Two solids X and Y dissociate into gaseous products at a certain temperature as follows:
i. `X(s) hArr A(g)+C(g)` and
ii. `Y(s) hArr B(g)+C(g)`
At a given temperature, pressure over excess solid 'X' is `40` mm of Hg and total pressure over solid 'Y(s)' is `60` mm of Hg.
Now, answer the following questions:
Ratio of `K_(p)` for reaction (i) to that of reaction (ii), is:

Options

A. `4:9`
B. `2:3`
C. `4:9`
D. `2:1`

Explanation

To solve the problem, we need to determine the ratio of the equilibrium constants \( K_p \) for the two reactions given. Let's break down the steps:

### Step 1: Understand the Reactions
We have two reactions:
1. \( X(s) \rightleftharpoons A(g) + C(g) \)
2. \( Y(s) \rightleftharpoons B(g) + C(g) \)

### Step 2: Analyze Reaction 1
For the first reaction, we start with 1 mole of solid \( X \) and none of \( A \) or \( C \). At equilibrium, let \( \alpha \) be the degree of dissociation of \( X \):
- Moles of \( X \) at equilibrium = \( 1 - \alpha \)
- Moles of \( A \) at equilibrium = \( \alpha \)
- Moles of \( C \) at equilibrium = \( \alpha \)

The total number of moles of gas at equilibrium = \( \alpha + \alpha = 2\alpha \).

### Step 3: Calculate Partial Pressures for Reaction 1
The total pressure over excess solid \( X \) is given as \( 40 \) mm Hg. The partial pressures of \( A \) and \( C \) can be expressed as:
- \( P_A = \frac{P_T}{2} = \frac{40}{2} = 20 \) mm Hg
- \( P_C = \frac{P_T}{2} = \frac{40}{2} = 20 \) mm Hg

### Step 4: Calculate \( K_{p1} \) for Reaction 1
The equilibrium constant \( K_{p1} \) for the first reaction is given by:
\[ K_{p1} = P_A \cdot P_C = 20 \cdot 20 = 400 \text{ mm}^2 \text{Hg}^2 \]

### Step 5: Analyze Reaction 2
For the second reaction, we start with 1 mole of solid \( Y \) and none of \( B \) or \( C \). At equilibrium, let \( \alpha' \) be the degree of dissociation of \( Y \):
- Moles of \( Y \) at equilibrium = \( 1 - \alpha' \)
- Moles of \( B \) at equilibrium = \( \alpha' \)
- Moles of \( C \) at equilibrium = \( \alpha' \)

The total number of moles of gas at equilibrium = \( \alpha' + \alpha' = 2\alpha' \).

### Step 6: Calculate Partial Pressures for Reaction 2
The total pressure over solid \( Y \) is given as \( 60 \) mm Hg. The partial pressures of \( B \) and \( C \) can be expressed as:
- \( P_B = \frac{P_T'}{2} = \frac{60}{2} = 30 \) mm Hg
- \( P_C = \frac{P_T'}{2} = \frac{60}{2} = 30 \) mm Hg

### Step 7: Calculate \( K_{p2} \) for Reaction 2
The equilibrium constant \( K_{p2} \) for the second reaction is given by:
\[ K_{p2} = P_B \cdot P_C = 30 \cdot 30 = 900 \text{ mm}^2 \text{Hg}^2 \]

### Step 8: Calculate the Ratio of \( K_{p1} \) to \( K_{p2} \)
Now, we find the ratio of \( K_{p1} \) to \( K_{p2} \):
\[ \frac{K_{p1}}{K_{p2}} = \frac{400}{900} = \frac{4}{9} \]

### Final Answer
The ratio of \( K_{p} \) for reaction (i) to that of reaction (ii) is:
\[ \boxed{\frac{4}{9}} \]