NH_4 Hs(s) NH_3 (g) + H_2 S(g) is an endothermic process. A 6.1589-g sample of t

Question

NH_4 Hs(s) NH_3 (g) + H_2 S(g) is an endothermic process. A 6.1589-g sample of the solid is placed in an evacuated 4.000-L vessel at exactly 24 degree C. After equilibrium has been established, the total pressure inside is 0.709 atm. Some solid NH_4 HS remains in the vessel. (a) What is the K_P for the reaction? (b) What percentage of the solid has decomposed? % (c) 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

mol os NH4HS = mass/MW = 6.1589/51.1114 = 0.120499 mol

V = 4 L

T = 24°C = 297K

P-equilibrium = 0.709 atm

initially

PNH3 = 0

P-H2S = 0

in equlibrium

P-NH3 = 0 + x

P-H2S = 0 + x

total P = P-NH3 +P-H2S = x+x = 2x

so

x = 0.709/2 = 0.3545

a)

Kp = P-NH3 * P-H2S = 0.3545*0.3545 = 0.12567

b)

% solid decomposition:

mol of NH3 = 0.3545 atm

PV = nRT

n = PV/(RT) = (0.3545)(4)/(0.082*297) = 0.0582 mol

ratio is 1:1

0.0582 mol have decomposed

%decomposition = (0.120499 -0.0582) /0.120499 * 100% = 51.7

c)

if V is doubled, this decreases P, therefore, this favours more gas formation, so the solid must DECREASE

between 20-50% is the best answer

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