Kp & partial pressure question... PLEASE HELP!?
NH4HS(s) NH3(g) + H2S(g)
(a) 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?
(b) To this system, sufficient H2S(g) is injected until the pressure of H2S is four times that of the ammonia at equilibrium. What are the partial pressures of NH3 and H2S?
partial pressure of NH3=___________atm
partial pressure of H2S=___________atm
(c) In a different experiment, 0.75 atm of NH3 and 0.5 atm of H2S are introduced into a 1.00 L vessel at 25°C. How many moles of NH4HS are present when equilibrium is established?
______________mol
Answer:
In general the equilibrium constant is defined in terms of activities. It is given by the product oft the product activities raised to the power their stoichiometric coefficient, divided by the product of the reactants raised to the power of their stoichiometric coefficient.
For this reaction:
K = a(NH₃) · a(H₂S) / a(NH₄HS)
For a component which occurs only in a pure phase like NH₄HS as solid the activity equals unity:
a(NH₄HS)= 1 → K = a(NH₃) · a(H₂S)
Because for ideal gases the activity is proportional to the partial pressure, you can express the equilibrium in terms of partial pressures:
Kp = p(NH₃) · p(H₂S)
(a)
The gas phase of the equilibrium contains only NH₃, H₂S which are formed by the chemical reaction. From the reaction equation you can see the the amounts N(NH₃) and N(H₂S) are the same. Since the partial pressure is defined as follows (ideal gas law):
p(X) = N(X) · R · T / V
the partial pressure must be the same. According to Dalton's law the partial pressures of all components add up to the total pressure:p = p(NH₃) + p(H₂S)
Therefore
p(NH₃) = p(H₂S) = p / 2 = 0.33atm
and
Kp = (0.33atm)² = 0.1089 atm²
(b)
p(NH₃) = 4 p(H₂S)
→
Kp = p(NH₃) · p(H₂S) = 4 · p²(H₂S)
→
p(H₂S) = sqrt(Kp/4) = sqrt(0.1089atm²/4) = 0.165atm
→
p(NH₃) = 4 p(H₂S) = 0.66atm
(c)
x be the amount of NH₄HS formed in equilibrium. Then the amount of the other compounds are:
N(NH₃) = N₀(NH₃) - x
N(H₂S) = N₀(H₂S) - x
The partial pressures are:
p(NH₃) = N(NH₃) · R · T / V = p₀(NH₃) - x·R·T/V
p(H₂S) = N(H₂S) · R · T / V = p₀(H₂S) - x·R·T/V
(₀ denotes the initial amounts, pressures)
Therefore:
Kp = p(NH₃) · p(H₂S) = (p₀(NH₃) - x·R·T/V) · (p₀(H₂S) - x·R·T/V)
←→
(x·R·T/V)² - [(p₀(NH₃) + (p₀(H₂S)]· (x·R·T/V) + p₀(NH₃)· (p₀(H₂S) - Kp = 0
Solve this quadratic equation for (x·R·T/V):
(x·R·T/V) = [(p₀(NH₃) + (p₀(H₂S)] - sqrt([(p₀(NH₃) + (p₀(H₂S)]²/4 - p₀(NH₃) · (p₀(H₂S) + Kp) = 90.9006kPa
→
x = 90.9006kPa · 1L / (8.314472kPaL/(molK) · 298.15K) = 0.03667mol
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