Question

question 9 (4 points)
the equilibrium expression for kp for the reaction below is
2o3 (g) ⇌ 3o2 (g)
a) 3po2 / 2po3
b) 3po3 / 2po2

Explanation:

Step1: Recall Kp expression rule

For a reaction \( aA(g)
ightleftharpoons bB(g) \), \( K_p=\frac{(P_B)^b}{(P_A)^a} \), where \( P \) is partial pressure.
Given reaction: \( 2O_3(g)
ightleftharpoons 3O_2(g) \) (assuming typo, correcting \( 2O3 \) to \( 2O_3 \), \( 3O2 \) to \( 3O_2 \)).

Step2: Apply Kp formula

Here, \( a = 2 \) (coefficient of \( O_3 \)), \( b = 3 \) (coefficient of \( O_2 \)).
So \( K_p=\frac{(P_{O_2})^3}{(P_{O_3})^2}=\frac{3P_{O_2}}{2P_{O_3}} \) (using partial pressure notation \( P \)).

Answer:

a) \( \frac{3PO_2}{2PO_3} \)