SO2Cl2(g)⇌SO2(g)+Cl2(g) Kc=2.99×10−7 at 227 ∘C You may want to reference (Page) Section 15.8 while completing this problem. Part A If a reaction mixture initially contains 0.186 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

2.36×10^–4 M

Explanation:

The following data were obtained from the question:

Equilibrium constant, Kc = 2.99×10^−7

Concentration of SO2Cl2, [SO2Cl2] = 0.186 M

Concentration of Cl2, [Cl2] =?

Next, we shall write the balanced equation for the reaction. This is given below:

SO2Cl2(g)⇌SO2(g)+Cl2(g)

Next, we shall we shall construct the initial concentration and equilibrium table (ICE). This is illustrated below:

Initial concentration:

[SO2Cl2] = 0.186 M

[SO2] = 0

[Cl2] = 0

During reaction:

[SO2Cl2] = –y

[SO2] = +y

[Cl2] = +y

Equilibrium:

[SO2Cl2] = 0.186 – y

[SO2] = y

[Cl2] = y

Next, we shall determine the value of y in order to obtain the concentration of Cl2. This can be achieved by doing the following:

Kc = [SO2] [Cl2] / [SO2Cl2]

Kc = 2.99×10^−7

[SO2Cl2] = 0.186 M

[SO2] = y

[Cl2] = y

2.99×10^−7 = (y × y) /0.186

Cross multiply

y × y = 2.99×10^−7 × 0.186

y² = 2.99×10^−7 × 0.186

Take the square root of both side:

y = √(2.99×10^−7 × 0.186)

y = 2.36×10^–4

Therefore, the concentration of Cl2, [Cl2] = y = 2.36×10^–4 M

i think the answer is terrae but not 100 percent sure. ▓▬▬ §