1) the value of Kb after the decomposition of sulfur trioxide (S03). 2SO3 (g)--2SO2 (g) +O2 (g) is 1.79 x10^-5 at 350 degree celcius. if SO3 initially at 0.200 atm pressure comes to equilibrium in an otherwise empty tank at 350 degree celcius, determine the partial pressure of O2 at equilibrium.

Did you make a typo? Is Kb supposed to be Kp?

sorry it's Kp, it's a typo

To determine the partial pressure of O2 at equilibrium, we need to calculate the equilibrium constant (Kb) and use it in the equilibrium expression for the reaction.

Step 1: Calculate the equilibrium constant (Kb)
From the given reaction equation, 2SO3 (g) → 2SO2 (g) + O2 (g)
The balanced equation implies that Kb = [SO2]^2 * [O2] / [SO3]^2

Given Kb = 1.79 × 10^(-5)

Step 2: Set up an ICE table
We will set up an ICE (Initial, Change, Equilibrium) table to track the changes in the number of moles of each substance during the reaction.

Initial:
SO3: 0.200 atm
SO2: 0 atm (empty tank)
O2: 0 atm (empty tank)

Change:
SO3: -2x (since 2 moles of SO3 will react for every mole of O2 produced)
SO2: +2x (due to the reaction stoichiometry)
O2: +x

Equilibrium:
SO3: 0.200 - 2x
SO2: 2x
O2: x

Step 3: Substitute the equilibrium concentrations into Kb expression
Using the equilibrium concentrations from the ICE table, we substitute them into the equilibrium expression for Kb:

Kb = (SO2)^2 * (O2) / (SO3)^2
1.79 × 10^(-5) = (2x)^2 * x / (0.200 - 2x)^2

Step 4: Solve for x
Rearrange the equation to solve for x:

(2x)^2 * x / (0.200 - 2x)^2 = 1.79 × 10^(-5)

Now, we can solve this equation for x. This can be done either algebraically or using numerical methods such as iteration or graphical methods.

Step 5: Calculate the partial pressure of O2 at equilibrium
Once you have obtained the value of x, substitute it back into the equilibrium expressions to find the equilibrium concentrations of each species.

O2 = x

Substituting the calculated value of x will give you the partial pressure of O2 at equilibrium.