[ \text{Concentration of D} = \frac{n}{2} \]

### Final Answer
The concentration of D at equilibrium is \( \frac{n}{2} \).

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