Answer the following questions regarding the decomposition of arsenic pentafluoride, AsFs(g). (a) A 55.8 g sample of AsF3(g) is introduced into an evacuated 10.5 L container at 105°C. i. What is the initial molar concentration of AsFs(g) in the container? No. of males Malarity Thus At 105°C, AsFs(g) decomposed equation. ii. What is the initial pressure, in atmospheres, of the AsF,(g) in the container? Mais x RT ⇒P= MM V Volumu (int) Кс 55-88 169.91368/mal x10.SL =0-0313 M ⇒P= 55.88 169-91368/mal => P=0.970 atm = ·X 0-08206 L-atm x 378k K-mal 10-SL Initial pressure = 0.970 atm into AsF3(g) and F₂(g) according to the following chemical AsF5(g) < >AsF3(g) + F₂(g) (b) In terms of molar concentrations, write the equilibrium-constant expression for the decomposition of AsF,(g). [As F3] [F₂] [As F5] (c) When equilibrium is established, 27.7 percent of the original number of moles of AsF; (g) has decomposed. i. Calculate the molar concentrations of AsFs(g) at equilibrium. Moles of AsFs at equilibrium=(1-0.277) × molcs of AsFs Molar concentration of AsFs at eq- moles of AsP, at oq Volume = 0.723 x 0.328 mol -0.237 moles ii. Using molar concentrations, calculate the value of the equilibrium constant, Kat 105°C. 0.237 mol 10.5 L = 0.0226 M =
Answer the following questions regarding the decomposition of arsenic pentafluoride, AsFs(g). (a) A 55.8 g sample of AsF3(g) is introduced into an evacuated 10.5 L container at 105°C. i. What is the initial molar concentration of AsFs(g) in the container? No. of males Malarity Thus At 105°C, AsFs(g) decomposed equation. ii. What is the initial pressure, in atmospheres, of the AsF,(g) in the container? Mais x RT ⇒P= MM V Volumu (int) Кс 55-88 169.91368/mal x10.SL =0-0313 M ⇒P= 55.88 169-91368/mal => P=0.970 atm = ·X 0-08206 L-atm x 378k K-mal 10-SL Initial pressure = 0.970 atm into AsF3(g) and F₂(g) according to the following chemical AsF5(g) < >AsF3(g) + F₂(g) (b) In terms of molar concentrations, write the equilibrium-constant expression for the decomposition of AsF,(g). [As F3] [F₂] [As F5] (c) When equilibrium is established, 27.7 percent of the original number of moles of AsF; (g) has decomposed. i. Calculate the molar concentrations of AsFs(g) at equilibrium. Moles of AsFs at equilibrium=(1-0.277) × molcs of AsFs Molar concentration of AsFs at eq- moles of AsP, at oq Volume = 0.723 x 0.328 mol -0.237 moles ii. Using molar concentrations, calculate the value of the equilibrium constant, Kat 105°C. 0.237 mol 10.5 L = 0.0226 M =
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter14: Chemical Equilibrium
Section: Chapter Questions
Problem 92AP
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I have been asking this question in parts. The solutions I have received from other Bartleby asks are included. I only need help with the last part. I am assuming you substitute into the Kc expression, but I don't understand where the other concentrations come from.
![Answer the following questions regarding the decomposition of arsenic pentafluoride,
AsFs(g).
(a) A 55.8 g sample of AsF3(g) is introduced into an evacuated 10.5 L container at
105°C.
i. What is the initial molar concentration of AsF-(g) in the container?
No. of males
Malarity
=
Thus
Valumu (int)
55-88
169.91368/mal x10.5L
= 0.0313 M
ii. What is the initial pressure, in atmospheres, of the AsF,(g) in the
container?
=> P= Mass x RT
·X
=> P=
Kc =
55.88
169-91368/mal
=> P= 0.970 atm
Initial pressure = 0.970 atm
At 105°C, AsF5(g) decomposed into AsF3(g) and F₂(g) according to the following chemical
equation.
0-08206 L-atm x 378k
k-mal
10-SL
AsF5(g) < >AsF3(g) + F₂(g)
(b) In terms of molar concentrations, write the equilibrium-constant expression for
the decomposition of AsF;(g).
[As F3] [F₂]
[As F5 ]
(c) When equilibrium is established, 27.7 percent of the original number of moles of
AsF,(g) has decomposed.
i. Calculate the molar concentrations of AsF5(g) at equilibrium.
Moles of AsFs at equilibrium= (1 -0.277) x molcs of AsFs
= 0.723 x 0.328 mol
= 0.237 moles
Molar concentration of AsFs at eq- moles of AsF, at oq
Volume
0.237 mol
10.5 L
= 0.0226 M
ii. Using molar concentrations, calculate the value of the equilibrium
constant, K₂, at 105°C.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F279bdcd0-8ae3-4db7-9d9a-edb02bba6281%2F13121023-93be-46e7-9f73-49e7d1fa8c03%2Fjtr1ezp_processed.png&w=3840&q=75)
Transcribed Image Text:Answer the following questions regarding the decomposition of arsenic pentafluoride,
AsFs(g).
(a) A 55.8 g sample of AsF3(g) is introduced into an evacuated 10.5 L container at
105°C.
i. What is the initial molar concentration of AsF-(g) in the container?
No. of males
Malarity
=
Thus
Valumu (int)
55-88
169.91368/mal x10.5L
= 0.0313 M
ii. What is the initial pressure, in atmospheres, of the AsF,(g) in the
container?
=> P= Mass x RT
·X
=> P=
Kc =
55.88
169-91368/mal
=> P= 0.970 atm
Initial pressure = 0.970 atm
At 105°C, AsF5(g) decomposed into AsF3(g) and F₂(g) according to the following chemical
equation.
0-08206 L-atm x 378k
k-mal
10-SL
AsF5(g) < >AsF3(g) + F₂(g)
(b) In terms of molar concentrations, write the equilibrium-constant expression for
the decomposition of AsF;(g).
[As F3] [F₂]
[As F5 ]
(c) When equilibrium is established, 27.7 percent of the original number of moles of
AsF,(g) has decomposed.
i. Calculate the molar concentrations of AsF5(g) at equilibrium.
Moles of AsFs at equilibrium= (1 -0.277) x molcs of AsFs
= 0.723 x 0.328 mol
= 0.237 moles
Molar concentration of AsFs at eq- moles of AsF, at oq
Volume
0.237 mol
10.5 L
= 0.0226 M
ii. Using molar concentrations, calculate the value of the equilibrium
constant, K₂, at 105°C.
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