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 step by step, we need to analyze the given chemical reaction and the information provided about the concentrations of the reactants and products at equilibrium.

### Step 1: Write the balanced chemical equation
The balanced chemical equation is:
\[ A + 2B \overset{K}{\rightleftharpoons} 2C + D \]

### Step 2: Define initial concentrations
Let the initial concentration of \( A \) be \( [A]_0 = A \). According to the problem, the initial concentration of \( B \) is 1.5 times that of \( A \):
\[ [B]_0 = 1.5A \]

### Step 3: Define changes in concentrations
Let \( X \) be the amount of \( A \) that reacts at equilibrium. According to the stoichiometry of the reaction:
- The change in concentration of \( A \) will be \( -X \)
- The change in concentration of \( B \) will be \( -2X \) (since 2 moles of \( B \) are consumed for every mole of \( A \))
- The change in concentration of \( C \) will be \( +2X \) (since 2 moles of \( C \) are produced for every mole of \( A \))
- The change in concentration of \( D \) will be \( +X \)

### Step 4: Write equilibrium concentrations
At equilibrium, the concentrations will be:
- \( [A]_{eq} = A - X \)
- \( [B]_{eq} = 1.5A - 2X \)
- \( [C]_{eq} = 2X \)
- \( [D]_{eq} = X \)

### Step 5: Use the information given
We know that at equilibrium, the concentrations of \( A \) and \( B \) are equal:
\[ A - X = 1.5A - 2X \]

### Step 6: Solve for \( X \)
Rearranging the equation gives:
\[ A - X + 2X = 1.5A \]
\[ A + X = 1.5A \]
\[ X = 0.5A \]

### Step 7: Substitute \( X \) back to find equilibrium concentrations
Now substituting \( X = 0.5A \) back into the equilibrium concentrations:
- \( [A]_{eq} = A - 0.5A = 0.5A \)
- \( [B]_{eq} = 1.5A - 2(0.5A) = 1.5A - A = 0.5A \)
- \( [C]_{eq} = 2(0.5A) = A \)
- \( [D]_{eq} = 0.5A \)

### Step 8: Write the expression for the equilibrium constant \( K \)
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 9: 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:
\[ K = 4 \]

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