To solve the problem step by step, we will analyze the equilibrium reaction and the changes that occur when the volume is doubled.
### Step 1: Write the Reaction and Initial Conditions
The equilibrium reaction is given as:
\[ A + B \rightleftharpoons C \]
At equilibrium, the concentrations of A and B are both 15 mol/L. Let’s denote the concentration of C at equilibrium as \( x \).
### Step 2: Establish Initial Equilibrium Concentrations
At the initial equilibrium:
- \([A] = 15 \, \text{mol/L}\)
- \([B] = 15 \, \text{mol/L}\)
- \([C] = x \, \text{mol/L}\)
The equilibrium constant \( K_c \) can be expressed as:
\[ K_c = \frac{[C]}{[A][B]} = \frac{x}{15 \times 15} = \frac{x}{225} \]
### Step 3: Analyze the Effect of Doubling the Volume
When the volume is doubled, the concentrations of A and B will be halved:
- New \([A] = \frac{15}{2} = 7.5 \, \text{mol/L}\)
- New \([B] = \frac{15}{2} = 7.5 \, \text{mol/L}\)
Let’s denote the new concentration of C at this point as \( y \). Since the system is not at equilibrium after the volume change, we can write:
\[ K_c = \frac{y}{(7.5)(7.5)} = \frac{y}{56.25} \]
### Step 4: Establish the New Equilibrium Concentration
We are given that after the volume is doubled, the equilibrium concentration of A becomes 10 mol/L. This means that the reaction will shift to the left (backward direction) to reach a new equilibrium.
Let’s denote the change in concentration of A due to the backward reaction as \( z \):
- New \([A] = 7.5 + z = 10 \Rightarrow z = 2.5 \, \text{mol/L}\)
Thus, at equilibrium:
- \([A] = 10 \, \text{mol/L}\)
- \([B] = 7.5 + z = 7.5 + 2.5 = 10 \, \text{mol/L}\)
- \([C] = y - z = y - 2.5\)
### Step 5: Relate the Concentrations to Find \( y \)
Now we have:
- \([A] = 10 \, \text{mol/L}\)
- \([B] = 10 \, \text{mol/L}\)
Using the equilibrium constant expression:
\[ K_c = \frac{y - 2.5}{(10)(10)} = \frac{y - 2.5}{100} \]
### Step 6: Equate the Two Expressions for \( K_c \)
Since \( K_c \) remains constant, we can set the two expressions equal:
\[ \frac{x}{225} = \frac{y - 2.5}{100} \]
### Step 7: Solve for \( y \)
From the previous steps, we know that:
- \( y = 10 + 2.5 = 12.5 \)
Now substituting \( y \) back into the equation:
\[ \frac{x}{225} = \frac{12.5 - 2.5}{100} \]
\[ \frac{x}{225} = \frac{10}{100} \]
\[ x = 225 \times 0.1 = 22.5 \]
### Step 8: Find the Original Concentration of C
The original concentration of C at equilibrium is \( x = 22.5 \, \text{mol/L} \).
### Step 9: Calculate \( K_c \)
Now substituting \( x \) back into the original \( K_c \) expression:
\[ K_c = \frac{22.5}{225} = 0.1 \]
### Conclusion
The equilibrium concentrations and the equilibrium constant have been determined.
