In the reversible reaction A + B `hArr` C + D, the concentration of each C and D at equilobrium was `0.8` mole/litre, then the equilibrium constant `K_(c)` will be

Options

A. 6.4
B. 0.64
C. 1.6
D. Unpredictable

Answer

To find the equilibrium constant \( K_c \) for the reaction \( A + B \rightleftharpoons C + D \), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant
The equilibrium constant \( K_c \) for the reaction can be expressed as:
\[ K_c = \frac{[C][D]}{[A][B]} \]
where \([C]\), \([D]\), \([A]\), and \([B]\) are the concentrations of the respective species at equilibrium.

### Step 2: Identify the equilibrium concentrations
According to the problem, at equilibrium, the concentrations of \( C \) and \( D \) are given as:
\[ [C] = 0.8 \, \text{mol/L} \quad \text{and} \quad [D] = 0.8 \, \text{mol/L} \]

### Step 3: Determine the initial concentrations
Initially, we assume the concentrations of \( A \) and \( B \) are both \( 1 \, \text{mol/L} \):
\[ [A]_0 = 1 \, \text{mol/L} \quad \text{and} \quad [B]_0 = 1 \, \text{mol/L} \]

### Step 4: Calculate the change in concentrations
Let \( x \) be the amount of \( A \) and \( B \) that reacts to form \( C \) and \( D \). Since at equilibrium, the concentrations of \( C \) and \( D \) are \( 0.8 \, \text{mol/L} \), we can say:
\[ x = 0.8 \]

### Step 5: Calculate the equilibrium concentrations of \( A \) and \( B \)
At equilibrium, the concentrations of \( A \) and \( B \) will be:
\[ [A] = [A]_0 - x = 1 - 0.8 = 0.2 \, \text{mol/L} \]
\[ [B] = [B]_0 - x = 1 - 0.8 = 0.2 \, \text{mol/L} \]

### Step 6: Substitute the equilibrium concentrations into the \( K_c \) expression
Now we can substitute the equilibrium concentrations into the \( K_c \) expression:
\[ K_c = \frac{[C][D]}{[A][B]} = \frac{(0.8)(0.8)}{(0.2)(0.2)} \]

### Step 7: Calculate \( K_c \)
Calculating the above expression gives:
\[ K_c = \frac{0.64}{0.04} = 16 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 16 \]

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