In a chemical reaction , `A+2Boverset(K)hArr2c+D,` the initial concentration of B was 1.5 times of the concentrations of A , but the equilibrium concentrations of A and B were found to be equal . The equilibrium constant (K) for the aforesaid chemical reaction is :

Options

A. 1
B. 16
C. `1/4`
D. 4

Explanation

To solve the problem, we need to determine the equilibrium constant \( K \) for the reaction:

\[ A + 2B \rightleftharpoons 2C + D \]

Given that the initial concentration of \( B \) is 1.5 times that of \( A \), and at equilibrium, the concentrations of \( A \) and \( B \) are equal.

### Step 1: Define Initial Concentrations
Let the initial concentration of \( A \) be \( [A]_0 = a \).
Then, the initial concentration of \( B \) will be:
\[ [B]_0 = 1.5a \]

### Step 2: Set Up Changes at Equilibrium
Let \( x \) be the change in concentration of \( A \) at equilibrium. Therefore, the changes in concentrations will be:
- For \( A \): \( [A] = a - x \)
- For \( B \): \( [B] = 1.5a - 2x \) (since 2 moles of \( B \) are consumed for every mole of \( A \) consumed)
- For \( C \): \( [C] = 2x \) (since 2 moles of \( C \) are produced for every mole of \( A \) consumed)
- For \( D \): \( [D] = x \) (since 1 mole of \( D \) is produced for every mole of \( A \) consumed)

### Step 3: Set Up the Equation for Equal Concentrations
According to the problem, at equilibrium, the concentrations of \( A \) and \( B \) are equal:
\[ a - x = 1.5a - 2x \]

### Step 4: Solve for \( x \)
Rearranging the equation gives:
\[ a - x + 2x = 1.5a \]
\[ a + x = 1.5a \]
\[ x = 1.5a - a \]
\[ x = 0.5a \]

### Step 5: Substitute \( x \) Back into Concentrations
Now substituting \( x \) back into the equilibrium concentrations:
- For \( A \):
\[ [A] = a - 0.5a = 0.5a \]
- For \( B \):
\[ [B] = 1.5a - 2(0.5a) = 1.5a - a = 0.5a \]
- For \( C \):
\[ [C] = 2(0.5a) = a \]
- For \( D \):
\[ [D] = 0.5a \]

### Step 6: Write the Expression for the Equilibrium Constant \( K \)
The expression for the equilibrium constant \( K \) is given by:
\[ K = \frac{[C]^2[D]}{[A][B]^2} \]

Substituting the equilibrium concentrations:
\[ K = \frac{(a)^2(0.5a)}{(0.5a)(0.5a)^2} \]

### Step 7: Simplify the Expression
Calculating the numerator:
\[ K = \frac{a^2 \cdot 0.5a}{0.5a \cdot (0.25a^2)} \]
\[ K = \frac{0.5a^3}{0.125a^3} \]
\[ K = \frac{0.5}{0.125} = 4 \]

### Conclusion
Thus, the equilibrium constant \( K \) for the reaction is:
\[ K = 4 \]