Equilibrium constant `K_c` for the equilibrium `A(g) Leftrightarrow B (g) +C(g)` is 0.45 at `200^@C.`
One litre of a container contains 0.2 mole of A, 0.3 mole of B and 0.3 mole of C at equilibrium. Calculate the new equilibrium concentrations of A, B and C if the volume of the container is (a) doubled (b) halved at `200^@C.`

Answer

To solve the problem, we will go through the following steps:

### Step 1: Write the equilibrium expression
For the reaction:
\[ A(g) \leftrightarrow B(g) + C(g) \]
The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[B][C]}{[A]} \]
Given \( K_c = 0.45 \).

### Step 2: Determine initial concentrations
Initially, we have:
- Concentration of A, \([A] = 0.2 \, \text{mol/L}\)
- Concentration of B, \([B] = 0.3 \, \text{mol/L}\)
- Concentration of C, \([C] = 0.3 \, \text{mol/L}\)

### Step 3: Calculate the reaction quotient \( Q_c \)
Using the initial concentrations:
\[ Q_c = \frac{[B][C]}{[A]} = \frac{(0.3)(0.3)}{0.2} = \frac{0.09}{0.2} = 0.45 \]

### Step 4: Analyze the system when the volume is doubled
When the volume is doubled, the concentrations will be halved:
- New \([A] = \frac{0.2}{2} = 0.1 \, \text{mol/L}\)
- New \([B] = \frac{0.3}{2} = 0.15 \, \text{mol/L}\)
- New \([C] = \frac{0.3}{2} = 0.15 \, \text{mol/L}\)

Now calculate the new \( Q_c \):
\[ Q_c = \frac{(0.15)(0.15)}{0.1} = \frac{0.0225}{0.1} = 0.225 \]

### Step 5: Compare \( Q_c \) with \( K_c \)
Since \( Q_c < K_c \), the reaction will shift to the right to reach equilibrium. Let \( x \) be the change in concentration:
- New \([A] = 0.1 - x\)
- New \([B] = 0.15 + x\)
- New \([C] = 0.15 + x\)

Setting up the equilibrium expression:
\[ K_c = \frac{(0.15 + x)(0.15 + x)}{0.1 - x} = 0.45 \]

### Step 6: Solve for \( x \)
Expanding and rearranging gives:
\[ (0.15 + x)^2 = 0.045(0.1 - x) \]
\[ 0.0225 + 0.3x + x^2 = 0.0045 - 0.045x \]
\[ x^2 + 0.345x + 0.018 = 0 \]

Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-0.345 \pm \sqrt{(0.345)^2 - 4(1)(0.018)}}{2(1)} \]
Calculating gives:
\[ x \approx 0.028 \]

### Step 7: Calculate new concentrations
- New \([A] = 0.1 - 0.028 = 0.072 \, \text{mol/L}\)
- New \([B] = 0.15 + 0.028 = 0.178 \, \text{mol/L}\)
- New \([C] = 0.15 + 0.028 = 0.178 \, \text{mol/L}\)

### Step 8: Analyze the system when the volume is halved
When the volume is halved, the concentrations will be doubled:
- New \([A] = 0.2 \times 2 = 0.4 \, \text{mol/L}\)
- New \([B] = 0.3 \times 2 = 0.6 \, \text{mol/L}\)
- New \([C] = 0.3 \times 2 = 0.6 \, \text{mol/L}\)

Now calculate the new \( Q_c \):
\[ Q_c = \frac{(0.6)(0.6)}{0.4} = \frac{0.36}{0.4} = 0.9 \]

### Step 9: Compare \( Q_c \) with \( K_c \)
Since \( Q_c > K_c \), the reaction will shift to the left to reach equilibrium. Let \( y \) be the change in concentration:
- New \([A] = 0.4 + y\)
- New \([B] = 0.6 - y\)
- New \([C] = 0.6 - y\)

Setting up the equilibrium expression:
\[ K_c = \frac{(0.6 - y)(0.6 - y)}{0.4 + y} = 0.45 \]

### Step 10: Solve for \( y \)
Expanding and rearranging gives:
\[ (0.6 - y)^2 = 0.45(0.4 + y) \]
\[ 0.36 - 1.2y + y^2 = 0.18 + 0.45y \]
\[ y^2 - 1.65y + 0.18 = 0 \]

Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1.65 \pm \sqrt{(1.65)^2 - 4(1)(0.18)}}{2(1)} \]
Calculating gives:
\[ y \approx 0.12 \]

### Step 11: Calculate new concentrations
- New \([A] = 0.4 + 0.12 = 0.52 \, \text{mol/L}\)
- New \([B] = 0.6 - 0.12 = 0.48 \, \text{mol/L}\)
- New \([C] = 0.6 - 0.12 = 0.48 \, \text{mol/L}\)

### Final Results
1. **When the volume is doubled:**
- \([A] = 0.072 \, \text{mol/L}\)
- \([B] = 0.178 \, \text{mol/L}\)
- \([C] = 0.178 \, \text{mol/L}\)

2. **When the volume is halved:**
- \([A] = 0.52 \, \text{mol/L}\)
- \([B] = 0.48 \, \text{mol/L}\)
- \([C] = 0.48 \, \text{mol/L}\)