To calculate the equilibrium constant $K_p$ for the reaction:

$2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g)$

We use the following formula for $K_p$:

$K_p = \frac{P_{\text{SO}_3}^2}{P_{\text{SO}_2}^2 \cdot P_{\text{O}_2}}$

where:

• $P_{\text{SO}_2} = 90.1 \, \text{atm}$
• $P_{\text{O}_2} = 30.8 \, \text{atm}$
• $P_{\text{SO}_3} = 16.7 \, \text{atm}$

Substituting these values:

$K_p = \frac{(16.7)^2}{(90.1)^2 \cdot (30.8)}$

Now, let's calculate the value:

1. $(16.7)^2 = 278.89$
2. $(90.1)^2 = 8118.01$
3. Multiply $8118.01 \times 30.8 = 249234.71$

Now, divide:

$K_p = \frac{278.89}{249234.71} = 0.001118$

Rounding to 2 significant digits:

$K_p \approx 0.0011$

Would you like any further clarification on the process?

Here are 5 related questions to extend this understanding:

1. What is the difference between $K_p$ and $K_c$?
2. How does temperature affect the value of $K_p$?
3. What would happen to $K_p$ if the pressure of $SO_2$ were increased?
4. Can $K_p$ be used to predict the direction of the reaction?
5. How does the stoichiometry of the reaction influence the equilibrium expression?

Tip: Ensure you always check the units of pressure when calculating $K_p$, as it must be consistent across all gases.