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. `16.0`

Explanation

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

### Step 1: Write the balanced equation
The balanced equation for the reaction is:
\[ A + B \rightleftharpoons C + D \]

### Step 2: Define initial concentrations
Let's assume that initially, we have:
- Concentration of \( A = 1 \, \text{mol/L} \)
- Concentration of \( B = 1 \, \text{mol/L} \)
- Concentration of \( C = 0 \, \text{mol/L} \)
- Concentration of \( D = 0 \, \text{mol/L} \)

### Step 3: Determine equilibrium concentrations
At equilibrium, we are given that:
- Concentration of \( C = 0.8 \, \text{mol/L} \)
- Concentration of \( D = 0.8 \, \text{mol/L} \)

### Step 4: Calculate the change in concentration for A and B
Since \( C \) and \( D \) are produced from \( A \) and \( B \), the change in concentration for \( A \) and \( B \) can be calculated as follows:
- For every mole of \( C \) and \( D \) produced, one mole of \( A \) and \( B \) is consumed.
- Therefore, the change in concentration for \( A \) and \( B \) is \( 0.8 \, \text{mol/L} \).

Thus, the equilibrium concentrations for \( A \) and \( B \) will be:
- Concentration of \( A = 1 - 0.8 = 0.2 \, \text{mol/L} \)
- Concentration of \( B = 1 - 0.8 = 0.2 \, \text{mol/L} \)

### Step 5: Write the expression for \( K_c \)
The equilibrium constant \( K_c \) is given by the formula:
\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} \]
For our reaction, this becomes:
\[ K_c = \frac{[C][D]}{[A][B]} \]

### Step 6: Substitute the equilibrium concentrations into the \( K_c \) expression
Substituting the values we have:
\[ K_c = \frac{(0.8)(0.8)}{(0.2)(0.2)} \]

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

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

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