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 step by step, we will analyze the reaction and the given information carefully.

### 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 = \( 3n \) moles
- Initial concentration of B = \( n \) moles

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

### Step 4: Use the information given in the problem
We are told that the concentration of C at equilibrium is equivalent to that 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 concentration of D
Since the concentration of D at equilibrium is also \( x \):
\[ \text{Concentration of D} = x = \frac{n}{2} \]

### Final Answer
Therefore, the concentration of D at equilibrium is:
\[ \frac{n}{2} \]

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