NH_4HS(s) NH_3(g) + H_2S(g) is an endothermic process. A 6.1589-g sample of the

Question

NH_4HS(s) NH_3(g) + H_2S(g) is an endothermic process. A 6.1589-g sample of the solid is placed in an evacuated 4.000-L vessel at exactly 24degreeAfter equilibrium has been established, the total pressure inside is 0.709 atm. Some solid NH_4HS remains in the vessel. What is the K_P for the reaction? What percentage of the solid has decomposed? If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel? More than 50% would remain. Between 20% and 50% would remain. Between 10% and 20% would remain. Less than 10% would remain.

Explanation / Answer

consider the given reaction

NH4HS (s) ---. NH3 + H2S

Kp = [pNH3] [pH2S]

we know that solids are not considered for the expression of Kp

now

initially no NH3 and H2S are present

now

using ICE table

initial pressure of NH3 , H2S are 0 , 0

change in pressure of NH3 , H2S are +y , +y

equilibrium pressure of NH3 , H2S are y , y

now

total pressure = pNH3 + pH2S

0.709 = y + y = 2y

y = 0.3545

now

Kp = [NH3] [H2S]

Kp = [y] [y]

Kp = [0.3545] [0.3545]

Kp = 0.12567

so

the value of Kp is 0.12567

b)

now

for gases

PV = nRT

so for NH3

0.3545 x 4 = n x 0.0821 x 297

n = 0.058153602

so

moles of NH3 at equilibrium = 0.058153602

now

NH4HS ---> NH3 + H2S

we can see that

moles of NH3 formed = moles of NH4HS decomposed = 0.058153602

now

initial moles of NH4HS = mass / molar mass

= 6.1589 / 51

initial moles of NH4HS = 0.120762745

now

% NH4HS decomposed = 0.058153602 x 100 / 0.120762745 = 48.15

so

48.15 % of the solids has decomposed

c)

according to Le chatlier principle

the equilibrium will shift in a direction to counter the change

now

volume is doubled

so the equilibrium will shift in a direction to reduce the volume

NH4HS (s) ---> NH3 + H2S

to reduced , the equilibrium should shift towards the reactants

because the products NH3, H2S are gases and gases have higher volume

So

more NH4HS will be formed and the % decomposition will be less than 48%

so

more than 50% of solid would remain

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