2. if initially there is 38.0 mol/l of so₃(g), at equilibrium there is 14.0 mol/l of so₂(g). calculate…

2. if initially there is 38.0 mol/l of so₃(g), at equilibrium there is 14.0 mol/l of so₂(g). calculate kc.\n2so₂(g) + o₂(g) ⇌ 2so₃(g)

2. if initially there is 38.0 mol/l of so₃(g), at equilibrium there is 14.0 mol/l of so₂(g). calculate kc.\n2so₂(g) + o₂(g) ⇌ 2so₃(g)

Answer

Explanation:

Step1: Determine the change in concentration of (SO_3)

The initial concentration of (SO_3) is (38.0\ mol/L) and at equilibrium the concentration of (SO_2) is (14.0\ mol/L). From the balanced - equation (2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)), for every 2 moles of (SO_2) produced, 2 moles of (SO_3) are consumed. The change in concentration of (SO_3) ((\Delta[SO_3])) is equal to the concentration of (SO_2) produced since the stoichiometric coefficients of (SO_2) and (SO_3) are the same. So, (\Delta[SO_3]=14.0\ mol/L). The equilibrium concentration of (SO_3) is ([SO_3]=38.0 - 14.0=24.0\ mol/L).

Step2: Determine the equilibrium concentration of (O_2)

From the balanced - equation, the stoichiometric ratio of (SO_2) to (O_2) is (2:1). So, if ([SO_2] = 14.0\ mol/L) at equilibrium, then ([O_2]=\frac{14.0}{2}=7.0\ mol/L) at equilibrium.

Step3: Write the expression for (K_c)

The equilibrium constant expression for the reaction (2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)) is (K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}).

Step4: Substitute the equilibrium concentrations into the (K_c) expression

Substitute ([SO_3]=24.0\ mol/L), ([SO_2]=14.0\ mol/L), and ([O_2]=7.0\ mol/L) into the (K_c) expression: [K_c=\frac{(24.0)^2}{(14.0)^2\times7.0}=\frac{576}{196\times7.0}=\frac{576}{1372}\approx0.42]

Answer:

(K_c\approx0.42)