Question

At a certain temperature, 0.900 mol of SO3 is placed in a 2.00-L container.

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

At equilibrium, 0.130 mol of O2 is present. Calculate Kc.

Answer 1

The equilibrium constant (Kc) for the reaction 2SO3(g) <--> 2SO2(g) + O2(g) is calculated using the equilibrium concentrations of the reactants and products, resulting in Kc = 0.0107.

Step-by-step explanation:

To calculate the equilibrium constant (Kc) for the reaction 2SO3(g) <--> 2SO2(g) + O2(g), we first need to determine the concentrations of all species at equilibrium. Given that initially 0.900 mol of SO3 is placed in a 2.00-L container and at equilibrium, 0.130 mol of O2 is present, we can calculate the changes in molar amounts for all substances using stoichiometry.

At equilibrium, the amount of SO3 that has dissociated is equal to the amount of O2 formed (since the stoichiometry ratio between O2 and SO3 is 1:2), which is 0.130 mol. Thus, the molar amount of SO3 at equilibrium is 0.900 mol - 2(0.130 mol) = 0.900 mol - 0.260 mol = 0.640 mol. The molar amount of SO2 formed will be twice that of O2 since the stoichiometry between SO2 and O2 is also 2:1, which is 2(0.130 mol) = 0.260 mol SO2.

Now we can find the concentrations at equilibrium:

[SO3] = 0.640 mol / 2.00 L = 0.320 M

[SO2] = 0.260 mol / 2.00 L = 0.130 M

[O2] = 0.130 mol / 2.00 L = 0.065 M

The equilibrium constant expression for the reaction is:
Kc = ([SO2]^2 * [O2]) / [SO3]^2

Substituting the concentrations we have:
Kc = (0.130^2 * 0.065) / 0.320^2
Kc = (0.0169 * 0.065) / 0.1024
Kc = 0.0010985 / 0.1024
Kc = 0.0107

Thus, the equilibrium constant for this reaction is 0.0107.

Answer 2

Kc = 2.34 mol*L

Step-by-step explanation:

The calculation of the Kc of a reaction is performed using the values of the concentrations of the participants in the equilibrium.

A + B ⇄ C + D

Kc = [C] * [D] / [A] * [B]

According to the reaction

Kc = [SO2]^2 * [O2]^2 / [SO3]^2

Knowing the 0.900 mol of SO3 is placed in a 2.00-L it means we have a 0.450 mol/L of SO3

0.450 --> 0 + 0 (Beginning of the reaction)

0.260 --> 0.260 + 0.130 (During the reaction)

0.190 --> 0.260 + 0.130 (Equilibrium of the reaction)

Kc = [0.260]^2 + [0.130]^2 / [0.190]^2

Kc = 2.34 mol*L