500 ml vessel contains 1.5 M each of A, B, C and D at equlibrium. If 0.5 M each of C and D are taken out, the value of `K_(c)` for `A+BhArrC+D` will be

Options

A. `(1)/(9)`
B. `(4)/(9)`
C. 1
D. `(5)/(9)`

Answer

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

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

Given that the initial concentrations of A, B, C, and D are all 1.5 M in a 500 ml vessel, we can follow these steps:

### Step 1: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by the formula:

\[ K_c = \frac{[C][D]}{[A][B]} \]

### Step 2: Substitute the initial concentrations
At equilibrium, the concentrations of A, B, C, and D are all 1.5 M. Therefore, substituting these values into the \( K_c \) expression gives:

\[ K_c = \frac{(1.5)(1.5)}{(1.5)(1.5)} = \frac{2.25}{2.25} = 1 \]

### Step 3: Consider the change in concentrations
Now, if we remove 0.5 M each of C and D, the new concentrations will be:

- Concentration of C = \( 1.5 - 0.5 = 1.0 \) M
- Concentration of D = \( 1.5 - 0.5 = 1.0 \) M
- Concentration of A = \( 1.5 \) M (unchanged)
- Concentration of B = \( 1.5 \) M (unchanged)

### Step 4: Calculate the new \( K_c \)
Now we can substitute the new concentrations back into the \( K_c \) expression:

\[ K_c = \frac{[C][D]}{[A][B]} = \frac{(1.0)(1.0)}{(1.5)(1.5)} = \frac{1.0}{2.25} \approx 0.444 \]

### Step 5: Conclusion
However, it is important to note that the value of \( K_c \) remains constant at a given temperature, regardless of the changes in concentrations. Therefore, the equilibrium constant \( K_c \) for the reaction remains:

\[ K_c = 1 \]

Thus, the final answer is:

\[ \boxed{1} \]