 Chemistry
30 August, 09:00

# 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?

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1. 30 August, 10:57
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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