Match the following :
`{:("Reaction (Homogeneous gaseous phase)","Degree of dissociation in terms of equilibrium constant"),((A)A(g)+B(g)hArr2C(g),(p)(sqrtk)//(1+sqrtk)),((B)2A(g)hArr+B(g)+C(g),(q)(sqrtk)//(2+sqrtk)),((C )A(g)+B(g)hArrC(g)+D(g),(r)2k//(1+2k)),((D)AB(g)hArrA/2(g)+B/2(g),(s)(2sqrtk)/(1+2sqrtk)),(,(t)k):}`

Explanation

To solve the problem of matching the reactions with their corresponding degree of dissociation in terms of equilibrium constant, we will analyze each reaction step by step.

### Step 1: Analyze Reaction A
**Reaction:** A(g) + B(g) ⇌ 2C(g)

1. **Initial Concentrations:**
- [A] = C₀
- [B] = C₀
- [C] = 0

2. **At Equilibrium:**
- [A] = C₀(1 - α)
- [B] = C₀(1 - α)
- [C] = 2C₀α

3. **Equilibrium Constant (K):**
\[ K = \frac{[C]^2}{[A][B]} = \frac{(2C₀α)^2}{C₀(1 - α)C₀(1 - α)} = \frac{4C₀^2α^2}{C₀^2(1 - α)^2} \]
Simplifying gives:
\[ K = \frac{4α^2}{(1 - α)^2} \]

4. **Expressing α in terms of K:**
\[ 2α = \sqrt{K}(1 - α) \implies 2α + α\sqrt{K} = \sqrt{K} \implies α(2 + \sqrt{K}) = \sqrt{K} \]
\[ α = \frac{\sqrt{K}}{2 + \sqrt{K}} \]

**Match:** Reaction A matches with (q) \(\frac{\sqrt{K}}{2 + \sqrt{K}}\).

### Step 2: Analyze Reaction B
**Reaction:** 2A(g) ⇌ B(g) + C(g)

1. **Initial Concentrations:**
- [A] = C₀
- [B] = 0
- [C] = 0

2. **At Equilibrium:**
- [A] = C₀(1 - 2α)
- [B] = C₀α
- [C] = C₀α

3. **Equilibrium Constant (K):**
\[ K = \frac{[B][C]}{[A]^2} = \frac{C₀α \cdot C₀α}{(C₀(1 - 2α))^2} = \frac{C₀^2α^2}{C₀^2(1 - 2α)^2} \]
Simplifying gives:
\[ K = \frac{α^2}{(1 - 2α)^2} \]

4. **Expressing α in terms of K:**
\[ \sqrt{K} = \frac{α}{1 - 2α} \implies α = \sqrt{K}(1 - 2α) \implies α + 2α\sqrt{K} = \sqrt{K} \]
\[ α(1 + 2\sqrt{K}) = \sqrt{K} \implies α = \frac{\sqrt{K}}{1 + 2\sqrt{K}} \]

**Match:** Reaction B matches with (s) \(\frac{2\sqrt{K}}{1 + 2\sqrt{K}}\).

### Step 3: Analyze Reaction C
**Reaction:** A(g) + B(g) ⇌ C(g) + D(g)

1. **Initial Concentrations:**
- [A] = C₀
- [B] = C₀
- [C] = 0
- [D] = 0

2. **At Equilibrium:**
- [A] = C₀(1 - α)
- [B] = C₀(1 - α)
- [C] = C₀α
- [D] = C₀α

3. **Equilibrium Constant (K):**
\[ K = \frac{[C][D]}{[A][B]} = \frac{C₀α \cdot C₀α}{C₀(1 - α)C₀(1 - α)} = \frac{α^2}{(1 - α)^2} \]

4. **Expressing α in terms of K:**
\[ \sqrt{K} = \frac{α}{1 - α} \implies α = \sqrt{K}(1 - α) \implies α + α\sqrt{K} = \sqrt{K} \]
\[ α(1 + \sqrt{K}) = \sqrt{K} \implies α = \frac{\sqrt{K}}{1 + \sqrt{K}} \]

**Match:** Reaction C matches with (p) \(\frac{α}{1 + α}\).

### Step 4: Analyze Reaction D
**Reaction:** AB(g) ⇌ A/2(g) + B/2(g)

1. **Initial Concentrations:**
- [AB] = C₀
- [A] = 0
- [B] = 0

2. **At Equilibrium:**
- [AB] = C₀(1 - α)
- [A] = C₀α/2
- [B] = C₀α/2

3. **Equilibrium Constant (K):**
\[ K = \frac{[A/2][B/2]}{[AB]} = \frac{(C₀α/2)(C₀α/2)}{C₀(1 - α)} = \frac{C₀^2α^2/4}{C₀(1 - α)} = \frac{C₀α^2}{4(1 - α)} \]

4. **Expressing α in terms of K:**
\[ K = \frac{α^2}{4(1 - α)} \implies 4K(1 - α) = α^2 \implies α^2 + 4Kα - 4K = 0 \]
Using the quadratic formula:
\[ α = \frac{-4K \pm \sqrt{(4K)^2 + 16K}}{2} = \frac{-4K \pm 4\sqrt{K}}{2} \]
Simplifying gives:
\[ α = \frac{2\sqrt{K}}{1 + 2K} \]

**Match:** Reaction D matches with (r) \(\frac{2K}{1 + 2K}\).

### Final Matches:
- A matches with (q)
- B matches with (s)
- C matches with (p)
- D matches with (r)