The decomposition of solid ammonium hydrogen sulfide to form ammonia gas and hyd

Question

The decomposition of solid ammonium hydrogen sulfide to form ammonia gas and hydrogen sulfide gas is an endothermic process.A 6.1589-g sample of the solid is placed in an evacuated 4.000-L vessel at 24.00C.After equilibrium is established, the total pressure inside the vessel is 0.709 bar and some of the solid remains in the vessel.

a.Calculate KP for the reaction.

b.Calculate the percent of the solid that has decomposed.

c.If the system was at equilibrium and the volume of the container were to double under isothermal conditions, what would happen to the amount of solid remaining in the vessel?Explain in mathematical terms using the reaction quotient, Q.

d.What would happen to the system at equilibrium if the temperature were to increase?Explain.

Explanation / Answer

NH4SH(s) <----> NH3(g) + H2S(g)

No of moles of NH4SH = 6.1589/51.111 = 0.12 mole.

total pressure Ptotal = 0.709 bar

at equilibrium

No of moles of NH3,H2S is in same.so that

partial pressure of each gas = 0.709/2 = 0.3545 bar

Kp = PNH3*PH2S

    = 0.3545*0.3545

    = 0.126 bar^2

b.

No of moles of NH3 liberated = PV/RT

P = 0.3545 bar = 0.35 atm

V = 4 L

T = 297 k

R = 0.0821 l.atm.k-1.mol-1

n = 0.35*4/(0.0821*297)

= 0.057 mole

mass of NH4SH decomposed = 0.057*51.11 = 2.913 grams

so that,

No of moles of NH4SH undecomposed = 0.12-0.057 = 0.063 mol

mass of NH4SH undecomposed = 0.063*51.11 = 3.22 grams

percent of solid decomposed = 2.913/6.1589*100 = 47.3%

c. if the volume doubled , pressure becomes half at equilibrium.

so that , solid NH4SH decomposes to NH3,H2S.

D. AS the process is endothermic,by increasing temperature decomposition of

NH4SH increases, so that equilibrium shifts to right.

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