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

Question

NH_4HS(s) NH_3(g) + H_2S(g) is an exothermic process. A 6.1589-g sample of the solid is placed in an evacuated 4.000-L vessel at exactly 24degree C. After 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

a) The total pressure inside the vessel is = 0.709atmosphere

NH4HS(s) <-------à NH3 (g) + H2S (g)

NH4HS is a solid .So it does not appear in the Keq expression.

KP = (PNH3)(PH2S)

PNH3 = PH2S = 0.709/2 atm = 0.3545atm

KP = (0.3545)(0.3545)

= 0.125

Kp for the reaction is 0.125

b) % of solid has decomposed

Molar mass of NH4HS = 51

Initial moles of NH4HS = 6.1589/51 = 0.1207

Moles of NH3

PV = nRT

(0.3545 atm) (4.000L) = n (0.0821 L-atm/mol-K) (297 K)

n = 0.05815

0.05815 moles of NH3 was produced from an equal number of moles of NH4HS

(0.05815 moles had decomposed) / (0.1207 mol NH3HS initially) = 0.4818

% of solid has decomposed = 48.18%

c) If the volume of vessel were double at constant T

Then moles ofNH3 (0.3545 atm) (8.000L) = n (0.0821 L-atm/mol-K) (297 K)

n= 0.116moles

0.116 moles of NH3 was produced from an equal number of moles of NH4HS

(0.116 moles had decomposed) / (0.1207 mol NH3HS initially) = 0.9637

% of solid has decomposed = 96.37

The amount of solid remain in the vessel = 100 – 96.37 = 3.63%

Less than 10% is correct option

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