Sulfur trioxide decomposes at high temperature in a sealed container: 2 SO3(g) 2 SO2(g) + O2(g) Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate K, at 1000 K. SO3 (atm) 0.500 SO₂ (atm) 0.00 02 (atm) Initial Change Equilibrium - 2x 0.200 + 2x 2x 0.00 + X X Initial Since we're given the equilibrium partial pressure of SO3, WE THUS KNOW: 0.500 - 2x = 0.200. Therefore, x = 0.150 Fill in the table accordingly! SO3 (atm) 0.500 Change - 0.300 SO2 (atm) 0.00 + 0.300 O2 (atm) 0.00 + 0.150 Equilibrium 0.200 0.300 0.150 Thus: Kp = Ps02 P02 PS03 2 K = (0.300)2(0.150) (0.200) 2 = 0.338
Sulfur trioxide decomposes at high temperature in a sealed container: 2 SO3(g) 2 SO2(g) + O2(g) Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate K, at 1000 K. SO3 (atm) 0.500 SO₂ (atm) 0.00 02 (atm) Initial Change Equilibrium - 2x 0.200 + 2x 2x 0.00 + X X Initial Since we're given the equilibrium partial pressure of SO3, WE THUS KNOW: 0.500 - 2x = 0.200. Therefore, x = 0.150 Fill in the table accordingly! SO3 (atm) 0.500 Change - 0.300 SO2 (atm) 0.00 + 0.300 O2 (atm) 0.00 + 0.150 Equilibrium 0.200 0.300 0.150 Thus: Kp = Ps02 P02 PS03 2 K = (0.300)2(0.150) (0.200) 2 = 0.338
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter12: Gaseous Chemical Equilibrium
Section: Chapter Questions
Problem 56QAP: Sulfur oxychloride, SO2Cl2, decomposes to sulfur dioxide and chlorine gases. SO2Cl2(g)SO2(g)+Cl2(g)...
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Question
Can you explain how they get x=150 and the rest of the problem? Please
![Sulfur trioxide decomposes at high temperature in a sealed container:
2 SO3(g) 2 SO2(g) + O2(g)
Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm.
At equilibrium the SO3 partial pressure is 0.200 atm. Calculate K, at 1000 K.
SO3 (atm)
0.500
SO₂ (atm)
0.00
02 (atm)
Initial
Change
Equilibrium
- 2x
0.200
+ 2x
2x
0.00
+ X
X
Initial
Since we're given the equilibrium partial pressure of SO3, WE THUS KNOW:
0.500 - 2x = 0.200. Therefore, x = 0.150 Fill in the table accordingly!
SO3 (atm)
0.500
Change
- 0.300
SO2 (atm)
0.00
+ 0.300
O2 (atm)
0.00
+ 0.150
Equilibrium
0.200
0.300
0.150
Thus: Kp
=
Ps02 P02
PS03
2
K = (0.300)2(0.150)
(0.200) 2
= 0.338](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F637a8243-98d2-45ca-8dd1-4e9d7d0f97ca%2F89fa9ba1-f49e-4d8d-8a6e-3c4af523641f%2Fdgh6rc.jpeg&w=3840&q=75)
Transcribed Image Text:Sulfur trioxide decomposes at high temperature in a sealed container:
2 SO3(g) 2 SO2(g) + O2(g)
Initially, the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm.
At equilibrium the SO3 partial pressure is 0.200 atm. Calculate K, at 1000 K.
SO3 (atm)
0.500
SO₂ (atm)
0.00
02 (atm)
Initial
Change
Equilibrium
- 2x
0.200
+ 2x
2x
0.00
+ X
X
Initial
Since we're given the equilibrium partial pressure of SO3, WE THUS KNOW:
0.500 - 2x = 0.200. Therefore, x = 0.150 Fill in the table accordingly!
SO3 (atm)
0.500
Change
- 0.300
SO2 (atm)
0.00
+ 0.300
O2 (atm)
0.00
+ 0.150
Equilibrium
0.200
0.300
0.150
Thus: Kp
=
Ps02 P02
PS03
2
K = (0.300)2(0.150)
(0.200) 2
= 0.338
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