The equilibrium constant of a reaction `A + B hArr 2C` if the concentrations of A and B together is 0.8 moles `L^(-1)` and that of C is 0.6 mol `L^(-1)` is ______

Answer

To find the equilibrium constant \( K_c \) for the reaction \( A + B \rightleftharpoons 2C \), 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:
\[ K_c = \frac{[C]^2}{[A][B]} \]
where \([C]\), \([A]\), and \([B]\) are the equilibrium concentrations of the respective substances.

### Step 2: Identify the given concentrations
From the problem, we know:
- The combined concentration of \( A \) and \( B \) is \( 0.8 \, \text{mol L}^{-1} \).
- The concentration of \( C \) is \( 0.6 \, \text{mol L}^{-1} \).

### Step 3: Express the concentrations of \( A \) and \( B \)
Let the concentration of \( A \) be \( [A] \) and the concentration of \( B \) be \( [B] \). Since the total concentration of \( A \) and \( B \) is \( 0.8 \, \text{mol L}^{-1} \), we can write:
\[ [A] + [B] = 0.8 \]

### Step 4: Substitute the values into the equilibrium constant expression
We can express \( K_c \) using the known concentrations:
\[ K_c = \frac{(0.6)^2}{[A][B]} \]
Substituting \( [C] = 0.6 \):
\[ K_c = \frac{0.36}{[A][B]} \]

### Step 5: Express \( [A][B] \) in terms of the total concentration
From the equation \( [A] + [B] = 0.8 \), we can express \( [B] \) as:
\[ [B] = 0.8 - [A] \]
Now, substituting this into the product \( [A][B] \):
\[ [A][B] = [A](0.8 - [A]) = 0.8[A] - [A]^2 \]

### Step 6: Substitute \( [A][B] \) back into the \( K_c \) expression
Now we can substitute this back into the \( K_c \) expression:
\[ K_c = \frac{0.36}{0.8[A] - [A]^2} \]

### Step 7: Solve for \( K_c \)
To find \( K_c \), we need to find a suitable value for \( [A] \) and \( [B] \). Let's assume \( [A] = [B] \) for simplicity, then:
\[ 2[A] = 0.8 \Rightarrow [A] = [B] = 0.4 \]
Now substituting \( [A] \) and \( [B] \) back into the equation:
\[ [A][B] = 0.4 \times 0.4 = 0.16 \]
Thus,
\[ K_c = \frac{0.36}{0.16} = 2.25 \]

### Final Answer
The equilibrium constant \( K_c \) for the reaction is approximately \( 2.25 \).

---