1. Find my mistakes and fix the problem! In solving the following equilibrium problem, I made 4 mistakes. Identify each mistake and solve the problem correctly on page 2. Please ignore any math and/or sig fig errors. Thanks for your help! Problem: 2 mol AB3, 1 mol A2, and 3 mol B2 are sealed in a 10-L flask at 298 K. Calculate the total pressure (atm) in the flask at equilibrium. 2AB3(g) = A2(9) + 3B2(g) Kp = 1.750 at 298 K Problem Solution ZAB3 (g) = Azlg) + 3B2!8) T atm I 2 atm 3 atm ー× ー× 1-X 3-X E 2+X (2) PAes Pp (ital) = 0.666 %3D (Pas)(Pes) CD(3) Since Pp < Kp, the reaction left. Solve goos equatton for X Kip= (Paz) (Pas) (2+x) > |1.750 = (1-x)(3-X) X= 0, 45 atm Total pressure at equilibrium (2+x) + (I-x) + (3-x) - 6-X Prea + PAz + Paz Total pressure f atm - 0.45 atm -4.55 atm %3D %3D

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Please show the correct way to get total pressure and the 4 mistakes noticed 

1. Find my mistakes and fix the problem! In solving the following equilibrium problem, I made 4
mistakes. Identify each mistake and solve the, problem correctly on page 2. Please ignore any math
and/or sig fig errors. Thanks for your help!
Problem: 2 mol AB3, 1 mol A2, and 3 mol B2 are sealed in a 10-L flask at 298 K. Calculate the total
pressure (atm) in the flask at equilibrium.
2AB3(g) = A2(9) + 3B2(g)
Kp = 1.750 at 298 K
Problem Solution
ZAB3 (g) = Azlg) + 3B2lg)
T atm
エ
2 atm
3 atm
ー×
ー×
1-X
3-X
2+X
(2)
PAes
Pp (mital) =
0.666
(Pas)(Pes) CD(3)
left.
Solve
Since Pp < Kp, the reaction
goes
equatton
for X
Kip=
(Paz) (Pas)
(2+X)
(1-x)(3-X)
> |1.750 =
X= 0, 45 atm
Total pressure
at equilibrium
(2+x) + (1-x) + (3-x) = 6-x
Paz
PA +
Prea +
Total pressure
at equilibrium
6 atm
0.45 atm
4.55 atm
%3D
%3D
Transcribed Image Text:1. Find my mistakes and fix the problem! In solving the following equilibrium problem, I made 4 mistakes. Identify each mistake and solve the, problem correctly on page 2. Please ignore any math and/or sig fig errors. Thanks for your help! Problem: 2 mol AB3, 1 mol A2, and 3 mol B2 are sealed in a 10-L flask at 298 K. Calculate the total pressure (atm) in the flask at equilibrium. 2AB3(g) = A2(9) + 3B2(g) Kp = 1.750 at 298 K Problem Solution ZAB3 (g) = Azlg) + 3B2lg) T atm エ 2 atm 3 atm ー× ー× 1-X 3-X 2+X (2) PAes Pp (mital) = 0.666 (Pas)(Pes) CD(3) left. Solve Since Pp < Kp, the reaction goes equatton for X Kip= (Paz) (Pas) (2+X) (1-x)(3-X) > |1.750 = X= 0, 45 atm Total pressure at equilibrium (2+x) + (1-x) + (3-x) = 6-x Paz PA + Prea + Total pressure at equilibrium 6 atm 0.45 atm 4.55 atm %3D %3D
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