If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

see explanation below

Explanation:

The question is incomplete. Here's the complete question:

Consider the following reaction:

SO2Cl2 > SO2(g) + Cl2(g)

Kc= 2.99 x 10^-7 at 227 degrees celcius

If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:

       SOCl2 > SO2 + Cl2     Kc = 2.99x10⁻⁷

i)       0.168                  0        0

c)         -x                    +x       +x

e)     0.168-x                x         x

Writting the Kc expression:

Kc = [SO2] [Cl2] / [SOCl2]

Replacing the values from the chart:

2.99x10⁻⁷ = x² / 0.168 - x

However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:

2.99x10⁻⁷ = x²/0.168

2.99x10⁻⁷ * 0.168 = x²

√5.02x10⁻⁸ = x

x = 2.24x10⁻⁴ M

This means then, that the concentration of Cl2 in equilibrium would be:

[Cl₂] = 2.24x10⁻⁴ M

i think its 100

i hope this !

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