Asked by Anonymous
At 900 K the following reaction has Kp=0.345; 2 SO2(g) + O2 (g) -> 2 SO3 (g)
In an equilibrium mixture the partial pressures of SO2 and O2 are 0.145 atm and 0.455 atm, respectively.
What is the equilibrium partial pressure of SO3 in the mixture?
In an equilibrium mixture the partial pressures of SO2 and O2 are 0.145 atm and 0.455 atm, respectively.
What is the equilibrium partial pressure of SO3 in the mixture?
Answers
Answered by
DrBob222
Write the equilibrium constant expression, K, for the reaction.Then substitute the equilibrium pressures of SO2 and O2 and solve for SO3 partial pressure.
Answered by
Anonymous
So then:
.345 = [SO3]^2 / (.145)^2 (.455)
Equilibrium partial pressure of SO3 = .057
Thank you so much!
.345 = [SO3]^2 / (.145)^2 (.455)
Equilibrium partial pressure of SO3 = .057
Thank you so much!
Answered by
DrBob222
Same number I have. You're welcome.
Answered by
Mon
4.34 = 0.391^2 / 0.125 / [SO2]^2
what is the concentration of SO2?
what is the concentration of SO2?
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