Question

Ammonium hydrogen sulphide dissociates as follows: [tex]NH_{4}HS(s) \rightleftharpoons NH_{3}(g)+H_{2}S(g), if the observed pressure of mixture is 1.12 atm. at 106°C. Find out the value of Kp for it.

Answer 1

While calculating ${ K }_{ c }\quad or\quad { K }_{ p }$ the value of the concentration is taken as 1 for the solid.

“Dalton law” of “Partial Pressure” states that the Observed Pressure is equal to the “sum of Pressure” of the gases.

Given, Observed Pressure = 1.12 atm

“Pressure of Ammonia” + “Pressure of Hydrogen sulphide” = Observed Pressure

“Pressure of Ammonia” + “Pressure of Hydrogen sulphide” = 1.12 atm

$\therefore Pressure\quad of\quad Ammonia = Pressure\quad of\quad Hydrogen\quad sulfide =\quad \frac { 1.12 }{ 2 } \quad atm\quad =\quad 0.56\quad atm$

${ K }_{ p }\quad =\quad \left( { NH }_{ 3 } \right) \quad \times \quad \left( { H }_{ 2 }S \right)$

$\Rightarrow \quad { K }_{ p }\quad =\quad 0.56\quad \times \quad 0.56$

$\therefore \quad { K }_{ p }\quad =\quad 0.3136\quad { atm }^{ 2 }$