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Sulphur 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 the value of Kp at 1000 K to 3 decimal places.
Kp=0.225
Sulphur 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 the value of Kp at 1000 K to 3 decimal places.
Kp=0.225
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Sulphur trioxide decomposes at high temperature in a sealed container: 2 SO3(g)...
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2) For the equilibrium: 2 SO3(g) <=> 02(g) + 2 SO2(g) Kp = 0.269 at 625...
2) For the equilibrium: 2 SO3(g) <=> 02(g) + 2 SO2(g) Kp = 0.269 at 625 °C What is Ke at this temperature? Kp = K[R R = 0.08206 L-atm/mol K (5pts)
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At a certain temperature K = 0.500 for the following reaction: SO3 (g) + NO (g) ⇌ NO2 (g) + SO2 (g) If 0.200 mol of SO3 and 0.200 mol of NO are placed in a 2.000 L container and allowed to come to equilibrium, what will be the concentration of SO2?
9. -10.1 points 0/4 Submissions Used Gaseous SO3 is placed in a closed container at 560...
9. -10.1 points 0/4 Submissions Used Gaseous SO3 is placed in a closed container at 560 °C, where it partially decomposes to SO2 and 02: 2 503(9) = 2 502(9) + 1 02(9) At equilibrium it is found that p(SO3) = 0.006960 atm, P(SO2) = 0.006580 atm, and p(O2) = 0.004150 atm. What is the value of Kp at this temperature? Kp =
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Sulfur dioxide and oxygen react to form sulfur trioxide, like this: 2SO2(g)+O2(g)→2SO3(g) Also, a chemist finds...
Sulfur dioxide and oxygen react to form sulfur trioxide, like
this: 2SO2(g)+O2(g)→2SO3(g)
Also, a chemist finds that at a certain temperature the
equilibrium mixture of sulfur dioxide, oxygen, and sulfur trioxide
has the following composition:
Calculate the value of the equilibrium constant Kp for this
reaction. Round your answer to 2 significant digits.
compound pressure at equilibrium SO2 58.3 atm 02 84.7 atm SO3 66.3 atm
For the equilibrium: 2 SO3(g) < = > O2(g) + 2 SO2(g) Kp = 0.269 at...
For the equilibrium: 2 SO3(g) < = > O2(g) + 2 SO2(g) Kp = 0.269 at 625 oC What is Kc at this temperature? Kp = Kc[RT]Δn R = 0.08206 L-atm/mol K
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The equilibrium constant, Kp, for the following reaction is 2.74 at 1.15x103K. 2803(g) 22502(g) + O2(g) + If an equilibrium mixture of the three gases in a 10.9 L container at 1.15*10²K contains SO3 at a pressure of 1.77 atm and SO2 at a pressure of 0.926 atm, the equilibrium partial pressure of O2 is atm.
2) For the equilibrium: 2 SO3(g) <=> O2(g) + 2 SO2(g) Kp = 0.269 at 625 °C What is Ke at this temperature? Kp = Kc[RT]An R = 0.08206 L-atm/mol K
2) For the equilibrium: 2 SO3(g) <=> 02(g) + 2 SO2(g) Kp = 0.269 at 625 °C What is Ke at this temperature? Kp = K[R R = 0.08206 L-atm/mol K (5pts)
9. -10.1 points 0/4 Submissions Used Gaseous SO3 is placed in a closed container at 560 °C, where it partially decomposes to SO2 and 02: 2 503(9) = 2 502(9) + 1 02(9) At equilibrium it is found that p(SO3) = 0.006960 atm, P(SO2) = 0.006580 atm, and p(O2) = 0.004150 atm. What is the value of Kp at this temperature? Kp =
Sulfur dioxide and oxygen react to form sulfur trioxide, like
this: 2SO2(g)+O2(g)→2SO3(g)
Also, a chemist finds that at a certain temperature the
equilibrium mixture of sulfur dioxide, oxygen, and sulfur trioxide
has the following composition:
Calculate the value of the equilibrium constant Kp for this
reaction. Round your answer to 2 significant digits.
compound pressure at equilibrium SO2 58.3 atm 02 84.7 atm SO3 66.3 atm
The equilibrium constant, Kp, for the following reaction is 2.74 at 1.15x103K. 2803(g) 22502(g) + O2(g) + If an equilibrium mixture of the three gases in a 10.9 L container at 1.15*10²K contains SO3 at a pressure of 1.77 atm and SO2 at a pressure of 0.926 atm, the equilibrium partial pressure of O2 is atm.
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