In the reversible reaction `A+B hArr C+D`, the concentration of each C and D at equilibrium 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

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 expression for the equilibrium constant \( K_C \)
The equilibrium constant \( K_C \) for the reaction is given by the formula:
\[ K_C = \frac{[C][D]}{[A][B]} \]
where \([C]\), \([D]\), \([A]\), and \([B]\) are the molar concentrations of the respective substances at equilibrium.

### Step 2: Identify the equilibrium concentrations
From the problem statement, we know that at equilibrium:
\[ [C] = 0.8 \, \text{mol/L} \]
\[ [D] = 0.8 \, \text{mol/L} \]
We need to find the equilibrium concentrations of \( A \) and \( B \).

### Step 3: Determine the initial concentrations
Assuming that initially, the concentrations of \( C \) and \( D \) were zero, we can let the initial concentrations of \( A \) and \( B \) be \( a \) and \( b \) respectively. As the reaction proceeds to equilibrium, \( C \) and \( D \) are formed from \( A \) and \( B \).

### Step 4: Set up the changes in concentration
Let’s denote the change in concentration of \( A \) and \( B \) as \( -x \) (since they are consumed) and the change for \( C \) and \( D \) as \( +x \) (since they are produced). At equilibrium:
\[ [A] = a - x \]
\[ [B] = b - x \]
\[ [C] = x \]
\[ [D] = x \]

Given that \([C] = 0.8\) and \([D] = 0.8\), we have:
\[ x = 0.8 \]

### Step 5: Substitute \( x \) into the equilibrium concentrations of \( A \) and \( B \)
Now substituting \( x = 0.8 \):
\[ [A] = a - 0.8 \]
\[ [B] = b - 0.8 \]

### Step 6: Assume initial concentrations
If we assume that the initial concentrations of \( A \) and \( B \) were both 1.0 mol/L (this is a common assumption unless stated otherwise), then:
\[ [A] = 1.0 - 0.8 = 0.2 \, \text{mol/L} \]
\[ [B] = 1.0 - 0.8 = 0.2 \, \text{mol/L} \]

### Step 7: Substitute all values 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 8: Calculate \( K_C \)
Calculating the above expression:
\[ K_C = \frac{0.64}{0.04} = 16 \]

### Final Answer
Thus, the equilibrium constant \( K_C \) is:
\[ \boxed{16} \]

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