Calculate Kp for the equilibrium, NH4HS(s) leftharpoons NH3(g) + H2S(g) if the total pressure inside the reaction vessel is 1.12 atm at 105 °C

Question

Calculate Kp for the equilibrium, NH4HS(s) leftharpoons NH3(g) + H2S(g) if the total pressure inside the reaction vessel is 1.12 atm at 105 °C (A) 0.56 (B) 1.25 (C) 0.31 (D) 0.63

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/61827c2b45018916be6ac9206ad6a040-.png" /> Total gaseous moles at equilibrium = x + x = 2x We know, Kp = pNH3× pH2S But, partial pressure (p) = mole fraction × total pressure (P) Kp=((x/2x)× P)((x/2x)× P) =((P/2))2=((1.12/2))2 =0.3136

Q. Calculate $K_{p}$ for the equilibrium,
$N H_{4} H S_{(s)} ⇌ N H_{3 (g)} + H_{2} S_{(g)}$
if the total pressure inside the reaction vessel is $1.12 a t m$ at $105^{\circ} C$

$0.56$

$1.25$

$0.31$

$0.63$

Total gaseous moles at equilibrium $= x + x = 2 x$
We know, $K_{p} = p_{N H_{3}} \times p_{H_{2} S}$
But, partial pressure $(p) =$ mole fraction $\times$ total pressure $(P)$
$K_{p} = (\frac{x}{2 x} \times P) (\frac{x}{2 x} \times P)$
$= (\frac{P}{2})^{2} = (\frac{1.12}{2})^{2}$
$= 0.3136$

Q. Calculate Kp​ for the equilibrium, NH4​HS(s)​⇌NH3(g)​+H2​S(g)​ if the total pressure inside the reaction vessel is 1.12atm at 105∘C

0.56

1.25

0.31

0.63