For `A+B hArr C`, the equilibrium concentration of A and B at a temperature are `15 mol L^(-1)`. When volume is doubled the reaction has equilibrium concentration of A as `10 mol L^(-1)`, calculate
a. `K_(c)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial equilibrium conditions
Given the reaction:
\[ A + B \rightleftharpoons C \]

At the initial equilibrium, the concentrations of A and B are both \( 15 \, \text{mol L}^{-1} \). We can denote the concentration of C as \( x \).

### Step 2: Write the equilibrium expression
The equilibrium constant \( K_c \) is defined as:
\[ K_c = \frac{[C]}{[A][B]} \]

At the initial equilibrium (let's call it Equilibrium 1):
- \([A] = 15 \, \text{mol L}^{-1}\)
- \([B] = 15 \, \text{mol L}^{-1}\)
- \([C] = x\)

Thus, the expression for \( K_c \) at this point is:
\[ K_{c1} = \frac{x}{15 \times 15} = \frac{x}{225} \]

### Step 3: Analyze the effect of doubling the volume
When the volume is doubled, the concentrations of A and B will be halved:
- New \([A] = \frac{15}{2} = 7.5 \, \text{mol L}^{-1}\)
- New \([B] = \frac{15}{2} = 7.5 \, \text{mol L}^{-1}\)

Let the new concentration of C at this equilibrium (Equilibrium 2) be \( y \).

### Step 4: Establish the new equilibrium conditions
At the new equilibrium:
- \([A] = 7.5 + x\)
- \([B] = 7.5 + x\)
- \([C] = y\)

From the problem, we know that after the volume is doubled, the equilibrium concentration of A is \( 10 \, \text{mol L}^{-1} \). Therefore:
\[ 7.5 + x = 10 \]
Thus, solving for \( x \):
\[ x = 10 - 7.5 = 2.5 \]

### Step 5: Calculate the new concentration of C
Now, substituting \( x \) into the concentration of C:
\[ y = \frac{x}{2} - x = \frac{A}{2} - x = \frac{15}{2} - 2.5 = 7.5 - 2.5 = 5 \, \text{mol L}^{-1} \]

### Step 6: Calculate \( K_c \) at the new equilibrium
Now we can calculate \( K_c \) at Equilibrium 2:
\[ K_{c2} = \frac{y}{(7.5 + 2.5)(7.5 + 2.5)} = \frac{5}{(10)(10)} = \frac{5}{100} = 0.05 \]

### Final Step: Conclusion
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 0.05 \]