The reaction SO2Cl2(g)--> SO2(g) + Cl2(g) has Kc = 2.99 x 10-7 at 227 degrees C. If a reaction mixture initially contains 0.166 M SO2Cl2, what is the equilibrium concentration of Cl2 at 227 degrees C?

The equilibrium concentration of Cl2 at 227 degrees C is 0.332 M.

To find the equilibrium concentration of Cl2, we need to set up an ICE table and use the equilibrium constant (Kc) expression.

The balanced equation for the reaction is:
SO2Cl2(g) --> SO2(g) + Cl2(g)

The initial concentration of SO2Cl2 is 0.166 M. At equilibrium, let's assume the concentration of Cl2 is x M.

Using the stoichiometry of the reaction, the concentration of Cl2 at equilibrium will be the same as the concentration of SO2Cl2 that has reacted. Therefore, the concentration of Cl2 at equilibrium is also x M.

The concentration of SO2 at equilibrium is also x M, as they have a 1:1 stoichiometric ratio.

To set up the ICE table:

| SO2Cl2 | SO2 | Cl2 |
---------------------------------
Initial | 0.166 | 0 | 0 |
Change | -x | x | x |
Equilibrium| 0.166-x | x | x |

The equilibrium constant expression for the reaction is:
Kc = [SO2][Cl2] / [SO2Cl2]

Plugging in the equilibrium concentrations, we get:
2.99 x 10^-7 = x * x / (0.166 - x)

Since the value of Kc is very small, we can assume that the change in concentration of SO2Cl2 is negligible compared to 0.166 M. So we can approximate the equation to:
2.99 x 10^-7 = x * x / 0.166

Rearranging the equation, we have:
x^2 = 2.99 x 10^-7 * 0.166

x^2 = 4.9734 x 10^-8

Taking the square root of both sides, we find:
x = 7.05 x 10^-5

Therefore, the equilibrium concentration of Cl2 at 227 degrees C is approximately 7.05 x 10^-5 M.

To find the equilibrium concentration of Cl2 at 227 degrees C, we can use the given equilibrium constant (Kc) and the initial concentration of SO2Cl2.

First, let's write the balanced chemical equation for the reaction:

SO2Cl2(g) --> SO2(g) + Cl2(g)

According to the stoichiometry of the reaction, for every 1 mole of SO2Cl2 that reacts, 1 mole of Cl2 is produced. Therefore, at equilibrium, the concentration of Cl2 will be equal to the concentration of SO2Cl2 that reacts.

Let's denote the equilibrium concentration of Cl2 as [Cl2]eq and the initial concentration of SO2Cl2 as [SO2Cl2]initial.

Given:
Kc = 2.99 x 10^(-7)
[SO2Cl2]initial = 0.166 M

Since the reaction is in equilibrium, we can assume that the concentration of SO2Cl2 will decrease by x M, and the concentration of Cl2 will increase by x M.

Therefore,
[SO2Cl2]eq = [SO2Cl2]initial - x
[Cl2]eq = x

Applying these concentrations to the expression for Kc:

Kc = [SO2]^eq * [Cl2]^eq / [SO2Cl2]^eq

Since the stoichiometry shows that the coefficient of Cl2 is 1, we can substitute [Cl2]^eq with x:

Kc = 1 * x / ([SO2Cl2]initial - x)

Now, we can plug in the given values:

2.99 x 10^(-7) = x / (0.166 - x)

To solve for x, we need to manipulate the equation algebraically:

2.99 x 10^(-7) * (0.166 - x) = x
5.93534 x 10^(-8) - 2.99 x 10^(-7) * x = x
5.93534 x 10^(-8) = (2.99 x 10^(-7) + 1) * x
x = 5.93534 x 10^(-8) / (2.99 x 10^(-7) + 1)

Using a calculator to do the calculation:

x ≈ 0.0197 M

Therefore, the equilibrium concentration of Cl2 at 227 degrees C is approximately 0.0197 M.