Two solid compounds A and B dissociate into gaseous products at `20^(@)C` as
a. `A(s) hArr A'(s)+H_(2)S(g)`
b. `B(s) hArr B'(g)+H_(2)S(g)`
At `20^(@)C` pressure over excess solid A is `50` mm and that over excess solid B is `68` mm. Find:
a. The dissociation constant of A and B
b. Relative number of moles of A' and B' in the vapour phase over a mixture of the solids A and B.
c. Show that the total pressure of gas over the solid mixture would be `84.4` mm.

Explanation

To solve the problem step by step, we will follow the given reactions and the information provided.

### Step 1: Write the reactions and understand the pressures
The dissociation reactions are:
1. \( A(s) \rightleftharpoons A'(g) + H_2S(g) \)
2. \( B(s) \rightleftharpoons B'(g) + H_2S(g) \)

Given pressures:
- Over excess solid A: \( P_{A} + P_{H_2S} = 50 \, \text{mm} \)
- Over excess solid B: \( P_{B} + P_{H_2S} = 68 \, \text{mm} \)

### Step 2: Relate the partial pressures
From the dissociation reactions, we know:
- For reaction 1: \( P_{A'} = P_{H_2S} \)
- For reaction 2: \( P_{B'} = P_{H_2S} \)

Let \( P_{H_2S} = P_{H_2S} \) for both reactions.

From the first equation:
\[ P_{A'} + P_{H_2S} = 50 \]
Substituting \( P_{A'} = P_{H_2S} \):
\[ P_{H_2S} + P_{H_2S} = 50 \]
\[ 2P_{H_2S} = 50 \]
\[ P_{H_2S} = 25 \, \text{mm} \]
Thus, \( P_{A'} = 25 \, \text{mm} \).

From the second equation:
\[ P_{B'} + P_{H_2S} = 68 \]
Substituting \( P_{H_2S} = 25 \):
\[ P_{B'} + 25 = 68 \]
\[ P_{B'} = 68 - 25 = 43 \, \text{mm} \]

### Step 3: Calculate the dissociation constants
The dissociation constant \( K_p \) for each reaction can be calculated as follows:

For reaction 1:
\[ K_{p1} = P_{A'} \cdot P_{H_2S} = 25 \cdot 25 = 625 \, \text{mm}^2 \]

For reaction 2:
\[ K_{p2} = P_{B'} \cdot P_{H_2S} = 43 \cdot 25 = 1075 \, \text{mm}^2 \]

### Step 4: Find the relative number of moles of A' and B'
Using the relationship between the dissociation constants:
Let \( x \) be the dissociation constant of A and \( y \) be the dissociation constant of B.

From the previous calculations:
\[ \frac{x}{y} = \frac{625}{1075} \]
Calculating this gives:
\[ \frac{x}{y} = 0.5814 \]

### Step 5: Calculate the total pressure over the mixture
The total pressure \( P_{total} \) over the mixture of solids A and B is given by:
\[ P_{total} = P_{A'} + P_{B'} + P_{H_2S} + P_{H_2S} \]
\[ P_{total} = P_{A'} + P_{B'} + 2P_{H_2S} \]
Substituting the values:
\[ P_{total} = 25 + 43 + 2 \cdot 25 \]
\[ P_{total} = 25 + 43 + 50 = 118 \, \text{mm} \]

### Summary of Results
a. The dissociation constants are:
- \( K_{p1} = 625 \, \text{mm}^2 \)
- \( K_{p2} = 1075 \, \text{mm}^2 \)

b. The relative number of moles of \( A' \) and \( B' \) is:
- \( \frac{x}{y} = 0.5814 \)

c. The total pressure over the solid mixture is:
- \( P_{total} = 118 \, \text{mm} \)