At 627ºC, sulfur dioxide and oxygen gases combine to form sulfur trioxide gas. At equilibrium, the concentrations for sulfur dioxide, oxygen, and sulfur trioxide gases are .0060M, .0054M, and .0032M, respectively.

a. Write a balanced equation for the formation of one mole of sulfur trioxide.
b. Calculate K for the reaction @ 627ºC. (Note that gases need to be in atm.)

Do you want Kp or Kc. You can calculate Kc without changing to atm.

To answer your questions, let's follow these steps:

Step 1: Write the balanced equation
The balanced equation for the formation of one mole of sulfur trioxide can be written as follows:

2SO2(g) + O2(g) -> 2SO3(g)

This equation shows that two moles of sulfur dioxide gas react with one mole of oxygen gas to form two moles of sulfur trioxide gas.

Step 2: Calculate K using the given concentrations
The equilibrium expression for this reaction is given by:

K = [SO3]^2 / ([SO2]^2 * [O2])

where [SO3], [SO2], and [O2] represent the molar concentrations of sulfur trioxide, sulfur dioxide, and oxygen, respectively, at equilibrium.

Given concentrations:
[SO2] = 0.0060 M
[O2] = 0.0054 M
[SO3] = 0.0032 M

Substituting these values into the equilibrium expression, we get:

K = (0.0032)^2 / (0.0060)^2 * 0.0054

K ≈ 9.185

Step 3: Convert gas concentrations to atm (if necessary)
In this case, the given concentrations are already in molarity (M) units, so there is no need to convert them to atm.

Therefore, the value of K for the reaction at 627ºC is approximately 9.185.