# Chemistry

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?

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1. Write the equilibrium constant expression, K, for the reaction.Then substitute the equilibrium pressures of SO2 and O2 and solve for SO3 partial pressure.

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2. So then:

.345 = [SO3]^2 / (.145)^2 (.455)

Equilibrium partial pressure of SO3 = .057

Thank you so much!

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3. Same number I have. You're welcome.

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4. 4.34 = 0.391^2 / 0.125 / [SO2]^2

what is the concentration of SO2?

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