Ammonium sulphide and ammonium selenide on heating dissociates as
`(NH_4)S(s) hArr 2NH_(3)(g) + H_(2)S(g) , k_(p1) = 9 xx 10^(-3) atm^(3)`
`(NH_4)_(2) Se(s) hArr 2NH_(3)(g) + H_(2)Se(g), K_(p2) = 4.5 xx 10^(-3) atm^(3)`.
The total pressure over the solid mixture at equilibrium is

Options

A. 0.15 atm
B. 0.3 atm
C. 0.45 atm
D. 0.6 atm

Answer

To solve the problem, we need to analyze the dissociation of ammonium sulfide \((NH_4)S\) and ammonium selenide \((NH_4)_2Se\) when heated, and calculate the total pressure at equilibrium.

### Step 1: Write the dissociation reactions and equilibrium expressions

1. For ammonium sulfide:
\[ (NH_4)S(s) \rightleftharpoons 2NH_3(g) + H_2S(g) \]
The equilibrium constant expression \(K_{p1}\) is given by:
\[ K_{p1} = \frac{(P_{NH_3})^2 \cdot P_{H_2S}}{1} = (P_{NH_3})^2 \cdot P_{H_2S} \]

2. For ammonium selenide:
\[ (NH_4)_2Se(s) \rightleftharpoons 2NH_3(g) + H_2Se(g) \]
The equilibrium constant expression \(K_{p2}\) is given by:
\[ K_{p2} = \frac{(P_{NH_3})^2 \cdot P_{H_2Se}}{1} = (P_{NH_3})^2 \cdot P_{H_2Se} \]

### Step 2: Set up the equations using the given \(K_p\) values

1. For ammonium sulfide:
\[ K_{p1} = 9 \times 10^{-3} = (P_1)^2 \cdot P_{H_2S} \]

2. For ammonium selenide:
\[ K_{p2} = 4.5 \times 10^{-3} = (P_2)^2 \cdot P_{H_2Se} \]

### Step 3: Assume the partial pressures at equilibrium

Let:
- \(P_1\) be the partial pressure of \(NH_3\) from \((NH_4)S\)
- \(P_{H_2S}\) be the partial pressure of \(H_2S\)
- \(P_2\) be the partial pressure of \(NH_3\) from \((NH_4)_2Se\)
- \(P_{H_2Se}\) be the partial pressure of \(H_2Se\)

### Step 4: Solve for \(P_1\) and \(P_2\)

1. For ammonium sulfide:
\[ K_{p1} = (P_1)^2 \cdot P_{H_2S} \]
Assume \(P_{H_2S} = P_1\) (since both are produced in the same ratio):
\[ 9 \times 10^{-3} = (P_1)^2 \cdot P_1 \implies 9 \times 10^{-3} = P_1^3 \]
\[ P_1 = (9 \times 10^{-3})^{1/3} = 0.3 \text{ atm} \]

2. For ammonium selenide:
\[ K_{p2} = (P_2)^2 \cdot P_{H_2Se} \]
Assume \(P_{H_2Se} = P_2\) (since both are produced in the same ratio):
\[ 4.5 \times 10^{-3} = (P_2)^2 \cdot P_2 \implies 4.5 \times 10^{-3} = P_2^3 \]
\[ P_2 = (4.5 \times 10^{-3})^{1/3} = 0.15 \text{ atm} \]

### Step 5: Calculate the total pressure at equilibrium

The total pressure \(P_{total}\) is the sum of the partial pressures:
\[ P_{total} = P_1 + P_2 = 0.3 \text{ atm} + 0.15 \text{ atm} = 0.45 \text{ atm} \]

### Final Answer
The total pressure over the solid mixture at equilibrium is:
\[ \boxed{0.45 \text{ atm}} \]