At 25°C in a closed system, ammonium hydrogen sulfide exists as the following equilibrium. NH4HS(s)<--->NH3(g) + H2S(g) When a sample of pure NH4HS(s) is placed in an evacuated reaction vessel and allowed to come to equilibrium at 25°C, total pressure is 0.660 atm. What is the value of Kp?

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Answer 1 · 2014-09-28T00:27:17

$K_{p} = 0.11 A t m^{2}$ $K_p=0.11Atm^2$

Start by writing down the system which is at equilibrium:

$N H_{4} H S_{(s)} ⇌ N H_{3 (g)} + H_{2} S_{(g)}$ $NH_4HS_((s))rightleftharpoonsNH_(3(g))+H_2S_((g))$

The solid reactant has zero partial pressure

So $K_{p}$ $K_p$ = $P_{N H_{3}} . P_{H_{2} S}$ $P_(NH_(3)).P_(H_2S)$

Where $P$ $P$ denotes the partial pressure of each gas.

$P_{N H_{3}} + P_{H_{2} S} = 0.66 A t m$ $P_(NH_(3))+P_(H_2S)=0.66Atm$

So $P_{N H_{3}} = 0.33 A t m$ $P_(NH_(3))=0.33Atm$

and $P_{H_{2} S} = 0.33 A t m$ $P_(H_2S)=0.33Atm$

since they are formed in a 1:1 ratio

So

$K_{p} = {(0.33)}^{2} = 0.11 A t m^{2}$ $K_p=(0.33)^2=0.11Atm^2$